Research ArticleTangent-Impulse Interception for a Hyperbolic Target

Dongzhe Wang Gang Zhang and Xibin Cao

Research Center of Satellite Technology Harbin Institute of Technology Harbin 150080 China

Correspondence should be addressed to Gang Zhang zhanggang3110163com

Received 30 November 2013 Accepted 30 March 2014 Published 28 April 2014

Academic Editor Antonio F Bertachini A Prado

Copyright copy 2014 Dongzhe Wang et alThis is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

The two-body interception problem with an upper-bounded tangent impulse for the interceptor on an elliptic parking orbit tocollide with a nonmaneuvering target on a hyperbolic orbit is studied Firstly four special initial true anomalies whose velocityvectors are parallel to either of the lines of asymptotes for the target hyperbolic orbit are obtained by using Newton-Raphsonmethod For different impulse points the solution-existence ranges of the target true anomaly for any conic transfer are discussedin detail Then the time-of-flight equation is solved by the secant method for a single-variable piecewise function about the targettrue anomaly Considering the sphere of influence of the Earth and the upper bound on the fuel all feasible solutions are obtainedfor different impulse points Finally a numerical example is provided to apply the proposed technique for all feasible solutions andthe global minimum-time solution with initial coasting time

1 Introduction

The earth-crossing comets asteroids space debris and otherspacecraft are potentially hazardous objects (PHOs) sincethey may closely encounter with the Earth Usually an activespacecraft is required to intercept impact or deflect the PHO[1ndash4]The impact and deflectionmission ismainly usedwhenthe PHO is beyond the sphere of influence of the Earth If thismission is failed the PHO will be on a hyperbolic trajectoryduring the Earth flyby thus the interception problem from aparking orbit for a hyperbolic target needs to be solved

Assume the nonmaneuvering target is subject only to aNewtonian gravitational attraction and the Lambert app-roach is the most fundamental and the simplest method toaccomplish the two-body interception problem with impulsemaneuvers [5ndash7] If the impulse point and the interceptionpoint are both assigned the time of flight (TOF) for the targetcan be obtained by solving Keplerrsquos equation of hyperbolicorbit and then the interception problem is equivalent toLambertrsquos problem [8ndash11] The required velocity vector ofthe transfer trajectory at the impulse point can be obtainedwith the Lambert approach It should be notified that thetransfer trajectory can be elliptic parabolic and hyperbolic

When the initial coasting time is considered the impulsepoint may be not at the initial time then interceptionoptimization problems to save time and fuel are studied bymany researchers [12ndash14]

However no constraints are imposed on the impulsedirection for the Lambert approach when both impulse andinterception points are assigned If two coplanar orbits aretangent and the velocity vectors at their common point arein the same direction then the task of nulling the relativevelocity will be simple [8] The cotangent transfer indicatesthat the transfer orbit is tangent to both the initial orbit andthe final orbit The classic Hohmann transfer [15] is a two-impulse cotangent transfer between two coplanar circularorbits and it is the minimum-energy one among all the two-impulse transfersThis result is also satisfied for two coplanarelliptic orbits sharing the same apsidal line For two coplanarelliptic orbits with different apsidal lines the numericalsolution [16] and the closed-form solution [17 18] have beenobtained for the cotangent transfer problem In additionZhang et al solved the two-impulse cotangent rendezvousproblem between two coplanar elliptic orbits [19] and bet-ween elliptic and hyperbolic orbits [20] respectively Dif-ferent from two impulses are required for the rendezvous

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2014 Article ID 837849 10 pageshttpdxdoiorg1011552014837849

2 Mathematical Problems in Engineering

problem a single impulse can fulfill the interceptionmissionThus the tangent-impulse interception problem requiresthe same TOF for two spacecraft and a transfer orbit tangentto the initial orbit Then for a given impulse point theinterception point is not arbitrary and needs to be solved

This paper studies the coplanar tangent-impulse intercep-tion problem from an interceptor on elliptic parking orbits toa target on hyperbolic orbits Based on four special initial trueanomalies the solution-existence ranges of the target trueanomaly are obtained for different impulse points Then theone-tangent-impulse interception trajectories are obtained bysolving a single-variable piecewise function

2 Problem Statement

Assume that the interceptor moves on the initial ellipticparking orbit and the target moves on the target hyperbolicorbit and the initial and target orbits are coplanar withdifferent apsidal lines At the initial time the interceptor isat 11987510

and the target is at 11987520

(see Figure 1) Let the motiondirection of both orbits be counterclockwiseThe coasting arcfor the interceptor is from 119875

10to 1198751 and the initial coasting

time before the tangent impulse is 1199051 The coasting arc for the

target is from 11987520

to 1198752 and the coasting time of the target

is 1199052 The tangent impulse indicates that the transfer orbit

is tangent to the initial elliptic orbit at 1198751 and the transfer

time is 1199053 For the orbital interception problem the flight time

of the interceptor and that of the target are the same Forconvenience let

120578 ≜1

1198791

(1199052minus 1199051minus 1199053) (1)

where 119905119895(119895 = 1 2 3) denotes the coasting time with zero

revolution and119879119895(119895 = 1 3) denotes the orbital period where

the subscripts ldquo1rdquo ldquo2rdquo and ldquo3rdquo denote the initial target andtransfer orbits respectively For the upper-bounded impulseinterception the minimum-time solution is interesting thenthe revolutionnumber for the transfer orbit is set to be 0Thenthe orbital interception problem with a tangent impulse is tosolve the impulse point 119875

1and the interception point 119875

2such

that the following expression is satisfied

120578 isin N = 0 1 2 (2)

in which 120578 is a positive integer indicates that the interceptormoves multiple revolutions before the first tangent impulseon point 119875

1

For the two-impulse cotangent rendezvous problem ifthe final true anomaly 119891

2is given there is a single closed-

form solution for the initial true anomaly 1198911 then 120578 in (1) is

a function only of 1198912for zero revolution However for the

one-tangent-impulse interception problem from the aboveanalysis it is known that 120578 is a function of two variables theinitial true anomaly 119891

1of the impulse pointand the target

true anomaly 1198912of the interception point Usually there is an

upper-bound constraint on the fuel consumptionΔV then theinterception problem is to solve all values of 119891

1and 119891

2such

that (2) is satisfied and the fuel consumption is less than agiven value In this paper firstly the interception problemwill

Transfer orbit

Initial elliptic orbit

Target hyperbolic orbit

120579

P10

P20

F1

r1r2

P1P2

1205741 1

Figure 1 Geometrical interpretation of tangent-impulse intercep-tion for a hyperbolic target

be solved at a given 1198911of the impulse point then all feasible

solutions will be obtained at different 1198911within one initial

orbital period

3 Time-of-Flight Equations

31 Coasting Time for the InitialTarget Orbit For the initialelliptic orbit the coasting time from119875

10to 1198751can be obtained

by Keplerrsquos equation For the target hyperbolic orbit thecoasting time from 119875

20to 1198752is [16]

1199052=119903201199032sin (119891

2minus 11989120)

radic1205831199012

+ radicminus1198863

2

120583[sinh (119867

2minus 11986720) minus (119867

2minus 11986720)]

= 11990320radic

1199012

120583

sin (1198912minus 11989120)

1 + 1198902cos1198912

+ radicminus1198863

2

120583[sinh (119867

2minus 11986720) minus (119867

2minus 11986720)]

(3)

where 120583 is the gravitational parameter 119903 is the radius of119875119891 isthe true anomaly119901 = 119886(1minus119890

2) is the semilatus rectum 119886 is the

semimajor axis 119890 is the eccentricity and119867 is the hyperbolicanomaly which is related to the true anomaly by

sinh1198672=

radic1198902

2minus 1 sin119891

2

1 + 1198902cos1198912

cosh1198672=

1198902+ cos119891

2

1 + 1198902cos1198912

(4)

Mathematical Problems in Engineering 3

from which it is known that1198672= log (sinh119867

2+ cosh119867

2)

= log(radic1198902

2minus 1 sin119891

2+ 1198902+ cos119891

2

1 + 1198902cos1198912

)

(5)

Then the derivative of1198672with respect to 119891

2is

d1198672

d1198912

=

radic1198902

2minus 1

1 + 1198902cos1198912

(6)

Furthermore the derivative of 1199052with respect to 119891

2is

d1199052

d1198912

= 11990320radic

1199012

120583

cos (1198912minus 11989120) + 1198902cos11989120

(1 + 1198902cos1198912)2

+ radicminus1198863

2

120583[cosh (119867

2minus 11986720) minus 1]

radic1198902

2minus 1

1 + 1198902cos1198912

(7)

32 Transfer Time between the Initial and Target Orbits Theinitial flight-path angle at 119875

1can be written as [8]

120574 = arctan(1198901sin1198911

1 + 1198901cos1198911

) (8)

Let the transfer angle be 120579 and [9]

120582 =1199032(1 minus cos 120579)

1199031cos2120574 minus 119903

2cos (120579 + 120574) cos 120574

(9)

V1119905= radic

120583120582

1199031

(10)

where V1119905

denotes the magnitude of the velocity vector atpoint 119875

1on the transfer orbit For different values of 120582 there

are three cases for the transfer time [9 19](1) If 120582 isin (0 2) the transfer trajectory is an elliptic orbit

and the transfer time is

1199053=

2

radic120583(

1199031

2 minus 120582)

32

tanminus1 (radic2120582 minus 1

cos 120574cot (1205792) minus sin 120574)

+ 1199032radic

1199031

120583120582

tan 120574 (1 minus cos 120579) + (1 minus 120582) sin 120579(2 minus 120582) cos 120574

(11)

(2) If 120582 gt 2 the transfer trajectory is a hyperbolic orbitand the transfer time is

1199053= 1199032radic

1199031

120583120582

timestan 120574 (1 minus cos 120579) + (1 minus 120582) sin 120579

(2 minus 120582) cos 120574

minus1

radic120583(

1199031

120582 minus 2)

32

times lnsin 120574 minus cos 120574cot (1205792) minus (1 minus 2120582)12

sin 120574 minus cos 120574cot (1205792) + (1 minus 2120582)12

(12)

(3) If 120582 = 2 the transfer trajectory is a parabolic orbit andthe transfer time is

1199053=

2

3

radic1199033

1

120583120582[

3 cos 120574cot (1205792)[cos 120574cot (1205792) minus sin 120574]2

+1

[cos 120574cot (1205792) minus sin 120574]3]

(13)

It should be notified that ((11)ndash(13)) are functions indepen-dent of V

1119905

The derivative of the transfer time 1199053with respect to the

true anomaly 1198912is

d1199053

d1198912

= (1205971199053

1205971199032

+1205971199053

120597120582

120597120582

1205971199032

)d1199032

d1198912

+ (1205971199053

120597120579+1205971199053

120597120582

120597120582

120597120579)

d120579d1198912

(14)

where the partial derivatives 12059711990531205971199032 1205971199053120597120582 120597119905

3120597120579 120597120582120597119903

2

and 120597120582120597120579 and the full derivative d1199032d1198912are obtained in

[19 20] andd120579d1198912= 1Thepartial derivatives 120597119905

3120597120582 120597119905

3120597120579

are different for elliptic transfers and hyperbolic transfersbut the partial derivatives 120597119905

31205971199032 120597120582120597119903

2 and 120597120582120597120579 and

the full derivative d1199032d1198912are the same for elliptic orbits

and hyperbolic orbits Note that the target true anomalies 1198912

for parabolic orbits are not continuous thus the expressiond1199053d1198912is not derived for 120582 = 2

For a given impulse point the initial coasting time 1199051for

the interceptor is independent of 1198912 Then the derivative of 120578

with respect to 1198912is

1205781015840≜

d120578d1198912

=1

1198791

(d1199052

d1198912

minusd1199053

d1198912

) (15)

4 Conditions of Solution Existence

The above section gives the transfer time for any conic orbitA necessary condition is 120582 gt 0 Since the initial flight-pathangle 120574 isin (minus1205872 1205872) the initial flight-direction angle is

120574 =120587

2minus 120574 (16)

For the coplanar orbits the transfer angle is

120579 = 1198912+ 120594 where 120594 ≜ 120596

2minus 1205961minus 1198911 (17)

where 120596 is the argument of perigeeLet

1198881= sin (120594 minus 120574) + 119890

2

1199031

1199012

sin 120574

1198882= cos (120594 minus 120574)

1198883= minus

1199031

1199012

sin 120574

(18)

where 119901 is the semilatus rectum There are three cases for120582 gt 0 [18] (1) if 119888

3radic1198882

1+ 1198882

2lt minus1 119891

2isin [0 2120587) (2) if

4 Mathematical Problems in Engineering

1198883radic1198882

1+ 1198882

2gt 1 there is no solution for 119891

2 (3) if minus1 le

1198883radic1198882

1+ 1198882

2le 1 then the range of 119891

2for 120582 gt 0 is

arcsin(1198883

radic1198882

1+ 1198882

2

)minus atan2 (1198881 1198882)

lt 1198912lt 120587 minus arcsin(

1198883

radic1198882

1+ 1198882

2

)minus atan2 (1198881 1198882)

(19)

where the four-quadrant inverse tangent function 120572 =

atan2(1198871 1198872) is defined as the angle satisfying both sin120572 =

1198871radic1198872

1+ 1198872

2and cos120572 = 119887

2radic1198872

1+ 1198872

2

For any elliptical transfer orbit 0 lt 120582 lt 2 is required Let

1198884=1199012

1199031

cos (2120574) + 1198902(1 minus cos (2120574)) cos120594

1198885=1199012

1199031

sin (2120574) + 1198902(1 minus cos (2120574)) sin120594

1198886=1199012

1199031

+ cos (2120574) minus 1

(20)

Then there are also three cases to be discussed [18] (1) if1198886radic1198882

4+ 1198882

5lt minus1 119891

2isin [0 2120587) (2) if 119888

6radic1198882

4+ 1198882

5gt 1 there is

no solution for 1198912 (3) if minus1 le 119888

6radic1198882

4+ 1198882

5le 1 then the range

of 1198912for 0 lt 120582 lt 2 can be expressed as

minus 120594 minus atan2 (1198884 1198885) + arcsin(

1198886

radic1198882

4+ 1198882

5

)

lt 1198912lt 120587 minus 120594 minus atan2 (119888

4 1198885) minus arcsin(

1198886

radic1198882

4+ 1198882

5

)

(21)

In addition the range of 1198911for existing elliptic transfers

needs to be solved If there exists some range of 1198912 for 0 lt

120582 lt 2 the following inequality must be satisfied

1198882

4+ 1198882

5minus 1198882

6ge 0 (22)

Substituting (20) into (22) yields

[1 minus cos (2120574)]

times (1198902

2minus 1) [1 minus cos (2120574)] + 2

1199012

1199011

(1 + 1198901cos1198911)

times [1 + 1198902cos (2120574 minus 120594)] ge 0

(23)

Note that the flight-direction angle 120574 isin (0 1205872) thencos(2120574) lt 1 is satisfied for any 119891

1 thus (23) can be rewritten

as

119865 ≜ (1198902

2minus 1) [1 minus cos (2120574)]

+ 21199012

1199011

(1 + 1198901cos1198911) [1 + 119890

2cos (2120574 minus 120594)] ge 0

(24)

From ((16) (17)) it is known that 120574 and 120594 are functions onlyof 1198911 then 119865 is a function only of 119891

1 The values of 119891

1for the

equality 119865 = 0 can be obtained by numerical methods finallythe range for inequality 119865 gt 0 in (24) is obtained

5 Special Initial True Anomalies andthe Solution-Existence Range

There are four special points on the initial orbit whosedirections of velocity vectors are parallel to either of the linesof asymptotes for the target hyperbolic orbit These initialtrue anomalies 119891

1119904119895(119895 = 1 2 3 4) are solved in the following

paragraphs Let 119906119904119895

= 1198911119904119895

+ 1205961 (119895 = 1 2 3 4) 119906

2max =

1198912max + 1205962 = arccos(minus1119890

2) + 1205962 and 119906

2 min = arccos(11198902) +

1205962 From the geometry of Figure 2(a) it is known that

1199061199041+ 1205741199041= 1199062max 997904rArr 119891

11199041+ 1205961+ 1205741199041

= arccos(minus 1

1198902

) + 1205962

(25)

Note that the flight-direction angle is a function only of thetrue anomaly then the above equation can be rewritten as

11989111199041

+ 1205961+120587

2minus arctan(

1198901sin11989111199041

1 + 1198901cos11989111199041

)

= arccos(minus 1

1198902

) + 1205962

(26)

which is a function only of 11989111199041

Thus the classic Newton-Raphson iterative algorithm can be used to obtain thenumerical solution for (26)

From the geometry of Figure 2(b) it is known that

11989111199042

+ 1205961+ 1205741199042minus [arccos( 1

1198902

) + 1205962] = 120587 (27)

which can be rewritten in a function only of 11989111199042

as

11989111199042

+ 1205961minus120587

2minus arctan(

1198901sin11989111199042

1 + 1198901cos11989111199042

)

= arccos( 1

1198902

) + 1205962

(28)

Similarly from the geometries of Figures 2(c) and 2(d) itis known that the function only of 119891

11199043is

11989111199043

+ 1205961minus120587

2minus arctan(

1198901sin11989111199043

1 + 1198901cos11989111199043

)

= arccos(minus 1

1198902

) + 1205962

(29)

Mathematical Problems in Engineering 5

Parallel line of asymptotes Target hyperbolic orbit

Initial elliptic orbit

F1

Ps1

us1

rs1 u2max

120574s1

(a) 11989111199041

Parallel line of asymptotesTarget hyperbolic orbit

Initial elliptic orbit

F1

Ps2

us2rs2

u2min120574s2

(b) 11989111199042

Parallel line of asymptotes

Target hyperbolic orbit

Initial elliptic orbit

F1

Ps3

us3

rs3u2max

120574s3

(c) 11989111199043

Parallel line of asymptotes

Target hyperbolic orbit

Initial elliptic orbit

F1

Ps4

us4

rs4u2min

120574s4

(d) 11989111199044

Figure 2 Geometry interpretation for special initial true anomalies

and the function only of 11989111199044

is

11989111199044

+ 1205961minus3120587

2minus arctan(

1198901sin11989111199044

1 + 1198901cos11989111199044

)

= arccos( 1

1198902

) + 1205962

(30)

All the values of 1198911119904119895

(119895 = 1 2 3 4) are obtained by Newton-Raphson iterative algorithm

Once the special initial true anomalies are obtained thesolution-existence range of target true anomaly 119891

2can be

discussed and obtained as follows

(1) If 1198911isin [11989111199041

11989111199042

] there is only one intersectionpoint for the velocity-vector line and the targethyperbolic orbit Assume that the true anomaly of

the intersection point on the target hyperbolic orbitis 1198912119902 which can be obtained by (19) then the

target true anomaly is in the range (minus1198912max 1198912119902)

The solution-existence range for any conic transferorbit is obtained as follows (a) If there is no rangefor elliptic transfer orbits there is also no range forany conic transfer orbit (b) If there exists a range(1198912119886 1198912119887) sub (minus119891

2max 1198912119902) for elliptic transfer orbitswhich is obtained by (21) then the range for any conictransfer orbit can be (119891

2119886 1198912119887) or (minus119891

2max 1198912119887) When1199053by (11)ndash(13) is positive for an arbitrarily selected

1198912isin (minus119891

2max 1198912119886) for example 1198912= (minus119891

2max +1198912119886)2 the solution-existence range is (minus119891

2max 1198912119887)otherwise the range is (119891

2119886 1198912119887)

(2) If 1198911

isin (11989111199042

11989111199043

) there are two intersectionpoints for the velocity-vector line and the target

6 Mathematical Problems in Engineering

hyperbolic orbit Assume that the true anomalies ofthe intersection points on the target hyperbolic orbitare 1198912119901

and 1198912119902 then the solution-existence range for

any conic transfer orbit is in the range (1198912119901 1198912119902) The

solution-existence range for elliptic transfer orbits isobtained as (119891

2119886 1198912119887) sub (119891

2119901 1198912119902) by (21) Then the

solution-existence range for any conic transfer orbitis (1198912119901 1198912119887)

(3) If 1198911isin [11989111199043

11989111199044

] there is only one intersectionpoint for the velocity-vector line and the targethyperbolic orbit Assume that the true anomaly ofthe intersection point on the target hyperbolic orbitis 1198912119901 which is obtained by (19) then the target true

anomaly is in the range (1198912119901 1198912max) The solution-

existence range for any conic transfer orbit is obtainedas follows (a) If there is no range for elliptic transferorbits then the solution-existence range for any conictransfer orbit is (119891

2119901 1198912max) (b) If there exists a range

(1198912119886 1198912119887) sub (119891

2119901 1198912max) for elliptic transfer orbits

then the range for any conic transfer orbit can be(1198912119901 1198912119887) or (119891

2119901 1198912max) When 119905

3by ((11)ndash(13)) is

positive for an arbitrarily selected 1198912isin (1198912119887 1198912max)

for example 1198912

= (1198912119887

+ 1198912max)2 the solution-

existence range is (1198912119901 1198912max) otherwise the range is

(1198912119901 1198912119887)

(4) If 1198911

isin (11989111199044

11989111199041

) there are two intersectionpoints for the velocity-vector line and the targethyperbolic orbit Assume that the true anomalies ofthe intersection points on the target hyperbolic orbitare 1198912119901

and 1198912119902 then the target true anomaly must be

in the range (1198912119902 1198912max) There is no range for elliptic

transfer orbits and the solution-existence range forany conic (hyperbolic) transfer orbit is (119891

2119902 1198912max)

From the above discussions it is known that there maybe no solution-existence range for any conic transfer orbitonly when 119891

1isin [119891

11199041 11989111199042

] and there is certainly nosolution-existence range for elliptic transfer orbits when 119891

1isin

[11989111199044

11989111199041

]Let the solution-existence range for any conic transfer

by the above method be (1198912119898 1198912119899) The target true anomaly

1198912119894at the impulse time is obtained by the hyperbolic orbit

propagationThen there are three cases (1) If 1198912119894lt 1198912119898 then

the final solution should be obtained in the range (1198912119898 1198912119899)

(2) If 1198912119894

isin [1198912119898 1198912119899] then the final solution should be

obtained in the range (1198912119894 1198912119899) (3) If 119891

2119894gt 1198912119899 then there

is no solution

6 Solving the Problem

For a given impulse point the solution-existence range forany conic transfer and that for elliptic transfers are obtainedin Section 5 Assume that the final solution-existence rangefor any conic transfer considering the target true anomalyat the impulse time is (119891

21198921 11989121198922

) The interception problemcan be solved by numerical methods

Thevalue of 1205781015840 can be calculated by (15) then the solutionsof 1198912for 1205781015840(119891

2) = 0 can be obtained by a numerical iterative

algorithm for example the golden section search If thereis no solution for 1205781015840(119891

2) = 0 then 120578 decreases or increases

monotonically Then for the range (11989121198921

11989121198922

) if lfloor120578(1198911198921

+

120575)rfloor = lfloor120578(1198911198922minus 120575)rfloor or max(lfloor120578(119891

1198921+ 120575)rfloor lfloor120578(119891

1198922minus 120575)rfloor) lt 0

where the function lfloor119910rfloor denotes the nearest integer less thanor equal to 119910 and 120575 is a small positive number then there isno solution for (2) Otherwise there are two cases

(1) If min(lfloor120578(1198911198921

+ 120575)rfloor lfloor120578(1198911198922

minus 120575)rfloor) ge 0 there are|lfloor120578(1198911198922minus 120575)rfloor minus lfloor120578(119891

1198921+ 120575)rfloor| solutions which can be

obtained by the secant method as

1198912119899+2

= 1198912119899+1

minus (1198912119899+1

minus 1198912119899)

120578 (1198912119899+1

) minus 1198731

120578 (1198912119899+1

) minus 120578 (1198912119899)

119899 isin N

(31)

where the initial guesses are11989120

= 1198911198921+12057511989121

= 1198911198922minus

1205751198731= lfloor120578(119891

1198921+120575)rfloor+1 lfloor120578(119891

1198921+120575)rfloor+2 lfloor120578(119891

1198922minus120575)rfloor

if lfloor120578(1198911198921+120575)rfloor lt lfloor120578(119891

1198922minus120575)rfloor and119873

1= lfloor120578(119891

1198922minus120575)rfloor+1

lfloor120578(1198911198922minus 120575)rfloor + 2 lfloor120578(119891

1198921+ 120575)rfloor if lfloor120578(119891

1198921+ 120575)rfloor gt

lfloor120578(1198911198922minus 120575)rfloor

(2) If min(lfloor120578(1198911198921

+ 120575)rfloor lfloor120578(1198911198922

minus 120575)rfloor) lt 0 there are[max(lfloor120578(119891

1198921+ 120575)rfloor lfloor120578(119891

1198922minus 120575)rfloor) + 1] solutions

which can be obtained by (31) with 1198731= 0 1

max(lfloor120578(1198911198921

+ 120575)rfloor lfloor120578(1198911198922

minus 120575)rfloor) Note that 120578(1198911198921)

instead of 120578(1198911198921

+ 120575) should be used for the range[1198911198921 1198911198922)

If there exist 119896119899ge 1 solutions for 1205781015840(119891

2) = 0 and these

119896119899extreme points for 120578(119891

2) are obtained as 119891

21198961lt 11989121198962

lt

sdot sdot sdot lt 1198912119896119899

then there are 119896119899+ 1 piecewise ranges that is

(11989121198921

11989121198961

) [11989121198961

11989121198962

) [1198912119896119899

11989121198922

) For each piecewiserange the function of 120578 decreases or increasesmonotonicallyThen the solution for each piecewise range can be obtainedby (31) with different initial guesses If lim

(1198912rarr11989121198922)120578 = +infin

(see Figure 3(d)) there are infinite solutions in the range[1198912119896119899

11989121198922

) and the minimum-time solution (or with theminimum 119891

2) will be obtained

In the patched-conic approximation method the radiusof the Earthrsquos sphere of influence is 119877SOI = 925 000 km thenthe true anomaly range of the solution is 119891

2isin (minus119891SOI 119891SOI)

where 119891SOI = arccos([1199012119877SOI minus 1]119890

2) Thus the solutions

beyond this range should be removedThe above section gives a method to obtain all solutions

for a given initial impulse point 1198911 For each solution the

corresponding fuel consumption is

ΔV = 1003816100381610038161003816V1119905 minus V1

1003816100381610038161003816

=

1003816100381610038161003816100381610038161003816100381610038161003816

radic120583

1199011

120582 (1 + 1198901cos1198911) minus radic

120583

1199011

(1 + 1198902

1+ 21198901cos1198911)

1003816100381610038161003816100381610038161003816100381610038161003816

(32)

The solution whose cost is greater than the given maxi-mal value should be removed Thus the feasible solutionneeds to satisfy two conditions including the radius of theEarthrsquos sphere of influence and the given maximal valueof fuel When considering the initial coasting time before

Mathematical Problems in Engineering 7

minus120 minus100 minus80 minus60 minus40 minus20 0 20 40 60minus1

minus08

minus06

minus04

minus02

0

02

True anomaly f2 (∘)

120578

(a) 1198911 = 160∘

minus150 minus100 minus50 0 50 100minus1

minus05

0

05

True anomaly f2 (∘)

120578

(b) 1198911 = 170∘

minus40 minus20 0 20 40 60 80 100 120 140minus5

minus4

minus3

minus2

minus1

0

1

True anomaly f2 (∘)

120578

(c) 1198911 = 240∘

60 70 80 90 100 110 120 1300

1

2

3

4

5

6

7

8

9

10

True anomaly f2 (∘)

120578

(d) 1198911 = 330∘

Figure 3 The curves of 120578 versus 1198912for different 119891

1

the impulse the impulse point is free then for each initialimpulse point all feasible solutions can be obtained and theminimum-time one is with the minimum value of 119891

2

Reference [20] dealt with the cotangent rendezvousbetween two coplanar elliptic and hyperbolic orbits (Problem1) whereas this paper is for the tangent-impulse interceptionproblem from an interceptor on elliptic parking orbits toa target on hyperbolic orbits (Problem 2) Three maindifferences between these two problems are listed as follows

(1) For Problem 1 there is a single closed-form solutionfor an arbitrary point on the target hyperbolic orbitthat is 119891

2isin (minus119891

2max 1198912max) where 1198912max =

arccos(minus11198902) thus there is only one degree of

freedom in (2) for zero revolution however in aspecified (eg counterclockwise) motion directionthe solution-existence range is not the whole rangeof target true anomaly For Problem 2 there isno certain relationship between the tangent-impulse

point1198911and the interception point119891

2 In other words

there are two degrees of freedom in (2) with zerorevolutions

(2) For Problem 1 with nonintersecting case (counter-clockwise) the solution-existence range for cotangentrendezvous is 119891

2isin (1198912119886 1198912max) where (1198912119886 1198912119887) is

obtained by the solution-existence range for elliptictransfers [20] But for Problem 2with nonintersectingcase (counterclockwise) the solution-existence rangeof 1198912can be obtained by the proposed method for

different1198911 which is dependent on four special initial

true anomalies whose velocity vectors are parallel tothe lines of asymptotes for the target hyperbolic orbitMoreover there may be no solution-existence rangefor any conic transfer orbit when 119891

1isin [11989111199041

11989111199042

](3) For Problem 1 with a given revolution number 119873

3

of the transfer orbit usually there is only one orbitsolution (see the examples in [20]) However for

8 Mathematical Problems in Engineering

Problem 2 with an upper-bounded impulse theminimum-time solution is interesting then 119873

3is set

to be 0 For a given impulse point 1198911 there may exist

multiple solutions of 1198912(see Figures 3(a) and 3(b))

With the initial coasting time all feasible solutionsfor different impulse points are obtained and theminimum-time one is with the minimum value of 119891

2

7 Numerical Examples

Assume that the interceptorrsquos orbit elements are 1198861= 2119877119864+

4000 km 1198901= 06 120596

1= 10∘ and 119891

10= 60∘ and the targetrsquos

orbit elements are 1198862= minus2119877

119864minus 10 000 km 119890

2= 16 120596

2= 0∘

and 11989120

= minus120∘ where 119877

119864= 637813 km is the Earthrsquos

radius The gravitational parameter used for the example is120583 = 3986004415 km3s2 The upper bound of the tangentimpulse is set to be 5 kms Four special initial true anomaliesare obtained as 119891

11199041= 454187

∘ 11989111199042

= 1581023∘ 11989111199043

=

1919457∘ and 119891

11199044= 2845334

∘ via (26) (28) (29) and(30) To obtain elliptic transfers the range of the initial trueanomaly of the impulse point is119891

1isin (799244

∘ 2499268

∘) via

(24) In addition considering the radius of the Earthrsquos sphereof influence that is 925000 km the range of true anomaly is1198912isin (minus1269425

∘ 1269425

∘)

If the impulse is imposed at the initial time that is 1198911=

11989110

= 60∘ then 119891

1isin [11989111199041

11989111199042

] is satisfied and there is norange for elliptic transfers thus no solution-existence rangeexists for 119891

2

If the impulse is imposed at 1198911

= 160∘ the target

true anomaly at the impulse time is 1198912

= minus1183547∘

by solving Keplerrsquos equation for hyperbolic orbits andthe plots of 120578 versus 119891

2can be obtained in Figure 3(a)

The final solution-existence range for any conic transferconsidering the target true anomaly at the impulse timeis (11989121198921

11989121198922

) = (minus1183547∘ 724802

∘) There are two

solutions that is minus928589∘ and 207296∘ the correspondingrevolution numbers are both 0 and the corresponding costsare 13182 kms and 10211 kms respectively The minimum-time solution is minus928589∘ the corresponding transfer time is219724 s and the total interception time 119905

2is 283104 s via (3)

The value of 120582 is obtained as 11178 by (9) thus the trajectoryis elliptic and it is plotted in Figure 4

If the impulse is imposed at 1198911= 170

∘ the target trueanomaly at the impulse time is 119891

2= minus1177828

∘ the plots of120578 versus 119891

2can be obtained in Figure 3(b)The final solution-

existence range is (11989121198921

11989121198922

) = (minus1163047∘ 883112

∘)

There are three solutions minus1158579∘ minus1092932

∘ and532432

∘ all the corresponding revolution numbers are 0and the corresponding costs are 211998 kms 39088 kmsand 11770 kms respectivelyThe energy cost ofminus1158579∘ ismuch greater than 5 kms thus it cannot be usedThe feasibleminimum-time solution is minus1092932∘ the correspondingtransfer time is 131883 s and the total interception time is212588 s The value of 120582 is 27636 thus the trajectory ishyperbolic and it is plotted in Figure 5

If the impulse is imposed at 1198911= 240

∘ the target trueanomaly at the impulse time is 119891

2= minus1125985

∘ the plots of120578 versus 119891

2can be obtained in Figure 3(c) There is a single

minus15 minus10 minus5 0 5

minus16

minus14

minus12

minus10

minus8

minus6

minus4

minus2

0

2

X (km)

Y (k

m)

InitialTargetTransfer

P1

P20

P10

P2

F1

times104

times104

Figure 4 Tangent-impulse interception with elliptic transfer trajec-tory

minus15 minus1 minus05 0 05 1minus16

minus14

minus12

minus10

minus8

minus6

minus4

minus2

0

2

4

Y (k

m)

X (km)

times104

times105

P1

P20

P10

P2

F1

InitialTargetTransfer

Figure 5 Tangent-impulse interception with hyperbolic transfertrajectory

solution 1152336∘ the corresponding revolution number is0 the cost is 17743 kms the transfer time is 363182 s andthe total interception time is 542081 sThe value of 120582 is 19316thus the trajectory is elliptic

If the impulse is imposed at 1198911= 330

∘ the target trueanomaly at the impulse time is119891

2= minus1102890

∘ the plots of 120578versus119891

2can be obtained in Figure 3(d) and there are infinite

solutions However when considering the sphere of influenceof the Earth there are three solutions The minimum-timesolution is 1186911∘ the corresponding revolution number

Mathematical Problems in Engineering 9

Table 1 Feasible solutions of1198912for different119891

1and the correspond-

ing interception time 1199052and ΔV

1198911(∘) Solutions of

(1198912(∘)119873

1) 119905

2(s) ΔV (kms)

150 (minus566897 0) 321084 09035(minus399134 0) 328443 08897

170 (minus1092932 0) 212588 39088(532432 1) 358345 11770

190 (891630 0) 390801 14596210 (1043983 0) 436651 16333230 (1123591 0) 499326 17338250 (1177130 0) 598472 18148270 (1219959 0) 802495 19082290 (1259555 0) 1602413 20322310 (1244383 1) 1114899 22976330 (1186911 1) 628954 29698

(1257512 2) 1506675 26872(1269266 3) 2368738 26464

350 (1142511 1) 525410 44692(1242882 2) 1085322 35407(1260145 3) 1632862 34192(1267542 4) 2175443 33698

10 (1261114 4) 1685896 49766(1266609 5) 2084768 48911

is 1 the corresponding costs is 29698 kms the correspond-ing transfer time is 209279 s and the total interception timeis 628954 s The value of 120582 is 27280 thus the trajectory ishyperbolic

For different impulse points all feasible solutions satisfy-ing 1198912isin (minus1269425

∘ 1269425

∘) and the fuel constraint and

the corresponding value of 1198732= 120578 for (2) can be obtained

by (31) and are listed in Table 1 The total interception time is1199052= 1199051+ 1199053 the corresponding costs are also listed in Table 1

It should be notified that there is no feasible solution when1198911= 30∘+ 119896119895sdot 20∘ 119896119895= 1 2 3 4 5 By using the numerical

optimization algorithm the minimum-time impulse point isat 1198911= 1705235

∘ the solution is 1198912= minus1108795

∘ the cost is49999 kms and the total interception time is 198106 s Theminimum-time interception trajectory is hyperbolic and it isnear to that in Figure 5

8 Conclusions

This paper studies the interception problem for a hyperbolictarget with an upper-bounded tangent impulse Based onfour special initial true anomalies the solution-existencerange of the target true anomaly for any conic transfer isobtained Then the problem is solved by the secant methodfor a time-of-flight equation For a given impulse pointthere may be zero one many or infinite solutions howeverwhen considering the sphere of influence of the Earth thereare always finite solutions For different impulse points allfeasible solutions and the global minimum-time solution are

obtained This proposed method is valid for an assignedimpulse point and the time optimization with the initialcoasting time The interception with a tangent-impulse tech-nique can be used in orbital interception task for potentiallyhazardous objects during the Earth flyby with simple control

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work is supported in part by the China Postdoctoral Sci-ence Foundation Funded Project (Grant no 2012M520753)theHeilongjiang Postdoctoral Science Foundation (Grant noLBH-Z12125) and theNational Natural Scientific Foundationof China (Grant no 61273096)

References

[1] J C Solem ldquoInterception of comets and asteroids on collisioncourse with Earthrdquo Journal of Spacecraft and Rockets vol 30 no2 pp 222ndash228 1993

[2] IMRoss S-Y Park and SDV Porter ldquoGravitational effects ofearth in optimizing ΔV for deflecting earth-crossing asteroidsrdquoJournal of Spacecraft and Rockets vol 38 no 5 pp 759ndash7642001

[3] B Dachwald and B Wie ldquoSolar sail trajectory optimization forintercepting impacting and deflecting near-earth asteroidsrdquoin Proceedings of the AIAA Guidance Navigation and ControlConference and Exhibit pp 3331ndash3348 San Francisco CalifUSA August 2005

[4] D Izzo ldquoOptimization of interplanetary trajectories for impul-sive and continuous asteroid deflectionrdquo Journal of GuidanceControl and Dynamics vol 30 no 2 pp 401ndash408 2007

[5] J Park R G Rhinehart and P T Kabamba ldquoMiss analysisin Lambert interceptions with application to a new guidancelawrdquo Mathematical Problems in Engineering Theory Methodsand Applications vol 6 no 2-3 pp 225ndash265 2000

[6] Y Ulybyshev ldquoDirect high-speed interception analytic solu-tions qualitative analysis and applicationsrdquo Journal of Space-craft and Rockets vol 38 no 3 pp 351ndash359 2001

[7] S P Burns and J J Scherock ldquoLambert guidance routinedesigned to match position and velocity of ballistic targetrdquoJournal of Guidance Control and Dynamics vol 27 no 6 pp989ndash996 2004

[8] R H Battin An Introduction to the Mathematics and Methodsof Astrodynamics American Institute of Aeronautics and Astro-nautics (AIAA) Reston Va USA Revised edition 1999

[9] S L Nelson and P Zarchan ldquoAlternative approach to thesolution of Lambertrsquos problemrdquo Journal of Guidance Controland Dynamics vol 15 no 4 pp 1003ndash1009 1992

[10] P V Arlulkar and S D Naik ldquoSolution based on dynamicalapproach for multiple-revolution lambert problemrdquo Journal ofGuidance Control and Dynamics vol 34 no 3 pp 920ndash9232011

[11] G Zhang D Mortari and D Zhou ldquoConstrained multiple-revolution Lambertrsquos problemrdquo Journal of Guidance Controland Dynamics vol 33 no 6 pp 1779ndash1786 2010

10 Mathematical Problems in Engineering

[12] C A Lembeck and J E Prussing ldquoOptimal impulsive interceptwith low-thrust rendezvous returnrdquo Journal of Guidance Con-trol and Dynamics vol 16 no 3 pp 426ndash433 1993

[13] D-R Taur V Coverstone-Carroll and J E Prussing ldquoOptimalimpulsive time-fixed orbital rendezvous and interception withpath constraintsrdquo Journal of Guidance Control and Dynamicsvol 18 no 1 pp 54ndash60 1995

[14] H Leeghim ldquoSpacecraft intercept using minimum controlenergy andwait timerdquoCelestialMechanics ampDynamical Astron-omy vol 115 no 1 pp 1ndash19 2013

[15] D A Vallado Fundamentals of Astrodynamics and ApplicationsMicrocosm Press El Segundo Calif USA 2nd edition 2001

[16] B F Thompson K K Choi S W Piggott and S R BeaverldquoOrbital targeting based on hodograph theory for improvedrendezvous safetyrdquo Journal of Guidance Control and Dynamicsvol 33 no 5 pp 1566ndash1576 2010

[17] V G Adamyan L V Adamyan G M Zaimtsyan and LT Manandyan ldquoDouble-pulse cotangential transfers betweencoplanar elliptic orbitsrdquo Journal of Applied Mathematics andMechanics vol 73 no 6 pp 664ndash672 2009

[18] G Zhang D Zhou D Mortari and T A Henderson ldquoAna-lytical study of tangent orbit and conditions for its solutionexistencerdquo Journal of Guidance Control and Dynamics vol 35no 1 pp 186ndash194 2012

[19] G Zhang D Zhou Z Sun and X Cao ldquoOptimal two-impulsecotangent rendezvous between coplanar elliptical orbitsrdquo Jour-nal of Guidance Control and Dynamics vol 36 no 3 pp 677ndash685 2013

[20] G Zhang X Cao and D Zhou ldquoTwo-impulse cotangent ren-dezvous between coplanar elliptic and hyperbolic orbitsrdquo Jour-nal of Guidance Control and Dynamics vol 37 no 3 pp 964ndash969 2014

Submit your manuscripts athttpwwwhindawicom

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Mathematical Problems in Engineering

problem a single impulse can fulfill the interceptionmissionThus the tangent-impulse interception problem requiresthe same TOF for two spacecraft and a transfer orbit tangentto the initial orbit Then for a given impulse point theinterception point is not arbitrary and needs to be solved

This paper studies the coplanar tangent-impulse intercep-tion problem from an interceptor on elliptic parking orbits toa target on hyperbolic orbits Based on four special initial trueanomalies the solution-existence ranges of the target trueanomaly are obtained for different impulse points Then theone-tangent-impulse interception trajectories are obtained bysolving a single-variable piecewise function

2 Problem Statement

Assume that the interceptor moves on the initial ellipticparking orbit and the target moves on the target hyperbolicorbit and the initial and target orbits are coplanar withdifferent apsidal lines At the initial time the interceptor isat 11987510

and the target is at 11987520

(see Figure 1) Let the motiondirection of both orbits be counterclockwiseThe coasting arcfor the interceptor is from 119875

10to 1198751 and the initial coasting

time before the tangent impulse is 1199051 The coasting arc for the

target is from 11987520

to 1198752 and the coasting time of the target

is 1199052 The tangent impulse indicates that the transfer orbit

is tangent to the initial elliptic orbit at 1198751 and the transfer

time is 1199053 For the orbital interception problem the flight time

of the interceptor and that of the target are the same Forconvenience let

120578 ≜1

1198791

(1199052minus 1199051minus 1199053) (1)

where 119905119895(119895 = 1 2 3) denotes the coasting time with zero

revolution and119879119895(119895 = 1 3) denotes the orbital period where

the subscripts ldquo1rdquo ldquo2rdquo and ldquo3rdquo denote the initial target andtransfer orbits respectively For the upper-bounded impulseinterception the minimum-time solution is interesting thenthe revolutionnumber for the transfer orbit is set to be 0Thenthe orbital interception problem with a tangent impulse is tosolve the impulse point 119875

1and the interception point 119875

2such

that the following expression is satisfied

120578 isin N = 0 1 2 (2)

in which 120578 is a positive integer indicates that the interceptormoves multiple revolutions before the first tangent impulseon point 119875

1

For the two-impulse cotangent rendezvous problem ifthe final true anomaly 119891

2is given there is a single closed-

form solution for the initial true anomaly 1198911 then 120578 in (1) is

a function only of 1198912for zero revolution However for the

one-tangent-impulse interception problem from the aboveanalysis it is known that 120578 is a function of two variables theinitial true anomaly 119891

1of the impulse pointand the target

true anomaly 1198912of the interception point Usually there is an

upper-bound constraint on the fuel consumptionΔV then theinterception problem is to solve all values of 119891

1and 119891

2such

that (2) is satisfied and the fuel consumption is less than agiven value In this paper firstly the interception problemwill

Transfer orbit

Initial elliptic orbit

Target hyperbolic orbit

120579

P10

P20

F1

r1r2

P1P2

1205741 1

Figure 1 Geometrical interpretation of tangent-impulse intercep-tion for a hyperbolic target

be solved at a given 1198911of the impulse point then all feasible

solutions will be obtained at different 1198911within one initial

orbital period

3 Time-of-Flight Equations

31 Coasting Time for the InitialTarget Orbit For the initialelliptic orbit the coasting time from119875

10to 1198751can be obtained

by Keplerrsquos equation For the target hyperbolic orbit thecoasting time from 119875

20to 1198752is [16]

1199052=119903201199032sin (119891

2minus 11989120)

radic1205831199012

+ radicminus1198863

2

120583[sinh (119867

2minus 11986720) minus (119867

2minus 11986720)]

= 11990320radic

1199012

120583

sin (1198912minus 11989120)

1 + 1198902cos1198912

+ radicminus1198863

2

120583[sinh (119867

2minus 11986720) minus (119867

2minus 11986720)]

(3)

where 120583 is the gravitational parameter 119903 is the radius of119875119891 isthe true anomaly119901 = 119886(1minus119890

2) is the semilatus rectum 119886 is the

semimajor axis 119890 is the eccentricity and119867 is the hyperbolicanomaly which is related to the true anomaly by

sinh1198672=

radic1198902

2minus 1 sin119891

2

1 + 1198902cos1198912

cosh1198672=

1198902+ cos119891

2

1 + 1198902cos1198912

(4)

Mathematical Problems in Engineering 3

from which it is known that1198672= log (sinh119867

2+ cosh119867

2)

= log(radic1198902

2minus 1 sin119891

2+ 1198902+ cos119891

2

1 + 1198902cos1198912

)

(5)

Then the derivative of1198672with respect to 119891

2is

d1198672

d1198912

=

radic1198902

2minus 1

1 + 1198902cos1198912

(6)

Furthermore the derivative of 1199052with respect to 119891

2is

d1199052

d1198912

= 11990320radic

1199012

120583

cos (1198912minus 11989120) + 1198902cos11989120

(1 + 1198902cos1198912)2

+ radicminus1198863

2

120583[cosh (119867

2minus 11986720) minus 1]

radic1198902

2minus 1

1 + 1198902cos1198912

(7)

32 Transfer Time between the Initial and Target Orbits Theinitial flight-path angle at 119875

1can be written as [8]

120574 = arctan(1198901sin1198911

1 + 1198901cos1198911

) (8)

Let the transfer angle be 120579 and [9]

120582 =1199032(1 minus cos 120579)

1199031cos2120574 minus 119903

2cos (120579 + 120574) cos 120574

(9)

V1119905= radic

120583120582

1199031

(10)

where V1119905

denotes the magnitude of the velocity vector atpoint 119875

1on the transfer orbit For different values of 120582 there

are three cases for the transfer time [9 19](1) If 120582 isin (0 2) the transfer trajectory is an elliptic orbit

and the transfer time is

1199053=

2

radic120583(

1199031

2 minus 120582)

32

tanminus1 (radic2120582 minus 1

cos 120574cot (1205792) minus sin 120574)

+ 1199032radic

1199031

120583120582

tan 120574 (1 minus cos 120579) + (1 minus 120582) sin 120579(2 minus 120582) cos 120574

(11)

(2) If 120582 gt 2 the transfer trajectory is a hyperbolic orbitand the transfer time is

1199053= 1199032radic

1199031

120583120582

timestan 120574 (1 minus cos 120579) + (1 minus 120582) sin 120579

(2 minus 120582) cos 120574

minus1

radic120583(

1199031

120582 minus 2)

32

times lnsin 120574 minus cos 120574cot (1205792) minus (1 minus 2120582)12

sin 120574 minus cos 120574cot (1205792) + (1 minus 2120582)12

(12)

(3) If 120582 = 2 the transfer trajectory is a parabolic orbit andthe transfer time is

1199053=

2

3

radic1199033

1

120583120582[

3 cos 120574cot (1205792)[cos 120574cot (1205792) minus sin 120574]2

+1

[cos 120574cot (1205792) minus sin 120574]3]

(13)

It should be notified that ((11)ndash(13)) are functions indepen-dent of V

1119905

The derivative of the transfer time 1199053with respect to the

true anomaly 1198912is

d1199053

d1198912

= (1205971199053

1205971199032

+1205971199053

120597120582

120597120582

1205971199032

)d1199032

d1198912

+ (1205971199053

120597120579+1205971199053

120597120582

120597120582

120597120579)

d120579d1198912

(14)

where the partial derivatives 12059711990531205971199032 1205971199053120597120582 120597119905

3120597120579 120597120582120597119903

2

and 120597120582120597120579 and the full derivative d1199032d1198912are obtained in

[19 20] andd120579d1198912= 1Thepartial derivatives 120597119905

3120597120582 120597119905

3120597120579

are different for elliptic transfers and hyperbolic transfersbut the partial derivatives 120597119905

31205971199032 120597120582120597119903

2 and 120597120582120597120579 and

the full derivative d1199032d1198912are the same for elliptic orbits

and hyperbolic orbits Note that the target true anomalies 1198912

for parabolic orbits are not continuous thus the expressiond1199053d1198912is not derived for 120582 = 2

For a given impulse point the initial coasting time 1199051for

the interceptor is independent of 1198912 Then the derivative of 120578

with respect to 1198912is

1205781015840≜

d120578d1198912

=1

1198791

(d1199052

d1198912

minusd1199053

d1198912

) (15)

4 Conditions of Solution Existence

The above section gives the transfer time for any conic orbitA necessary condition is 120582 gt 0 Since the initial flight-pathangle 120574 isin (minus1205872 1205872) the initial flight-direction angle is

120574 =120587

2minus 120574 (16)

For the coplanar orbits the transfer angle is

120579 = 1198912+ 120594 where 120594 ≜ 120596

2minus 1205961minus 1198911 (17)

where 120596 is the argument of perigeeLet

1198881= sin (120594 minus 120574) + 119890

2

1199031

1199012

sin 120574

1198882= cos (120594 minus 120574)

1198883= minus

1199031

1199012

sin 120574

(18)

where 119901 is the semilatus rectum There are three cases for120582 gt 0 [18] (1) if 119888

3radic1198882

1+ 1198882

2lt minus1 119891

2isin [0 2120587) (2) if

4 Mathematical Problems in Engineering

1198883radic1198882

1+ 1198882

2gt 1 there is no solution for 119891

2 (3) if minus1 le

1198883radic1198882

1+ 1198882

2le 1 then the range of 119891

2for 120582 gt 0 is

arcsin(1198883

radic1198882

1+ 1198882

2

)minus atan2 (1198881 1198882)

lt 1198912lt 120587 minus arcsin(

1198883

radic1198882

1+ 1198882

2

)minus atan2 (1198881 1198882)

(19)

where the four-quadrant inverse tangent function 120572 =

atan2(1198871 1198872) is defined as the angle satisfying both sin120572 =

1198871radic1198872

1+ 1198872

2and cos120572 = 119887

2radic1198872

1+ 1198872

2

For any elliptical transfer orbit 0 lt 120582 lt 2 is required Let

1198884=1199012

1199031

cos (2120574) + 1198902(1 minus cos (2120574)) cos120594

1198885=1199012

1199031

sin (2120574) + 1198902(1 minus cos (2120574)) sin120594

1198886=1199012

1199031

+ cos (2120574) minus 1

(20)

Then there are also three cases to be discussed [18] (1) if1198886radic1198882

4+ 1198882

5lt minus1 119891

2isin [0 2120587) (2) if 119888

6radic1198882

4+ 1198882

5gt 1 there is

no solution for 1198912 (3) if minus1 le 119888

6radic1198882

4+ 1198882

5le 1 then the range

of 1198912for 0 lt 120582 lt 2 can be expressed as

minus 120594 minus atan2 (1198884 1198885) + arcsin(

1198886

radic1198882

4+ 1198882

5

)

lt 1198912lt 120587 minus 120594 minus atan2 (119888

4 1198885) minus arcsin(

1198886

radic1198882

4+ 1198882

5

)

(21)

In addition the range of 1198911for existing elliptic transfers

needs to be solved If there exists some range of 1198912 for 0 lt

120582 lt 2 the following inequality must be satisfied

1198882

4+ 1198882

5minus 1198882

6ge 0 (22)

Substituting (20) into (22) yields

[1 minus cos (2120574)]

times (1198902

2minus 1) [1 minus cos (2120574)] + 2

1199012

1199011

(1 + 1198901cos1198911)

times [1 + 1198902cos (2120574 minus 120594)] ge 0

(23)

Note that the flight-direction angle 120574 isin (0 1205872) thencos(2120574) lt 1 is satisfied for any 119891

1 thus (23) can be rewritten

as

119865 ≜ (1198902

2minus 1) [1 minus cos (2120574)]

+ 21199012

1199011

(1 + 1198901cos1198911) [1 + 119890

2cos (2120574 minus 120594)] ge 0

(24)

From ((16) (17)) it is known that 120574 and 120594 are functions onlyof 1198911 then 119865 is a function only of 119891

1 The values of 119891

1for the

equality 119865 = 0 can be obtained by numerical methods finallythe range for inequality 119865 gt 0 in (24) is obtained

5 Special Initial True Anomalies andthe Solution-Existence Range

There are four special points on the initial orbit whosedirections of velocity vectors are parallel to either of the linesof asymptotes for the target hyperbolic orbit These initialtrue anomalies 119891

1119904119895(119895 = 1 2 3 4) are solved in the following

paragraphs Let 119906119904119895

= 1198911119904119895

+ 1205961 (119895 = 1 2 3 4) 119906

2max =

1198912max + 1205962 = arccos(minus1119890

2) + 1205962 and 119906

2 min = arccos(11198902) +

1205962 From the geometry of Figure 2(a) it is known that

1199061199041+ 1205741199041= 1199062max 997904rArr 119891

11199041+ 1205961+ 1205741199041

= arccos(minus 1

1198902

) + 1205962

(25)

Note that the flight-direction angle is a function only of thetrue anomaly then the above equation can be rewritten as

11989111199041

+ 1205961+120587

2minus arctan(

1198901sin11989111199041

1 + 1198901cos11989111199041

)

= arccos(minus 1

1198902

) + 1205962

(26)

which is a function only of 11989111199041

Thus the classic Newton-Raphson iterative algorithm can be used to obtain thenumerical solution for (26)

From the geometry of Figure 2(b) it is known that

11989111199042

+ 1205961+ 1205741199042minus [arccos( 1

1198902

) + 1205962] = 120587 (27)

which can be rewritten in a function only of 11989111199042

as

11989111199042

+ 1205961minus120587

2minus arctan(

1198901sin11989111199042

1 + 1198901cos11989111199042

)

= arccos( 1

1198902

) + 1205962

(28)

Similarly from the geometries of Figures 2(c) and 2(d) itis known that the function only of 119891

11199043is

11989111199043

+ 1205961minus120587

2minus arctan(

1198901sin11989111199043

1 + 1198901cos11989111199043

)

= arccos(minus 1

1198902

) + 1205962

(29)

Mathematical Problems in Engineering 5

Parallel line of asymptotes Target hyperbolic orbit

Initial elliptic orbit

F1

Ps1

us1

rs1 u2max

120574s1

(a) 11989111199041

Parallel line of asymptotesTarget hyperbolic orbit

Initial elliptic orbit

F1

Ps2

us2rs2

u2min120574s2

(b) 11989111199042

Parallel line of asymptotes

Target hyperbolic orbit

Initial elliptic orbit

F1

Ps3

us3

rs3u2max

120574s3

(c) 11989111199043

Parallel line of asymptotes

Target hyperbolic orbit

Initial elliptic orbit

F1

Ps4

us4

rs4u2min

120574s4

(d) 11989111199044

Figure 2 Geometry interpretation for special initial true anomalies

and the function only of 11989111199044

is

11989111199044

+ 1205961minus3120587

2minus arctan(

1198901sin11989111199044

1 + 1198901cos11989111199044

)

= arccos( 1

1198902

) + 1205962

(30)

All the values of 1198911119904119895

(119895 = 1 2 3 4) are obtained by Newton-Raphson iterative algorithm

Once the special initial true anomalies are obtained thesolution-existence range of target true anomaly 119891

2can be

discussed and obtained as follows

(1) If 1198911isin [11989111199041

11989111199042

] there is only one intersectionpoint for the velocity-vector line and the targethyperbolic orbit Assume that the true anomaly of

the intersection point on the target hyperbolic orbitis 1198912119902 which can be obtained by (19) then the

target true anomaly is in the range (minus1198912max 1198912119902)

The solution-existence range for any conic transferorbit is obtained as follows (a) If there is no rangefor elliptic transfer orbits there is also no range forany conic transfer orbit (b) If there exists a range(1198912119886 1198912119887) sub (minus119891

2max 1198912119902) for elliptic transfer orbitswhich is obtained by (21) then the range for any conictransfer orbit can be (119891

2119886 1198912119887) or (minus119891

2max 1198912119887) When1199053by (11)ndash(13) is positive for an arbitrarily selected

1198912isin (minus119891

2max 1198912119886) for example 1198912= (minus119891

2max +1198912119886)2 the solution-existence range is (minus119891

2max 1198912119887)otherwise the range is (119891

2119886 1198912119887)

(2) If 1198911

isin (11989111199042

11989111199043

) there are two intersectionpoints for the velocity-vector line and the target

6 Mathematical Problems in Engineering

hyperbolic orbit Assume that the true anomalies ofthe intersection points on the target hyperbolic orbitare 1198912119901

and 1198912119902 then the solution-existence range for

any conic transfer orbit is in the range (1198912119901 1198912119902) The

solution-existence range for elliptic transfer orbits isobtained as (119891

2119886 1198912119887) sub (119891

2119901 1198912119902) by (21) Then the

solution-existence range for any conic transfer orbitis (1198912119901 1198912119887)

(3) If 1198911isin [11989111199043

11989111199044

] there is only one intersectionpoint for the velocity-vector line and the targethyperbolic orbit Assume that the true anomaly ofthe intersection point on the target hyperbolic orbitis 1198912119901 which is obtained by (19) then the target true

anomaly is in the range (1198912119901 1198912max) The solution-

existence range for any conic transfer orbit is obtainedas follows (a) If there is no range for elliptic transferorbits then the solution-existence range for any conictransfer orbit is (119891

2119901 1198912max) (b) If there exists a range

(1198912119886 1198912119887) sub (119891

2119901 1198912max) for elliptic transfer orbits

then the range for any conic transfer orbit can be(1198912119901 1198912119887) or (119891

2119901 1198912max) When 119905

3by ((11)ndash(13)) is

positive for an arbitrarily selected 1198912isin (1198912119887 1198912max)

for example 1198912

= (1198912119887

+ 1198912max)2 the solution-

existence range is (1198912119901 1198912max) otherwise the range is

(1198912119901 1198912119887)

(4) If 1198911

isin (11989111199044

11989111199041

) there are two intersectionpoints for the velocity-vector line and the targethyperbolic orbit Assume that the true anomalies ofthe intersection points on the target hyperbolic orbitare 1198912119901

and 1198912119902 then the target true anomaly must be

in the range (1198912119902 1198912max) There is no range for elliptic

transfer orbits and the solution-existence range forany conic (hyperbolic) transfer orbit is (119891

2119902 1198912max)

From the above discussions it is known that there maybe no solution-existence range for any conic transfer orbitonly when 119891

1isin [119891

11199041 11989111199042

] and there is certainly nosolution-existence range for elliptic transfer orbits when 119891

1isin

[11989111199044

11989111199041

]Let the solution-existence range for any conic transfer

by the above method be (1198912119898 1198912119899) The target true anomaly

1198912119894at the impulse time is obtained by the hyperbolic orbit

propagationThen there are three cases (1) If 1198912119894lt 1198912119898 then

the final solution should be obtained in the range (1198912119898 1198912119899)

(2) If 1198912119894

isin [1198912119898 1198912119899] then the final solution should be

obtained in the range (1198912119894 1198912119899) (3) If 119891

2119894gt 1198912119899 then there

is no solution

6 Solving the Problem

For a given impulse point the solution-existence range forany conic transfer and that for elliptic transfers are obtainedin Section 5 Assume that the final solution-existence rangefor any conic transfer considering the target true anomalyat the impulse time is (119891

21198921 11989121198922

) The interception problemcan be solved by numerical methods

Thevalue of 1205781015840 can be calculated by (15) then the solutionsof 1198912for 1205781015840(119891

2) = 0 can be obtained by a numerical iterative

algorithm for example the golden section search If thereis no solution for 1205781015840(119891

2) = 0 then 120578 decreases or increases

monotonically Then for the range (11989121198921

11989121198922

) if lfloor120578(1198911198921

+

120575)rfloor = lfloor120578(1198911198922minus 120575)rfloor or max(lfloor120578(119891

1198921+ 120575)rfloor lfloor120578(119891

1198922minus 120575)rfloor) lt 0

where the function lfloor119910rfloor denotes the nearest integer less thanor equal to 119910 and 120575 is a small positive number then there isno solution for (2) Otherwise there are two cases

(1) If min(lfloor120578(1198911198921

+ 120575)rfloor lfloor120578(1198911198922

minus 120575)rfloor) ge 0 there are|lfloor120578(1198911198922minus 120575)rfloor minus lfloor120578(119891

1198921+ 120575)rfloor| solutions which can be

obtained by the secant method as

1198912119899+2

= 1198912119899+1

minus (1198912119899+1

minus 1198912119899)

120578 (1198912119899+1

) minus 1198731

120578 (1198912119899+1

) minus 120578 (1198912119899)

119899 isin N

(31)

where the initial guesses are11989120

= 1198911198921+12057511989121

= 1198911198922minus

1205751198731= lfloor120578(119891

1198921+120575)rfloor+1 lfloor120578(119891

1198921+120575)rfloor+2 lfloor120578(119891

1198922minus120575)rfloor

if lfloor120578(1198911198921+120575)rfloor lt lfloor120578(119891

1198922minus120575)rfloor and119873

1= lfloor120578(119891

1198922minus120575)rfloor+1

lfloor120578(1198911198922minus 120575)rfloor + 2 lfloor120578(119891

1198921+ 120575)rfloor if lfloor120578(119891

1198921+ 120575)rfloor gt

lfloor120578(1198911198922minus 120575)rfloor

(2) If min(lfloor120578(1198911198921

+ 120575)rfloor lfloor120578(1198911198922

minus 120575)rfloor) lt 0 there are[max(lfloor120578(119891

1198921+ 120575)rfloor lfloor120578(119891

1198922minus 120575)rfloor) + 1] solutions

which can be obtained by (31) with 1198731= 0 1

max(lfloor120578(1198911198921

+ 120575)rfloor lfloor120578(1198911198922

minus 120575)rfloor) Note that 120578(1198911198921)

instead of 120578(1198911198921

+ 120575) should be used for the range[1198911198921 1198911198922)

If there exist 119896119899ge 1 solutions for 1205781015840(119891

2) = 0 and these

119896119899extreme points for 120578(119891

2) are obtained as 119891

21198961lt 11989121198962

lt

sdot sdot sdot lt 1198912119896119899

then there are 119896119899+ 1 piecewise ranges that is

(11989121198921

11989121198961

) [11989121198961

11989121198962

) [1198912119896119899

11989121198922

) For each piecewiserange the function of 120578 decreases or increasesmonotonicallyThen the solution for each piecewise range can be obtainedby (31) with different initial guesses If lim

(1198912rarr11989121198922)120578 = +infin

(see Figure 3(d)) there are infinite solutions in the range[1198912119896119899

11989121198922

) and the minimum-time solution (or with theminimum 119891

2) will be obtained

In the patched-conic approximation method the radiusof the Earthrsquos sphere of influence is 119877SOI = 925 000 km thenthe true anomaly range of the solution is 119891

2isin (minus119891SOI 119891SOI)

where 119891SOI = arccos([1199012119877SOI minus 1]119890

2) Thus the solutions

beyond this range should be removedThe above section gives a method to obtain all solutions

for a given initial impulse point 1198911 For each solution the

corresponding fuel consumption is

ΔV = 1003816100381610038161003816V1119905 minus V1

1003816100381610038161003816

=

1003816100381610038161003816100381610038161003816100381610038161003816

radic120583

1199011

120582 (1 + 1198901cos1198911) minus radic

120583

1199011

(1 + 1198902

1+ 21198901cos1198911)

1003816100381610038161003816100381610038161003816100381610038161003816

(32)

The solution whose cost is greater than the given maxi-mal value should be removed Thus the feasible solutionneeds to satisfy two conditions including the radius of theEarthrsquos sphere of influence and the given maximal valueof fuel When considering the initial coasting time before

Mathematical Problems in Engineering 7

minus120 minus100 minus80 minus60 minus40 minus20 0 20 40 60minus1

minus08

minus06

minus04

minus02

0

02

True anomaly f2 (∘)

120578

(a) 1198911 = 160∘

minus150 minus100 minus50 0 50 100minus1

minus05

0

05

True anomaly f2 (∘)

120578

(b) 1198911 = 170∘

minus40 minus20 0 20 40 60 80 100 120 140minus5

minus4

minus3

minus2

minus1

0

1

True anomaly f2 (∘)

120578

(c) 1198911 = 240∘

60 70 80 90 100 110 120 1300

1

2

3

4

5

6

7

8

9

10

True anomaly f2 (∘)

120578

(d) 1198911 = 330∘

Figure 3 The curves of 120578 versus 1198912for different 119891

1

the impulse the impulse point is free then for each initialimpulse point all feasible solutions can be obtained and theminimum-time one is with the minimum value of 119891

2

Reference [20] dealt with the cotangent rendezvousbetween two coplanar elliptic and hyperbolic orbits (Problem1) whereas this paper is for the tangent-impulse interceptionproblem from an interceptor on elliptic parking orbits toa target on hyperbolic orbits (Problem 2) Three maindifferences between these two problems are listed as follows

(1) For Problem 1 there is a single closed-form solutionfor an arbitrary point on the target hyperbolic orbitthat is 119891

2isin (minus119891

2max 1198912max) where 1198912max =

arccos(minus11198902) thus there is only one degree of

freedom in (2) for zero revolution however in aspecified (eg counterclockwise) motion directionthe solution-existence range is not the whole rangeof target true anomaly For Problem 2 there isno certain relationship between the tangent-impulse

point1198911and the interception point119891

2 In other words

there are two degrees of freedom in (2) with zerorevolutions

(2) For Problem 1 with nonintersecting case (counter-clockwise) the solution-existence range for cotangentrendezvous is 119891

2isin (1198912119886 1198912max) where (1198912119886 1198912119887) is

obtained by the solution-existence range for elliptictransfers [20] But for Problem 2with nonintersectingcase (counterclockwise) the solution-existence rangeof 1198912can be obtained by the proposed method for

different1198911 which is dependent on four special initial

true anomalies whose velocity vectors are parallel tothe lines of asymptotes for the target hyperbolic orbitMoreover there may be no solution-existence rangefor any conic transfer orbit when 119891

1isin [11989111199041

11989111199042

](3) For Problem 1 with a given revolution number 119873

3

of the transfer orbit usually there is only one orbitsolution (see the examples in [20]) However for

8 Mathematical Problems in Engineering

Problem 2 with an upper-bounded impulse theminimum-time solution is interesting then 119873

3is set

to be 0 For a given impulse point 1198911 there may exist

multiple solutions of 1198912(see Figures 3(a) and 3(b))

With the initial coasting time all feasible solutionsfor different impulse points are obtained and theminimum-time one is with the minimum value of 119891

2

7 Numerical Examples

Assume that the interceptorrsquos orbit elements are 1198861= 2119877119864+

4000 km 1198901= 06 120596

1= 10∘ and 119891

10= 60∘ and the targetrsquos

orbit elements are 1198862= minus2119877

119864minus 10 000 km 119890

2= 16 120596

2= 0∘

and 11989120

= minus120∘ where 119877

119864= 637813 km is the Earthrsquos

radius The gravitational parameter used for the example is120583 = 3986004415 km3s2 The upper bound of the tangentimpulse is set to be 5 kms Four special initial true anomaliesare obtained as 119891

11199041= 454187

∘ 11989111199042

= 1581023∘ 11989111199043

=

1919457∘ and 119891

11199044= 2845334

∘ via (26) (28) (29) and(30) To obtain elliptic transfers the range of the initial trueanomaly of the impulse point is119891

1isin (799244

∘ 2499268

∘) via

(24) In addition considering the radius of the Earthrsquos sphereof influence that is 925000 km the range of true anomaly is1198912isin (minus1269425

∘ 1269425

∘)

If the impulse is imposed at the initial time that is 1198911=

11989110

= 60∘ then 119891

1isin [11989111199041

11989111199042

] is satisfied and there is norange for elliptic transfers thus no solution-existence rangeexists for 119891

2

If the impulse is imposed at 1198911

= 160∘ the target

true anomaly at the impulse time is 1198912

= minus1183547∘

by solving Keplerrsquos equation for hyperbolic orbits andthe plots of 120578 versus 119891

2can be obtained in Figure 3(a)

The final solution-existence range for any conic transferconsidering the target true anomaly at the impulse timeis (11989121198921

11989121198922

) = (minus1183547∘ 724802

∘) There are two

solutions that is minus928589∘ and 207296∘ the correspondingrevolution numbers are both 0 and the corresponding costsare 13182 kms and 10211 kms respectively The minimum-time solution is minus928589∘ the corresponding transfer time is219724 s and the total interception time 119905

2is 283104 s via (3)

The value of 120582 is obtained as 11178 by (9) thus the trajectoryis elliptic and it is plotted in Figure 4

If the impulse is imposed at 1198911= 170

∘ the target trueanomaly at the impulse time is 119891

2= minus1177828

∘ the plots of120578 versus 119891

2can be obtained in Figure 3(b)The final solution-

existence range is (11989121198921

11989121198922

) = (minus1163047∘ 883112

∘)

There are three solutions minus1158579∘ minus1092932

∘ and532432

∘ all the corresponding revolution numbers are 0and the corresponding costs are 211998 kms 39088 kmsand 11770 kms respectivelyThe energy cost ofminus1158579∘ ismuch greater than 5 kms thus it cannot be usedThe feasibleminimum-time solution is minus1092932∘ the correspondingtransfer time is 131883 s and the total interception time is212588 s The value of 120582 is 27636 thus the trajectory ishyperbolic and it is plotted in Figure 5

If the impulse is imposed at 1198911= 240

∘ the target trueanomaly at the impulse time is 119891

2= minus1125985

∘ the plots of120578 versus 119891

2can be obtained in Figure 3(c) There is a single

minus15 minus10 minus5 0 5

minus16

minus14

minus12

minus10

minus8

minus6

minus4

minus2

0

2

X (km)

Y (k

m)

InitialTargetTransfer

P1

P20

P10

P2

F1

times104

times104

Figure 4 Tangent-impulse interception with elliptic transfer trajec-tory

minus15 minus1 minus05 0 05 1minus16

minus14

minus12

minus10

minus8

minus6

minus4

minus2

0

2

4

Y (k

m)

X (km)

times104

times105

P1

P20

P10

P2

F1

InitialTargetTransfer

Figure 5 Tangent-impulse interception with hyperbolic transfertrajectory

solution 1152336∘ the corresponding revolution number is0 the cost is 17743 kms the transfer time is 363182 s andthe total interception time is 542081 sThe value of 120582 is 19316thus the trajectory is elliptic

If the impulse is imposed at 1198911= 330

∘ the target trueanomaly at the impulse time is119891

2= minus1102890

∘ the plots of 120578versus119891

2can be obtained in Figure 3(d) and there are infinite

solutions However when considering the sphere of influenceof the Earth there are three solutions The minimum-timesolution is 1186911∘ the corresponding revolution number

Mathematical Problems in Engineering 9

Table 1 Feasible solutions of1198912for different119891

1and the correspond-

ing interception time 1199052and ΔV

1198911(∘) Solutions of

(1198912(∘)119873

1) 119905

2(s) ΔV (kms)

150 (minus566897 0) 321084 09035(minus399134 0) 328443 08897

170 (minus1092932 0) 212588 39088(532432 1) 358345 11770

190 (891630 0) 390801 14596210 (1043983 0) 436651 16333230 (1123591 0) 499326 17338250 (1177130 0) 598472 18148270 (1219959 0) 802495 19082290 (1259555 0) 1602413 20322310 (1244383 1) 1114899 22976330 (1186911 1) 628954 29698

(1257512 2) 1506675 26872(1269266 3) 2368738 26464

350 (1142511 1) 525410 44692(1242882 2) 1085322 35407(1260145 3) 1632862 34192(1267542 4) 2175443 33698

10 (1261114 4) 1685896 49766(1266609 5) 2084768 48911

is 1 the corresponding costs is 29698 kms the correspond-ing transfer time is 209279 s and the total interception timeis 628954 s The value of 120582 is 27280 thus the trajectory ishyperbolic

For different impulse points all feasible solutions satisfy-ing 1198912isin (minus1269425

∘ 1269425

∘) and the fuel constraint and

the corresponding value of 1198732= 120578 for (2) can be obtained

by (31) and are listed in Table 1 The total interception time is1199052= 1199051+ 1199053 the corresponding costs are also listed in Table 1

It should be notified that there is no feasible solution when1198911= 30∘+ 119896119895sdot 20∘ 119896119895= 1 2 3 4 5 By using the numerical

optimization algorithm the minimum-time impulse point isat 1198911= 1705235

∘ the solution is 1198912= minus1108795

∘ the cost is49999 kms and the total interception time is 198106 s Theminimum-time interception trajectory is hyperbolic and it isnear to that in Figure 5

8 Conclusions

This paper studies the interception problem for a hyperbolictarget with an upper-bounded tangent impulse Based onfour special initial true anomalies the solution-existencerange of the target true anomaly for any conic transfer isobtained Then the problem is solved by the secant methodfor a time-of-flight equation For a given impulse pointthere may be zero one many or infinite solutions howeverwhen considering the sphere of influence of the Earth thereare always finite solutions For different impulse points allfeasible solutions and the global minimum-time solution are

obtained This proposed method is valid for an assignedimpulse point and the time optimization with the initialcoasting time The interception with a tangent-impulse tech-nique can be used in orbital interception task for potentiallyhazardous objects during the Earth flyby with simple control

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work is supported in part by the China Postdoctoral Sci-ence Foundation Funded Project (Grant no 2012M520753)theHeilongjiang Postdoctoral Science Foundation (Grant noLBH-Z12125) and theNational Natural Scientific Foundationof China (Grant no 61273096)

References

[1] J C Solem ldquoInterception of comets and asteroids on collisioncourse with Earthrdquo Journal of Spacecraft and Rockets vol 30 no2 pp 222ndash228 1993

[2] IMRoss S-Y Park and SDV Porter ldquoGravitational effects ofearth in optimizing ΔV for deflecting earth-crossing asteroidsrdquoJournal of Spacecraft and Rockets vol 38 no 5 pp 759ndash7642001

[3] B Dachwald and B Wie ldquoSolar sail trajectory optimization forintercepting impacting and deflecting near-earth asteroidsrdquoin Proceedings of the AIAA Guidance Navigation and ControlConference and Exhibit pp 3331ndash3348 San Francisco CalifUSA August 2005

[4] D Izzo ldquoOptimization of interplanetary trajectories for impul-sive and continuous asteroid deflectionrdquo Journal of GuidanceControl and Dynamics vol 30 no 2 pp 401ndash408 2007

[5] J Park R G Rhinehart and P T Kabamba ldquoMiss analysisin Lambert interceptions with application to a new guidancelawrdquo Mathematical Problems in Engineering Theory Methodsand Applications vol 6 no 2-3 pp 225ndash265 2000

[6] Y Ulybyshev ldquoDirect high-speed interception analytic solu-tions qualitative analysis and applicationsrdquo Journal of Space-craft and Rockets vol 38 no 3 pp 351ndash359 2001

[7] S P Burns and J J Scherock ldquoLambert guidance routinedesigned to match position and velocity of ballistic targetrdquoJournal of Guidance Control and Dynamics vol 27 no 6 pp989ndash996 2004

[8] R H Battin An Introduction to the Mathematics and Methodsof Astrodynamics American Institute of Aeronautics and Astro-nautics (AIAA) Reston Va USA Revised edition 1999

[9] S L Nelson and P Zarchan ldquoAlternative approach to thesolution of Lambertrsquos problemrdquo Journal of Guidance Controland Dynamics vol 15 no 4 pp 1003ndash1009 1992

[10] P V Arlulkar and S D Naik ldquoSolution based on dynamicalapproach for multiple-revolution lambert problemrdquo Journal ofGuidance Control and Dynamics vol 34 no 3 pp 920ndash9232011

[11] G Zhang D Mortari and D Zhou ldquoConstrained multiple-revolution Lambertrsquos problemrdquo Journal of Guidance Controland Dynamics vol 33 no 6 pp 1779ndash1786 2010

10 Mathematical Problems in Engineering

[12] C A Lembeck and J E Prussing ldquoOptimal impulsive interceptwith low-thrust rendezvous returnrdquo Journal of Guidance Con-trol and Dynamics vol 16 no 3 pp 426ndash433 1993

[13] D-R Taur V Coverstone-Carroll and J E Prussing ldquoOptimalimpulsive time-fixed orbital rendezvous and interception withpath constraintsrdquo Journal of Guidance Control and Dynamicsvol 18 no 1 pp 54ndash60 1995

[14] H Leeghim ldquoSpacecraft intercept using minimum controlenergy andwait timerdquoCelestialMechanics ampDynamical Astron-omy vol 115 no 1 pp 1ndash19 2013

[15] D A Vallado Fundamentals of Astrodynamics and ApplicationsMicrocosm Press El Segundo Calif USA 2nd edition 2001

[16] B F Thompson K K Choi S W Piggott and S R BeaverldquoOrbital targeting based on hodograph theory for improvedrendezvous safetyrdquo Journal of Guidance Control and Dynamicsvol 33 no 5 pp 1566ndash1576 2010

[17] V G Adamyan L V Adamyan G M Zaimtsyan and LT Manandyan ldquoDouble-pulse cotangential transfers betweencoplanar elliptic orbitsrdquo Journal of Applied Mathematics andMechanics vol 73 no 6 pp 664ndash672 2009

[18] G Zhang D Zhou D Mortari and T A Henderson ldquoAna-lytical study of tangent orbit and conditions for its solutionexistencerdquo Journal of Guidance Control and Dynamics vol 35no 1 pp 186ndash194 2012

[19] G Zhang D Zhou Z Sun and X Cao ldquoOptimal two-impulsecotangent rendezvous between coplanar elliptical orbitsrdquo Jour-nal of Guidance Control and Dynamics vol 36 no 3 pp 677ndash685 2013

[20] G Zhang X Cao and D Zhou ldquoTwo-impulse cotangent ren-dezvous between coplanar elliptic and hyperbolic orbitsrdquo Jour-nal of Guidance Control and Dynamics vol 37 no 3 pp 964ndash969 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 3

from which it is known that1198672= log (sinh119867

2+ cosh119867

2)

= log(radic1198902

2minus 1 sin119891

2+ 1198902+ cos119891

2

1 + 1198902cos1198912

)

(5)

Then the derivative of1198672with respect to 119891

2is

d1198672

d1198912

=

radic1198902

2minus 1

1 + 1198902cos1198912

(6)

Furthermore the derivative of 1199052with respect to 119891

2is

d1199052

d1198912

= 11990320radic

1199012

120583

cos (1198912minus 11989120) + 1198902cos11989120

(1 + 1198902cos1198912)2

+ radicminus1198863

2

120583[cosh (119867

2minus 11986720) minus 1]

radic1198902

2minus 1

1 + 1198902cos1198912

(7)

32 Transfer Time between the Initial and Target Orbits Theinitial flight-path angle at 119875

1can be written as [8]

120574 = arctan(1198901sin1198911

1 + 1198901cos1198911

) (8)

Let the transfer angle be 120579 and [9]

120582 =1199032(1 minus cos 120579)

1199031cos2120574 minus 119903

2cos (120579 + 120574) cos 120574

(9)

V1119905= radic

120583120582

1199031

(10)

where V1119905

denotes the magnitude of the velocity vector atpoint 119875

1on the transfer orbit For different values of 120582 there

are three cases for the transfer time [9 19](1) If 120582 isin (0 2) the transfer trajectory is an elliptic orbit

and the transfer time is

1199053=

2

radic120583(

1199031

2 minus 120582)

32

tanminus1 (radic2120582 minus 1

cos 120574cot (1205792) minus sin 120574)

+ 1199032radic

1199031

120583120582

tan 120574 (1 minus cos 120579) + (1 minus 120582) sin 120579(2 minus 120582) cos 120574

(11)

(2) If 120582 gt 2 the transfer trajectory is a hyperbolic orbitand the transfer time is

1199053= 1199032radic

1199031

120583120582

timestan 120574 (1 minus cos 120579) + (1 minus 120582) sin 120579

(2 minus 120582) cos 120574

minus1

radic120583(

1199031

120582 minus 2)

32

times lnsin 120574 minus cos 120574cot (1205792) minus (1 minus 2120582)12

sin 120574 minus cos 120574cot (1205792) + (1 minus 2120582)12

(12)

(3) If 120582 = 2 the transfer trajectory is a parabolic orbit andthe transfer time is

1199053=

2

3

radic1199033

1

120583120582[

3 cos 120574cot (1205792)[cos 120574cot (1205792) minus sin 120574]2

+1

[cos 120574cot (1205792) minus sin 120574]3]

(13)

It should be notified that ((11)ndash(13)) are functions indepen-dent of V

1119905

The derivative of the transfer time 1199053with respect to the

true anomaly 1198912is

d1199053

d1198912

= (1205971199053

1205971199032

+1205971199053

120597120582

120597120582

1205971199032

)d1199032

d1198912

+ (1205971199053

120597120579+1205971199053

120597120582

120597120582

120597120579)

d120579d1198912

(14)

where the partial derivatives 12059711990531205971199032 1205971199053120597120582 120597119905

3120597120579 120597120582120597119903

2

and 120597120582120597120579 and the full derivative d1199032d1198912are obtained in

[19 20] andd120579d1198912= 1Thepartial derivatives 120597119905

3120597120582 120597119905

3120597120579

are different for elliptic transfers and hyperbolic transfersbut the partial derivatives 120597119905

31205971199032 120597120582120597119903

2 and 120597120582120597120579 and

the full derivative d1199032d1198912are the same for elliptic orbits

and hyperbolic orbits Note that the target true anomalies 1198912

for parabolic orbits are not continuous thus the expressiond1199053d1198912is not derived for 120582 = 2

For a given impulse point the initial coasting time 1199051for

the interceptor is independent of 1198912 Then the derivative of 120578

with respect to 1198912is

1205781015840≜

d120578d1198912

=1

1198791

(d1199052

d1198912

minusd1199053

d1198912

) (15)

4 Conditions of Solution Existence

The above section gives the transfer time for any conic orbitA necessary condition is 120582 gt 0 Since the initial flight-pathangle 120574 isin (minus1205872 1205872) the initial flight-direction angle is

120574 =120587

2minus 120574 (16)

For the coplanar orbits the transfer angle is

120579 = 1198912+ 120594 where 120594 ≜ 120596

2minus 1205961minus 1198911 (17)

where 120596 is the argument of perigeeLet

1198881= sin (120594 minus 120574) + 119890

2

1199031

1199012

sin 120574

1198882= cos (120594 minus 120574)

1198883= minus

1199031

1199012

sin 120574

(18)

where 119901 is the semilatus rectum There are three cases for120582 gt 0 [18] (1) if 119888

3radic1198882

1+ 1198882

2lt minus1 119891

2isin [0 2120587) (2) if

4 Mathematical Problems in Engineering

1198883radic1198882

1+ 1198882

2gt 1 there is no solution for 119891

2 (3) if minus1 le

1198883radic1198882

1+ 1198882

2le 1 then the range of 119891

2for 120582 gt 0 is

arcsin(1198883

radic1198882

1+ 1198882

2

)minus atan2 (1198881 1198882)

lt 1198912lt 120587 minus arcsin(

1198883

radic1198882

1+ 1198882

2

)minus atan2 (1198881 1198882)

(19)

where the four-quadrant inverse tangent function 120572 =

atan2(1198871 1198872) is defined as the angle satisfying both sin120572 =

1198871radic1198872

1+ 1198872

2and cos120572 = 119887

2radic1198872

1+ 1198872

2

For any elliptical transfer orbit 0 lt 120582 lt 2 is required Let

1198884=1199012

1199031

cos (2120574) + 1198902(1 minus cos (2120574)) cos120594

1198885=1199012

1199031

sin (2120574) + 1198902(1 minus cos (2120574)) sin120594

1198886=1199012

1199031

+ cos (2120574) minus 1

(20)

Then there are also three cases to be discussed [18] (1) if1198886radic1198882

4+ 1198882

5lt minus1 119891

2isin [0 2120587) (2) if 119888

6radic1198882

4+ 1198882

5gt 1 there is

no solution for 1198912 (3) if minus1 le 119888

6radic1198882

4+ 1198882

5le 1 then the range

of 1198912for 0 lt 120582 lt 2 can be expressed as

minus 120594 minus atan2 (1198884 1198885) + arcsin(

1198886

radic1198882

4+ 1198882

5

)

lt 1198912lt 120587 minus 120594 minus atan2 (119888

4 1198885) minus arcsin(

1198886

radic1198882

4+ 1198882

5

)

(21)

In addition the range of 1198911for existing elliptic transfers

needs to be solved If there exists some range of 1198912 for 0 lt

120582 lt 2 the following inequality must be satisfied

1198882

4+ 1198882

5minus 1198882

6ge 0 (22)

Substituting (20) into (22) yields

[1 minus cos (2120574)]

times (1198902

2minus 1) [1 minus cos (2120574)] + 2

1199012

1199011

(1 + 1198901cos1198911)

times [1 + 1198902cos (2120574 minus 120594)] ge 0

(23)

Note that the flight-direction angle 120574 isin (0 1205872) thencos(2120574) lt 1 is satisfied for any 119891

1 thus (23) can be rewritten

as

119865 ≜ (1198902

2minus 1) [1 minus cos (2120574)]

+ 21199012

1199011

(1 + 1198901cos1198911) [1 + 119890

2cos (2120574 minus 120594)] ge 0

(24)

From ((16) (17)) it is known that 120574 and 120594 are functions onlyof 1198911 then 119865 is a function only of 119891

1 The values of 119891

1for the

equality 119865 = 0 can be obtained by numerical methods finallythe range for inequality 119865 gt 0 in (24) is obtained

5 Special Initial True Anomalies andthe Solution-Existence Range

There are four special points on the initial orbit whosedirections of velocity vectors are parallel to either of the linesof asymptotes for the target hyperbolic orbit These initialtrue anomalies 119891

1119904119895(119895 = 1 2 3 4) are solved in the following

paragraphs Let 119906119904119895

= 1198911119904119895

+ 1205961 (119895 = 1 2 3 4) 119906

2max =

1198912max + 1205962 = arccos(minus1119890

2) + 1205962 and 119906

2 min = arccos(11198902) +

1205962 From the geometry of Figure 2(a) it is known that

1199061199041+ 1205741199041= 1199062max 997904rArr 119891

11199041+ 1205961+ 1205741199041

= arccos(minus 1

1198902

) + 1205962

(25)

Note that the flight-direction angle is a function only of thetrue anomaly then the above equation can be rewritten as

11989111199041

+ 1205961+120587

2minus arctan(

1198901sin11989111199041

1 + 1198901cos11989111199041

)

= arccos(minus 1

1198902

) + 1205962

(26)

which is a function only of 11989111199041

Thus the classic Newton-Raphson iterative algorithm can be used to obtain thenumerical solution for (26)

From the geometry of Figure 2(b) it is known that

11989111199042

+ 1205961+ 1205741199042minus [arccos( 1

1198902

) + 1205962] = 120587 (27)

which can be rewritten in a function only of 11989111199042

as

11989111199042

+ 1205961minus120587

2minus arctan(

1198901sin11989111199042

1 + 1198901cos11989111199042

)

= arccos( 1

1198902

) + 1205962

(28)

Similarly from the geometries of Figures 2(c) and 2(d) itis known that the function only of 119891

11199043is

11989111199043

+ 1205961minus120587

2minus arctan(

1198901sin11989111199043

1 + 1198901cos11989111199043

)

= arccos(minus 1

1198902

) + 1205962

(29)

Mathematical Problems in Engineering 5

Parallel line of asymptotes Target hyperbolic orbit

Initial elliptic orbit

F1

Ps1

us1

rs1 u2max

120574s1

(a) 11989111199041

Parallel line of asymptotesTarget hyperbolic orbit

Initial elliptic orbit

F1

Ps2

us2rs2

u2min120574s2

(b) 11989111199042

Parallel line of asymptotes

Target hyperbolic orbit

Initial elliptic orbit

F1

Ps3

us3

rs3u2max

120574s3

(c) 11989111199043

Parallel line of asymptotes

Target hyperbolic orbit

Initial elliptic orbit

F1

Ps4

us4

rs4u2min

120574s4

(d) 11989111199044

Figure 2 Geometry interpretation for special initial true anomalies

and the function only of 11989111199044

is

11989111199044

+ 1205961minus3120587

2minus arctan(

1198901sin11989111199044

1 + 1198901cos11989111199044

)

= arccos( 1

1198902

) + 1205962

(30)

All the values of 1198911119904119895

(119895 = 1 2 3 4) are obtained by Newton-Raphson iterative algorithm

Once the special initial true anomalies are obtained thesolution-existence range of target true anomaly 119891

2can be

discussed and obtained as follows

(1) If 1198911isin [11989111199041

11989111199042

] there is only one intersectionpoint for the velocity-vector line and the targethyperbolic orbit Assume that the true anomaly of

the intersection point on the target hyperbolic orbitis 1198912119902 which can be obtained by (19) then the

target true anomaly is in the range (minus1198912max 1198912119902)

The solution-existence range for any conic transferorbit is obtained as follows (a) If there is no rangefor elliptic transfer orbits there is also no range forany conic transfer orbit (b) If there exists a range(1198912119886 1198912119887) sub (minus119891

2max 1198912119902) for elliptic transfer orbitswhich is obtained by (21) then the range for any conictransfer orbit can be (119891

2119886 1198912119887) or (minus119891

2max 1198912119887) When1199053by (11)ndash(13) is positive for an arbitrarily selected

1198912isin (minus119891

2max 1198912119886) for example 1198912= (minus119891

2max +1198912119886)2 the solution-existence range is (minus119891

2max 1198912119887)otherwise the range is (119891

2119886 1198912119887)

(2) If 1198911

isin (11989111199042

11989111199043

) there are two intersectionpoints for the velocity-vector line and the target

6 Mathematical Problems in Engineering

hyperbolic orbit Assume that the true anomalies ofthe intersection points on the target hyperbolic orbitare 1198912119901

and 1198912119902 then the solution-existence range for

any conic transfer orbit is in the range (1198912119901 1198912119902) The

solution-existence range for elliptic transfer orbits isobtained as (119891

2119886 1198912119887) sub (119891

2119901 1198912119902) by (21) Then the

solution-existence range for any conic transfer orbitis (1198912119901 1198912119887)

(3) If 1198911isin [11989111199043

11989111199044

] there is only one intersectionpoint for the velocity-vector line and the targethyperbolic orbit Assume that the true anomaly ofthe intersection point on the target hyperbolic orbitis 1198912119901 which is obtained by (19) then the target true

anomaly is in the range (1198912119901 1198912max) The solution-

existence range for any conic transfer orbit is obtainedas follows (a) If there is no range for elliptic transferorbits then the solution-existence range for any conictransfer orbit is (119891

2119901 1198912max) (b) If there exists a range

(1198912119886 1198912119887) sub (119891

2119901 1198912max) for elliptic transfer orbits

then the range for any conic transfer orbit can be(1198912119901 1198912119887) or (119891

2119901 1198912max) When 119905

3by ((11)ndash(13)) is

positive for an arbitrarily selected 1198912isin (1198912119887 1198912max)

for example 1198912

= (1198912119887

+ 1198912max)2 the solution-

existence range is (1198912119901 1198912max) otherwise the range is

(1198912119901 1198912119887)

(4) If 1198911

isin (11989111199044

11989111199041

) there are two intersectionpoints for the velocity-vector line and the targethyperbolic orbit Assume that the true anomalies ofthe intersection points on the target hyperbolic orbitare 1198912119901

and 1198912119902 then the target true anomaly must be

in the range (1198912119902 1198912max) There is no range for elliptic

transfer orbits and the solution-existence range forany conic (hyperbolic) transfer orbit is (119891

2119902 1198912max)

From the above discussions it is known that there maybe no solution-existence range for any conic transfer orbitonly when 119891

1isin [119891

11199041 11989111199042

] and there is certainly nosolution-existence range for elliptic transfer orbits when 119891

1isin

[11989111199044

11989111199041

]Let the solution-existence range for any conic transfer

by the above method be (1198912119898 1198912119899) The target true anomaly

1198912119894at the impulse time is obtained by the hyperbolic orbit

propagationThen there are three cases (1) If 1198912119894lt 1198912119898 then

the final solution should be obtained in the range (1198912119898 1198912119899)

(2) If 1198912119894

isin [1198912119898 1198912119899] then the final solution should be

obtained in the range (1198912119894 1198912119899) (3) If 119891

2119894gt 1198912119899 then there

is no solution

6 Solving the Problem

For a given impulse point the solution-existence range forany conic transfer and that for elliptic transfers are obtainedin Section 5 Assume that the final solution-existence rangefor any conic transfer considering the target true anomalyat the impulse time is (119891

21198921 11989121198922

) The interception problemcan be solved by numerical methods

Thevalue of 1205781015840 can be calculated by (15) then the solutionsof 1198912for 1205781015840(119891

2) = 0 can be obtained by a numerical iterative

algorithm for example the golden section search If thereis no solution for 1205781015840(119891

2) = 0 then 120578 decreases or increases

monotonically Then for the range (11989121198921

11989121198922

) if lfloor120578(1198911198921

+

120575)rfloor = lfloor120578(1198911198922minus 120575)rfloor or max(lfloor120578(119891

1198921+ 120575)rfloor lfloor120578(119891

1198922minus 120575)rfloor) lt 0

where the function lfloor119910rfloor denotes the nearest integer less thanor equal to 119910 and 120575 is a small positive number then there isno solution for (2) Otherwise there are two cases

(1) If min(lfloor120578(1198911198921

+ 120575)rfloor lfloor120578(1198911198922

minus 120575)rfloor) ge 0 there are|lfloor120578(1198911198922minus 120575)rfloor minus lfloor120578(119891

1198921+ 120575)rfloor| solutions which can be

obtained by the secant method as

1198912119899+2

= 1198912119899+1

minus (1198912119899+1

minus 1198912119899)

120578 (1198912119899+1

) minus 1198731

120578 (1198912119899+1

) minus 120578 (1198912119899)

119899 isin N

(31)

where the initial guesses are11989120

= 1198911198921+12057511989121

= 1198911198922minus

1205751198731= lfloor120578(119891

1198921+120575)rfloor+1 lfloor120578(119891

1198921+120575)rfloor+2 lfloor120578(119891

1198922minus120575)rfloor

if lfloor120578(1198911198921+120575)rfloor lt lfloor120578(119891

1198922minus120575)rfloor and119873

1= lfloor120578(119891

1198922minus120575)rfloor+1

lfloor120578(1198911198922minus 120575)rfloor + 2 lfloor120578(119891

1198921+ 120575)rfloor if lfloor120578(119891

1198921+ 120575)rfloor gt

lfloor120578(1198911198922minus 120575)rfloor

(2) If min(lfloor120578(1198911198921

+ 120575)rfloor lfloor120578(1198911198922

minus 120575)rfloor) lt 0 there are[max(lfloor120578(119891

1198921+ 120575)rfloor lfloor120578(119891

1198922minus 120575)rfloor) + 1] solutions

which can be obtained by (31) with 1198731= 0 1

max(lfloor120578(1198911198921

+ 120575)rfloor lfloor120578(1198911198922

minus 120575)rfloor) Note that 120578(1198911198921)

instead of 120578(1198911198921

+ 120575) should be used for the range[1198911198921 1198911198922)

If there exist 119896119899ge 1 solutions for 1205781015840(119891

2) = 0 and these

119896119899extreme points for 120578(119891

2) are obtained as 119891

21198961lt 11989121198962

lt

sdot sdot sdot lt 1198912119896119899

then there are 119896119899+ 1 piecewise ranges that is

(11989121198921

11989121198961

) [11989121198961

11989121198962

) [1198912119896119899

11989121198922

) For each piecewiserange the function of 120578 decreases or increasesmonotonicallyThen the solution for each piecewise range can be obtainedby (31) with different initial guesses If lim

(1198912rarr11989121198922)120578 = +infin

(see Figure 3(d)) there are infinite solutions in the range[1198912119896119899

11989121198922

) and the minimum-time solution (or with theminimum 119891

2) will be obtained

In the patched-conic approximation method the radiusof the Earthrsquos sphere of influence is 119877SOI = 925 000 km thenthe true anomaly range of the solution is 119891

2isin (minus119891SOI 119891SOI)

where 119891SOI = arccos([1199012119877SOI minus 1]119890

2) Thus the solutions

beyond this range should be removedThe above section gives a method to obtain all solutions

for a given initial impulse point 1198911 For each solution the

corresponding fuel consumption is

ΔV = 1003816100381610038161003816V1119905 minus V1

1003816100381610038161003816

=

1003816100381610038161003816100381610038161003816100381610038161003816

radic120583

1199011

120582 (1 + 1198901cos1198911) minus radic

120583

1199011

(1 + 1198902

1+ 21198901cos1198911)

1003816100381610038161003816100381610038161003816100381610038161003816

(32)

The solution whose cost is greater than the given maxi-mal value should be removed Thus the feasible solutionneeds to satisfy two conditions including the radius of theEarthrsquos sphere of influence and the given maximal valueof fuel When considering the initial coasting time before

Mathematical Problems in Engineering 7

minus120 minus100 minus80 minus60 minus40 minus20 0 20 40 60minus1

minus08

minus06

minus04

minus02

0

02

True anomaly f2 (∘)

120578

(a) 1198911 = 160∘

minus150 minus100 minus50 0 50 100minus1

minus05

0

05

True anomaly f2 (∘)

120578

(b) 1198911 = 170∘

minus40 minus20 0 20 40 60 80 100 120 140minus5

minus4

minus3

minus2

minus1

0

1

True anomaly f2 (∘)

120578

(c) 1198911 = 240∘

60 70 80 90 100 110 120 1300

1

2

3

4

5

6

7

8

9

10

True anomaly f2 (∘)

120578

(d) 1198911 = 330∘

Figure 3 The curves of 120578 versus 1198912for different 119891

1

the impulse the impulse point is free then for each initialimpulse point all feasible solutions can be obtained and theminimum-time one is with the minimum value of 119891

2

Reference [20] dealt with the cotangent rendezvousbetween two coplanar elliptic and hyperbolic orbits (Problem1) whereas this paper is for the tangent-impulse interceptionproblem from an interceptor on elliptic parking orbits toa target on hyperbolic orbits (Problem 2) Three maindifferences between these two problems are listed as follows

(1) For Problem 1 there is a single closed-form solutionfor an arbitrary point on the target hyperbolic orbitthat is 119891

2isin (minus119891

2max 1198912max) where 1198912max =

arccos(minus11198902) thus there is only one degree of

freedom in (2) for zero revolution however in aspecified (eg counterclockwise) motion directionthe solution-existence range is not the whole rangeof target true anomaly For Problem 2 there isno certain relationship between the tangent-impulse

point1198911and the interception point119891

2 In other words

there are two degrees of freedom in (2) with zerorevolutions

(2) For Problem 1 with nonintersecting case (counter-clockwise) the solution-existence range for cotangentrendezvous is 119891

2isin (1198912119886 1198912max) where (1198912119886 1198912119887) is

obtained by the solution-existence range for elliptictransfers [20] But for Problem 2with nonintersectingcase (counterclockwise) the solution-existence rangeof 1198912can be obtained by the proposed method for

different1198911 which is dependent on four special initial

true anomalies whose velocity vectors are parallel tothe lines of asymptotes for the target hyperbolic orbitMoreover there may be no solution-existence rangefor any conic transfer orbit when 119891

1isin [11989111199041

11989111199042

](3) For Problem 1 with a given revolution number 119873

3

of the transfer orbit usually there is only one orbitsolution (see the examples in [20]) However for

8 Mathematical Problems in Engineering

Problem 2 with an upper-bounded impulse theminimum-time solution is interesting then 119873

3is set

to be 0 For a given impulse point 1198911 there may exist

multiple solutions of 1198912(see Figures 3(a) and 3(b))

With the initial coasting time all feasible solutionsfor different impulse points are obtained and theminimum-time one is with the minimum value of 119891

2

7 Numerical Examples

Assume that the interceptorrsquos orbit elements are 1198861= 2119877119864+

4000 km 1198901= 06 120596

1= 10∘ and 119891

10= 60∘ and the targetrsquos

orbit elements are 1198862= minus2119877

119864minus 10 000 km 119890

2= 16 120596

2= 0∘

and 11989120

= minus120∘ where 119877

119864= 637813 km is the Earthrsquos

radius The gravitational parameter used for the example is120583 = 3986004415 km3s2 The upper bound of the tangentimpulse is set to be 5 kms Four special initial true anomaliesare obtained as 119891

11199041= 454187

∘ 11989111199042

= 1581023∘ 11989111199043

=

1919457∘ and 119891

11199044= 2845334

∘ via (26) (28) (29) and(30) To obtain elliptic transfers the range of the initial trueanomaly of the impulse point is119891

1isin (799244

∘ 2499268

∘) via

(24) In addition considering the radius of the Earthrsquos sphereof influence that is 925000 km the range of true anomaly is1198912isin (minus1269425

∘ 1269425

∘)

If the impulse is imposed at the initial time that is 1198911=

11989110

= 60∘ then 119891

1isin [11989111199041

11989111199042

] is satisfied and there is norange for elliptic transfers thus no solution-existence rangeexists for 119891

2

If the impulse is imposed at 1198911

= 160∘ the target

true anomaly at the impulse time is 1198912

= minus1183547∘

by solving Keplerrsquos equation for hyperbolic orbits andthe plots of 120578 versus 119891

2can be obtained in Figure 3(a)

The final solution-existence range for any conic transferconsidering the target true anomaly at the impulse timeis (11989121198921

11989121198922

) = (minus1183547∘ 724802

∘) There are two

solutions that is minus928589∘ and 207296∘ the correspondingrevolution numbers are both 0 and the corresponding costsare 13182 kms and 10211 kms respectively The minimum-time solution is minus928589∘ the corresponding transfer time is219724 s and the total interception time 119905

2is 283104 s via (3)

The value of 120582 is obtained as 11178 by (9) thus the trajectoryis elliptic and it is plotted in Figure 4

If the impulse is imposed at 1198911= 170

∘ the target trueanomaly at the impulse time is 119891

2= minus1177828

∘ the plots of120578 versus 119891

2can be obtained in Figure 3(b)The final solution-

existence range is (11989121198921

11989121198922

) = (minus1163047∘ 883112

∘)

There are three solutions minus1158579∘ minus1092932

∘ and532432

∘ all the corresponding revolution numbers are 0and the corresponding costs are 211998 kms 39088 kmsand 11770 kms respectivelyThe energy cost ofminus1158579∘ ismuch greater than 5 kms thus it cannot be usedThe feasibleminimum-time solution is minus1092932∘ the correspondingtransfer time is 131883 s and the total interception time is212588 s The value of 120582 is 27636 thus the trajectory ishyperbolic and it is plotted in Figure 5

If the impulse is imposed at 1198911= 240

∘ the target trueanomaly at the impulse time is 119891

2= minus1125985

∘ the plots of120578 versus 119891

2can be obtained in Figure 3(c) There is a single

minus15 minus10 minus5 0 5

minus16

minus14

minus12

minus10

minus8

minus6

minus4

minus2

0

2

X (km)

Y (k

m)

InitialTargetTransfer

P1

P20

P10

P2

F1

times104

times104

Figure 4 Tangent-impulse interception with elliptic transfer trajec-tory

minus15 minus1 minus05 0 05 1minus16

minus14

minus12

minus10

minus8

minus6

minus4

minus2

0

2

4

Y (k

m)

X (km)

times104

times105

P1

P20

P10

P2

F1

InitialTargetTransfer

Figure 5 Tangent-impulse interception with hyperbolic transfertrajectory

solution 1152336∘ the corresponding revolution number is0 the cost is 17743 kms the transfer time is 363182 s andthe total interception time is 542081 sThe value of 120582 is 19316thus the trajectory is elliptic

If the impulse is imposed at 1198911= 330

∘ the target trueanomaly at the impulse time is119891

2= minus1102890

∘ the plots of 120578versus119891

2can be obtained in Figure 3(d) and there are infinite

solutions However when considering the sphere of influenceof the Earth there are three solutions The minimum-timesolution is 1186911∘ the corresponding revolution number

Mathematical Problems in Engineering 9

Table 1 Feasible solutions of1198912for different119891

1and the correspond-

ing interception time 1199052and ΔV

1198911(∘) Solutions of

(1198912(∘)119873

1) 119905

2(s) ΔV (kms)

150 (minus566897 0) 321084 09035(minus399134 0) 328443 08897

170 (minus1092932 0) 212588 39088(532432 1) 358345 11770

190 (891630 0) 390801 14596210 (1043983 0) 436651 16333230 (1123591 0) 499326 17338250 (1177130 0) 598472 18148270 (1219959 0) 802495 19082290 (1259555 0) 1602413 20322310 (1244383 1) 1114899 22976330 (1186911 1) 628954 29698

(1257512 2) 1506675 26872(1269266 3) 2368738 26464

350 (1142511 1) 525410 44692(1242882 2) 1085322 35407(1260145 3) 1632862 34192(1267542 4) 2175443 33698

10 (1261114 4) 1685896 49766(1266609 5) 2084768 48911

is 1 the corresponding costs is 29698 kms the correspond-ing transfer time is 209279 s and the total interception timeis 628954 s The value of 120582 is 27280 thus the trajectory ishyperbolic

For different impulse points all feasible solutions satisfy-ing 1198912isin (minus1269425

∘ 1269425

∘) and the fuel constraint and

the corresponding value of 1198732= 120578 for (2) can be obtained

by (31) and are listed in Table 1 The total interception time is1199052= 1199051+ 1199053 the corresponding costs are also listed in Table 1

It should be notified that there is no feasible solution when1198911= 30∘+ 119896119895sdot 20∘ 119896119895= 1 2 3 4 5 By using the numerical

optimization algorithm the minimum-time impulse point isat 1198911= 1705235

∘ the solution is 1198912= minus1108795

∘ the cost is49999 kms and the total interception time is 198106 s Theminimum-time interception trajectory is hyperbolic and it isnear to that in Figure 5

8 Conclusions

This paper studies the interception problem for a hyperbolictarget with an upper-bounded tangent impulse Based onfour special initial true anomalies the solution-existencerange of the target true anomaly for any conic transfer isobtained Then the problem is solved by the secant methodfor a time-of-flight equation For a given impulse pointthere may be zero one many or infinite solutions howeverwhen considering the sphere of influence of the Earth thereare always finite solutions For different impulse points allfeasible solutions and the global minimum-time solution are

obtained This proposed method is valid for an assignedimpulse point and the time optimization with the initialcoasting time The interception with a tangent-impulse tech-nique can be used in orbital interception task for potentiallyhazardous objects during the Earth flyby with simple control

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work is supported in part by the China Postdoctoral Sci-ence Foundation Funded Project (Grant no 2012M520753)theHeilongjiang Postdoctoral Science Foundation (Grant noLBH-Z12125) and theNational Natural Scientific Foundationof China (Grant no 61273096)

References

[1] J C Solem ldquoInterception of comets and asteroids on collisioncourse with Earthrdquo Journal of Spacecraft and Rockets vol 30 no2 pp 222ndash228 1993

[2] IMRoss S-Y Park and SDV Porter ldquoGravitational effects ofearth in optimizing ΔV for deflecting earth-crossing asteroidsrdquoJournal of Spacecraft and Rockets vol 38 no 5 pp 759ndash7642001

[3] B Dachwald and B Wie ldquoSolar sail trajectory optimization forintercepting impacting and deflecting near-earth asteroidsrdquoin Proceedings of the AIAA Guidance Navigation and ControlConference and Exhibit pp 3331ndash3348 San Francisco CalifUSA August 2005

[4] D Izzo ldquoOptimization of interplanetary trajectories for impul-sive and continuous asteroid deflectionrdquo Journal of GuidanceControl and Dynamics vol 30 no 2 pp 401ndash408 2007

[5] J Park R G Rhinehart and P T Kabamba ldquoMiss analysisin Lambert interceptions with application to a new guidancelawrdquo Mathematical Problems in Engineering Theory Methodsand Applications vol 6 no 2-3 pp 225ndash265 2000

[6] Y Ulybyshev ldquoDirect high-speed interception analytic solu-tions qualitative analysis and applicationsrdquo Journal of Space-craft and Rockets vol 38 no 3 pp 351ndash359 2001

[7] S P Burns and J J Scherock ldquoLambert guidance routinedesigned to match position and velocity of ballistic targetrdquoJournal of Guidance Control and Dynamics vol 27 no 6 pp989ndash996 2004

[8] R H Battin An Introduction to the Mathematics and Methodsof Astrodynamics American Institute of Aeronautics and Astro-nautics (AIAA) Reston Va USA Revised edition 1999

[9] S L Nelson and P Zarchan ldquoAlternative approach to thesolution of Lambertrsquos problemrdquo Journal of Guidance Controland Dynamics vol 15 no 4 pp 1003ndash1009 1992

[10] P V Arlulkar and S D Naik ldquoSolution based on dynamicalapproach for multiple-revolution lambert problemrdquo Journal ofGuidance Control and Dynamics vol 34 no 3 pp 920ndash9232011

[11] G Zhang D Mortari and D Zhou ldquoConstrained multiple-revolution Lambertrsquos problemrdquo Journal of Guidance Controland Dynamics vol 33 no 6 pp 1779ndash1786 2010

10 Mathematical Problems in Engineering

[12] C A Lembeck and J E Prussing ldquoOptimal impulsive interceptwith low-thrust rendezvous returnrdquo Journal of Guidance Con-trol and Dynamics vol 16 no 3 pp 426ndash433 1993

[13] D-R Taur V Coverstone-Carroll and J E Prussing ldquoOptimalimpulsive time-fixed orbital rendezvous and interception withpath constraintsrdquo Journal of Guidance Control and Dynamicsvol 18 no 1 pp 54ndash60 1995

[14] H Leeghim ldquoSpacecraft intercept using minimum controlenergy andwait timerdquoCelestialMechanics ampDynamical Astron-omy vol 115 no 1 pp 1ndash19 2013

[15] D A Vallado Fundamentals of Astrodynamics and ApplicationsMicrocosm Press El Segundo Calif USA 2nd edition 2001

[16] B F Thompson K K Choi S W Piggott and S R BeaverldquoOrbital targeting based on hodograph theory for improvedrendezvous safetyrdquo Journal of Guidance Control and Dynamicsvol 33 no 5 pp 1566ndash1576 2010

[17] V G Adamyan L V Adamyan G M Zaimtsyan and LT Manandyan ldquoDouble-pulse cotangential transfers betweencoplanar elliptic orbitsrdquo Journal of Applied Mathematics andMechanics vol 73 no 6 pp 664ndash672 2009

[18] G Zhang D Zhou D Mortari and T A Henderson ldquoAna-lytical study of tangent orbit and conditions for its solutionexistencerdquo Journal of Guidance Control and Dynamics vol 35no 1 pp 186ndash194 2012

[19] G Zhang D Zhou Z Sun and X Cao ldquoOptimal two-impulsecotangent rendezvous between coplanar elliptical orbitsrdquo Jour-nal of Guidance Control and Dynamics vol 36 no 3 pp 677ndash685 2013

[20] G Zhang X Cao and D Zhou ldquoTwo-impulse cotangent ren-dezvous between coplanar elliptic and hyperbolic orbitsrdquo Jour-nal of Guidance Control and Dynamics vol 37 no 3 pp 964ndash969 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Mathematical Problems in Engineering

1198883radic1198882

1+ 1198882

2gt 1 there is no solution for 119891

2 (3) if minus1 le

1198883radic1198882

1+ 1198882

2le 1 then the range of 119891

2for 120582 gt 0 is

arcsin(1198883

radic1198882

1+ 1198882

2

)minus atan2 (1198881 1198882)

lt 1198912lt 120587 minus arcsin(

1198883

radic1198882

1+ 1198882

2

)minus atan2 (1198881 1198882)

(19)

where the four-quadrant inverse tangent function 120572 =

atan2(1198871 1198872) is defined as the angle satisfying both sin120572 =

1198871radic1198872

1+ 1198872

2and cos120572 = 119887

2radic1198872

1+ 1198872

2

For any elliptical transfer orbit 0 lt 120582 lt 2 is required Let

1198884=1199012

1199031

cos (2120574) + 1198902(1 minus cos (2120574)) cos120594

1198885=1199012

1199031

sin (2120574) + 1198902(1 minus cos (2120574)) sin120594

1198886=1199012

1199031

+ cos (2120574) minus 1

(20)

Then there are also three cases to be discussed [18] (1) if1198886radic1198882

4+ 1198882

5lt minus1 119891

2isin [0 2120587) (2) if 119888

6radic1198882

4+ 1198882

5gt 1 there is

no solution for 1198912 (3) if minus1 le 119888

6radic1198882

4+ 1198882

5le 1 then the range

of 1198912for 0 lt 120582 lt 2 can be expressed as

minus 120594 minus atan2 (1198884 1198885) + arcsin(

1198886

radic1198882

4+ 1198882

5

)

lt 1198912lt 120587 minus 120594 minus atan2 (119888

4 1198885) minus arcsin(

1198886

radic1198882

4+ 1198882

5

)

(21)

In addition the range of 1198911for existing elliptic transfers

needs to be solved If there exists some range of 1198912 for 0 lt

120582 lt 2 the following inequality must be satisfied

1198882

4+ 1198882

5minus 1198882

6ge 0 (22)

Substituting (20) into (22) yields

[1 minus cos (2120574)]

times (1198902

2minus 1) [1 minus cos (2120574)] + 2

1199012

1199011

(1 + 1198901cos1198911)

times [1 + 1198902cos (2120574 minus 120594)] ge 0

(23)

Note that the flight-direction angle 120574 isin (0 1205872) thencos(2120574) lt 1 is satisfied for any 119891

1 thus (23) can be rewritten

as

119865 ≜ (1198902

2minus 1) [1 minus cos (2120574)]

+ 21199012

1199011

(1 + 1198901cos1198911) [1 + 119890

2cos (2120574 minus 120594)] ge 0

(24)

From ((16) (17)) it is known that 120574 and 120594 are functions onlyof 1198911 then 119865 is a function only of 119891

1 The values of 119891

1for the

equality 119865 = 0 can be obtained by numerical methods finallythe range for inequality 119865 gt 0 in (24) is obtained

5 Special Initial True Anomalies andthe Solution-Existence Range

There are four special points on the initial orbit whosedirections of velocity vectors are parallel to either of the linesof asymptotes for the target hyperbolic orbit These initialtrue anomalies 119891

1119904119895(119895 = 1 2 3 4) are solved in the following

paragraphs Let 119906119904119895

= 1198911119904119895

+ 1205961 (119895 = 1 2 3 4) 119906

2max =

1198912max + 1205962 = arccos(minus1119890

2) + 1205962 and 119906

2 min = arccos(11198902) +

1205962 From the geometry of Figure 2(a) it is known that

1199061199041+ 1205741199041= 1199062max 997904rArr 119891

11199041+ 1205961+ 1205741199041

= arccos(minus 1

1198902

) + 1205962

(25)

Note that the flight-direction angle is a function only of thetrue anomaly then the above equation can be rewritten as

11989111199041

+ 1205961+120587

2minus arctan(

1198901sin11989111199041

1 + 1198901cos11989111199041

)

= arccos(minus 1

1198902

) + 1205962

(26)

which is a function only of 11989111199041

Thus the classic Newton-Raphson iterative algorithm can be used to obtain thenumerical solution for (26)

From the geometry of Figure 2(b) it is known that

11989111199042

+ 1205961+ 1205741199042minus [arccos( 1

1198902

) + 1205962] = 120587 (27)

which can be rewritten in a function only of 11989111199042

as

11989111199042

+ 1205961minus120587

2minus arctan(

1198901sin11989111199042

1 + 1198901cos11989111199042

)

= arccos( 1

1198902

) + 1205962

(28)

Similarly from the geometries of Figures 2(c) and 2(d) itis known that the function only of 119891

11199043is

11989111199043

+ 1205961minus120587

2minus arctan(

1198901sin11989111199043

1 + 1198901cos11989111199043

)

= arccos(minus 1

1198902

) + 1205962

(29)

Mathematical Problems in Engineering 5

Parallel line of asymptotes Target hyperbolic orbit

Initial elliptic orbit

F1

Ps1

us1

rs1 u2max

120574s1

(a) 11989111199041

Parallel line of asymptotesTarget hyperbolic orbit

Initial elliptic orbit

F1

Ps2

us2rs2

u2min120574s2

(b) 11989111199042

Parallel line of asymptotes

Target hyperbolic orbit

Initial elliptic orbit

F1

Ps3

us3

rs3u2max

120574s3

(c) 11989111199043

Parallel line of asymptotes

Target hyperbolic orbit

Initial elliptic orbit

F1

Ps4

us4

rs4u2min

120574s4

(d) 11989111199044

Figure 2 Geometry interpretation for special initial true anomalies

and the function only of 11989111199044

is

11989111199044

+ 1205961minus3120587

2minus arctan(

1198901sin11989111199044

1 + 1198901cos11989111199044

)

= arccos( 1

1198902

) + 1205962

(30)

All the values of 1198911119904119895

(119895 = 1 2 3 4) are obtained by Newton-Raphson iterative algorithm

Once the special initial true anomalies are obtained thesolution-existence range of target true anomaly 119891

2can be

discussed and obtained as follows

(1) If 1198911isin [11989111199041

11989111199042

] there is only one intersectionpoint for the velocity-vector line and the targethyperbolic orbit Assume that the true anomaly of

the intersection point on the target hyperbolic orbitis 1198912119902 which can be obtained by (19) then the

target true anomaly is in the range (minus1198912max 1198912119902)

The solution-existence range for any conic transferorbit is obtained as follows (a) If there is no rangefor elliptic transfer orbits there is also no range forany conic transfer orbit (b) If there exists a range(1198912119886 1198912119887) sub (minus119891

2max 1198912119902) for elliptic transfer orbitswhich is obtained by (21) then the range for any conictransfer orbit can be (119891

2119886 1198912119887) or (minus119891

2max 1198912119887) When1199053by (11)ndash(13) is positive for an arbitrarily selected

1198912isin (minus119891

2max 1198912119886) for example 1198912= (minus119891

2max +1198912119886)2 the solution-existence range is (minus119891

2max 1198912119887)otherwise the range is (119891

2119886 1198912119887)

(2) If 1198911

isin (11989111199042

11989111199043

) there are two intersectionpoints for the velocity-vector line and the target

6 Mathematical Problems in Engineering

hyperbolic orbit Assume that the true anomalies ofthe intersection points on the target hyperbolic orbitare 1198912119901

and 1198912119902 then the solution-existence range for

any conic transfer orbit is in the range (1198912119901 1198912119902) The

solution-existence range for elliptic transfer orbits isobtained as (119891

2119886 1198912119887) sub (119891

2119901 1198912119902) by (21) Then the

solution-existence range for any conic transfer orbitis (1198912119901 1198912119887)

(3) If 1198911isin [11989111199043

11989111199044

] there is only one intersectionpoint for the velocity-vector line and the targethyperbolic orbit Assume that the true anomaly ofthe intersection point on the target hyperbolic orbitis 1198912119901 which is obtained by (19) then the target true

anomaly is in the range (1198912119901 1198912max) The solution-

existence range for any conic transfer orbit is obtainedas follows (a) If there is no range for elliptic transferorbits then the solution-existence range for any conictransfer orbit is (119891

2119901 1198912max) (b) If there exists a range

(1198912119886 1198912119887) sub (119891

2119901 1198912max) for elliptic transfer orbits

then the range for any conic transfer orbit can be(1198912119901 1198912119887) or (119891

2119901 1198912max) When 119905

3by ((11)ndash(13)) is

positive for an arbitrarily selected 1198912isin (1198912119887 1198912max)

for example 1198912

= (1198912119887

+ 1198912max)2 the solution-

existence range is (1198912119901 1198912max) otherwise the range is

(1198912119901 1198912119887)

(4) If 1198911

isin (11989111199044

11989111199041

) there are two intersectionpoints for the velocity-vector line and the targethyperbolic orbit Assume that the true anomalies ofthe intersection points on the target hyperbolic orbitare 1198912119901

and 1198912119902 then the target true anomaly must be

in the range (1198912119902 1198912max) There is no range for elliptic

transfer orbits and the solution-existence range forany conic (hyperbolic) transfer orbit is (119891

2119902 1198912max)

From the above discussions it is known that there maybe no solution-existence range for any conic transfer orbitonly when 119891

1isin [119891

11199041 11989111199042

] and there is certainly nosolution-existence range for elliptic transfer orbits when 119891

1isin

[11989111199044

11989111199041

]Let the solution-existence range for any conic transfer

by the above method be (1198912119898 1198912119899) The target true anomaly

1198912119894at the impulse time is obtained by the hyperbolic orbit

propagationThen there are three cases (1) If 1198912119894lt 1198912119898 then

the final solution should be obtained in the range (1198912119898 1198912119899)

(2) If 1198912119894

isin [1198912119898 1198912119899] then the final solution should be

obtained in the range (1198912119894 1198912119899) (3) If 119891

2119894gt 1198912119899 then there

is no solution

6 Solving the Problem

For a given impulse point the solution-existence range forany conic transfer and that for elliptic transfers are obtainedin Section 5 Assume that the final solution-existence rangefor any conic transfer considering the target true anomalyat the impulse time is (119891

21198921 11989121198922

) The interception problemcan be solved by numerical methods

Thevalue of 1205781015840 can be calculated by (15) then the solutionsof 1198912for 1205781015840(119891

2) = 0 can be obtained by a numerical iterative

algorithm for example the golden section search If thereis no solution for 1205781015840(119891

2) = 0 then 120578 decreases or increases

monotonically Then for the range (11989121198921

11989121198922

) if lfloor120578(1198911198921

+

120575)rfloor = lfloor120578(1198911198922minus 120575)rfloor or max(lfloor120578(119891

1198921+ 120575)rfloor lfloor120578(119891

1198922minus 120575)rfloor) lt 0

where the function lfloor119910rfloor denotes the nearest integer less thanor equal to 119910 and 120575 is a small positive number then there isno solution for (2) Otherwise there are two cases

(1) If min(lfloor120578(1198911198921

+ 120575)rfloor lfloor120578(1198911198922

minus 120575)rfloor) ge 0 there are|lfloor120578(1198911198922minus 120575)rfloor minus lfloor120578(119891

1198921+ 120575)rfloor| solutions which can be

obtained by the secant method as

1198912119899+2

= 1198912119899+1

minus (1198912119899+1

minus 1198912119899)

120578 (1198912119899+1

) minus 1198731

120578 (1198912119899+1

) minus 120578 (1198912119899)

119899 isin N

(31)

where the initial guesses are11989120

= 1198911198921+12057511989121

= 1198911198922minus

1205751198731= lfloor120578(119891

1198921+120575)rfloor+1 lfloor120578(119891

1198921+120575)rfloor+2 lfloor120578(119891

1198922minus120575)rfloor

if lfloor120578(1198911198921+120575)rfloor lt lfloor120578(119891

1198922minus120575)rfloor and119873

1= lfloor120578(119891

1198922minus120575)rfloor+1

lfloor120578(1198911198922minus 120575)rfloor + 2 lfloor120578(119891

1198921+ 120575)rfloor if lfloor120578(119891

1198921+ 120575)rfloor gt

lfloor120578(1198911198922minus 120575)rfloor

(2) If min(lfloor120578(1198911198921

+ 120575)rfloor lfloor120578(1198911198922

minus 120575)rfloor) lt 0 there are[max(lfloor120578(119891

1198921+ 120575)rfloor lfloor120578(119891

1198922minus 120575)rfloor) + 1] solutions

which can be obtained by (31) with 1198731= 0 1

max(lfloor120578(1198911198921

+ 120575)rfloor lfloor120578(1198911198922

minus 120575)rfloor) Note that 120578(1198911198921)

instead of 120578(1198911198921

+ 120575) should be used for the range[1198911198921 1198911198922)

If there exist 119896119899ge 1 solutions for 1205781015840(119891

2) = 0 and these

119896119899extreme points for 120578(119891

2) are obtained as 119891

21198961lt 11989121198962

lt

sdot sdot sdot lt 1198912119896119899

then there are 119896119899+ 1 piecewise ranges that is

(11989121198921

11989121198961

) [11989121198961

11989121198962

) [1198912119896119899

11989121198922

) For each piecewiserange the function of 120578 decreases or increasesmonotonicallyThen the solution for each piecewise range can be obtainedby (31) with different initial guesses If lim

(1198912rarr11989121198922)120578 = +infin

(see Figure 3(d)) there are infinite solutions in the range[1198912119896119899

11989121198922

) and the minimum-time solution (or with theminimum 119891

2) will be obtained

In the patched-conic approximation method the radiusof the Earthrsquos sphere of influence is 119877SOI = 925 000 km thenthe true anomaly range of the solution is 119891

2isin (minus119891SOI 119891SOI)

where 119891SOI = arccos([1199012119877SOI minus 1]119890

2) Thus the solutions

beyond this range should be removedThe above section gives a method to obtain all solutions

for a given initial impulse point 1198911 For each solution the

corresponding fuel consumption is

ΔV = 1003816100381610038161003816V1119905 minus V1

1003816100381610038161003816

=

1003816100381610038161003816100381610038161003816100381610038161003816

radic120583

1199011

120582 (1 + 1198901cos1198911) minus radic

120583

1199011

(1 + 1198902

1+ 21198901cos1198911)

1003816100381610038161003816100381610038161003816100381610038161003816

(32)

The solution whose cost is greater than the given maxi-mal value should be removed Thus the feasible solutionneeds to satisfy two conditions including the radius of theEarthrsquos sphere of influence and the given maximal valueof fuel When considering the initial coasting time before

Mathematical Problems in Engineering 7

minus120 minus100 minus80 minus60 minus40 minus20 0 20 40 60minus1

minus08

minus06

minus04

minus02

0

02

True anomaly f2 (∘)

120578

(a) 1198911 = 160∘

minus150 minus100 minus50 0 50 100minus1

minus05

0

05

True anomaly f2 (∘)

120578

(b) 1198911 = 170∘

minus40 minus20 0 20 40 60 80 100 120 140minus5

minus4

minus3

minus2

minus1

0

1

True anomaly f2 (∘)

120578

(c) 1198911 = 240∘

60 70 80 90 100 110 120 1300

1

2

3

4

5

6

7

8

9

10

True anomaly f2 (∘)

120578

(d) 1198911 = 330∘

Figure 3 The curves of 120578 versus 1198912for different 119891

1

the impulse the impulse point is free then for each initialimpulse point all feasible solutions can be obtained and theminimum-time one is with the minimum value of 119891

2

Reference [20] dealt with the cotangent rendezvousbetween two coplanar elliptic and hyperbolic orbits (Problem1) whereas this paper is for the tangent-impulse interceptionproblem from an interceptor on elliptic parking orbits toa target on hyperbolic orbits (Problem 2) Three maindifferences between these two problems are listed as follows

(1) For Problem 1 there is a single closed-form solutionfor an arbitrary point on the target hyperbolic orbitthat is 119891

2isin (minus119891

2max 1198912max) where 1198912max =

arccos(minus11198902) thus there is only one degree of

freedom in (2) for zero revolution however in aspecified (eg counterclockwise) motion directionthe solution-existence range is not the whole rangeof target true anomaly For Problem 2 there isno certain relationship between the tangent-impulse

point1198911and the interception point119891

2 In other words

there are two degrees of freedom in (2) with zerorevolutions

(2) For Problem 1 with nonintersecting case (counter-clockwise) the solution-existence range for cotangentrendezvous is 119891

2isin (1198912119886 1198912max) where (1198912119886 1198912119887) is

obtained by the solution-existence range for elliptictransfers [20] But for Problem 2with nonintersectingcase (counterclockwise) the solution-existence rangeof 1198912can be obtained by the proposed method for

different1198911 which is dependent on four special initial

true anomalies whose velocity vectors are parallel tothe lines of asymptotes for the target hyperbolic orbitMoreover there may be no solution-existence rangefor any conic transfer orbit when 119891

1isin [11989111199041

11989111199042

](3) For Problem 1 with a given revolution number 119873

3

of the transfer orbit usually there is only one orbitsolution (see the examples in [20]) However for

8 Mathematical Problems in Engineering

Problem 2 with an upper-bounded impulse theminimum-time solution is interesting then 119873

3is set

to be 0 For a given impulse point 1198911 there may exist

multiple solutions of 1198912(see Figures 3(a) and 3(b))

With the initial coasting time all feasible solutionsfor different impulse points are obtained and theminimum-time one is with the minimum value of 119891

2

7 Numerical Examples

Assume that the interceptorrsquos orbit elements are 1198861= 2119877119864+

4000 km 1198901= 06 120596

1= 10∘ and 119891

10= 60∘ and the targetrsquos

orbit elements are 1198862= minus2119877

119864minus 10 000 km 119890

2= 16 120596

2= 0∘

and 11989120

= minus120∘ where 119877

119864= 637813 km is the Earthrsquos

radius The gravitational parameter used for the example is120583 = 3986004415 km3s2 The upper bound of the tangentimpulse is set to be 5 kms Four special initial true anomaliesare obtained as 119891

11199041= 454187

∘ 11989111199042

= 1581023∘ 11989111199043

=

1919457∘ and 119891

11199044= 2845334

∘ via (26) (28) (29) and(30) To obtain elliptic transfers the range of the initial trueanomaly of the impulse point is119891

1isin (799244

∘ 2499268

∘) via

(24) In addition considering the radius of the Earthrsquos sphereof influence that is 925000 km the range of true anomaly is1198912isin (minus1269425

∘ 1269425

∘)

If the impulse is imposed at the initial time that is 1198911=

11989110

= 60∘ then 119891

1isin [11989111199041

11989111199042

] is satisfied and there is norange for elliptic transfers thus no solution-existence rangeexists for 119891

2

If the impulse is imposed at 1198911

= 160∘ the target

true anomaly at the impulse time is 1198912

= minus1183547∘

by solving Keplerrsquos equation for hyperbolic orbits andthe plots of 120578 versus 119891

2can be obtained in Figure 3(a)

The final solution-existence range for any conic transferconsidering the target true anomaly at the impulse timeis (11989121198921

11989121198922

) = (minus1183547∘ 724802

∘) There are two

solutions that is minus928589∘ and 207296∘ the correspondingrevolution numbers are both 0 and the corresponding costsare 13182 kms and 10211 kms respectively The minimum-time solution is minus928589∘ the corresponding transfer time is219724 s and the total interception time 119905

2is 283104 s via (3)

The value of 120582 is obtained as 11178 by (9) thus the trajectoryis elliptic and it is plotted in Figure 4

If the impulse is imposed at 1198911= 170

∘ the target trueanomaly at the impulse time is 119891

2= minus1177828

∘ the plots of120578 versus 119891

2can be obtained in Figure 3(b)The final solution-

existence range is (11989121198921

11989121198922

) = (minus1163047∘ 883112

∘)

There are three solutions minus1158579∘ minus1092932

∘ and532432

∘ all the corresponding revolution numbers are 0and the corresponding costs are 211998 kms 39088 kmsand 11770 kms respectivelyThe energy cost ofminus1158579∘ ismuch greater than 5 kms thus it cannot be usedThe feasibleminimum-time solution is minus1092932∘ the correspondingtransfer time is 131883 s and the total interception time is212588 s The value of 120582 is 27636 thus the trajectory ishyperbolic and it is plotted in Figure 5

If the impulse is imposed at 1198911= 240

∘ the target trueanomaly at the impulse time is 119891

2= minus1125985

∘ the plots of120578 versus 119891

2can be obtained in Figure 3(c) There is a single

minus15 minus10 minus5 0 5

minus16

minus14

minus12

minus10

minus8

minus6

minus4

minus2

0

2

X (km)

Y (k

m)

InitialTargetTransfer

P1

P20

P10

P2

F1

times104

times104

Figure 4 Tangent-impulse interception with elliptic transfer trajec-tory

minus15 minus1 minus05 0 05 1minus16

minus14

minus12

minus10

minus8

minus6

minus4

minus2

0

2

4

Y (k

m)

X (km)

times104

times105

P1

P20

P10

P2

F1

InitialTargetTransfer

Figure 5 Tangent-impulse interception with hyperbolic transfertrajectory

solution 1152336∘ the corresponding revolution number is0 the cost is 17743 kms the transfer time is 363182 s andthe total interception time is 542081 sThe value of 120582 is 19316thus the trajectory is elliptic

If the impulse is imposed at 1198911= 330

∘ the target trueanomaly at the impulse time is119891

2= minus1102890

∘ the plots of 120578versus119891

2can be obtained in Figure 3(d) and there are infinite

solutions However when considering the sphere of influenceof the Earth there are three solutions The minimum-timesolution is 1186911∘ the corresponding revolution number

Mathematical Problems in Engineering 9

Table 1 Feasible solutions of1198912for different119891

1and the correspond-

ing interception time 1199052and ΔV

1198911(∘) Solutions of

(1198912(∘)119873

1) 119905

2(s) ΔV (kms)

150 (minus566897 0) 321084 09035(minus399134 0) 328443 08897

170 (minus1092932 0) 212588 39088(532432 1) 358345 11770

190 (891630 0) 390801 14596210 (1043983 0) 436651 16333230 (1123591 0) 499326 17338250 (1177130 0) 598472 18148270 (1219959 0) 802495 19082290 (1259555 0) 1602413 20322310 (1244383 1) 1114899 22976330 (1186911 1) 628954 29698

(1257512 2) 1506675 26872(1269266 3) 2368738 26464

350 (1142511 1) 525410 44692(1242882 2) 1085322 35407(1260145 3) 1632862 34192(1267542 4) 2175443 33698

10 (1261114 4) 1685896 49766(1266609 5) 2084768 48911

is 1 the corresponding costs is 29698 kms the correspond-ing transfer time is 209279 s and the total interception timeis 628954 s The value of 120582 is 27280 thus the trajectory ishyperbolic

For different impulse points all feasible solutions satisfy-ing 1198912isin (minus1269425

∘ 1269425

∘) and the fuel constraint and

the corresponding value of 1198732= 120578 for (2) can be obtained

by (31) and are listed in Table 1 The total interception time is1199052= 1199051+ 1199053 the corresponding costs are also listed in Table 1

It should be notified that there is no feasible solution when1198911= 30∘+ 119896119895sdot 20∘ 119896119895= 1 2 3 4 5 By using the numerical

optimization algorithm the minimum-time impulse point isat 1198911= 1705235

∘ the solution is 1198912= minus1108795

∘ the cost is49999 kms and the total interception time is 198106 s Theminimum-time interception trajectory is hyperbolic and it isnear to that in Figure 5

8 Conclusions

This paper studies the interception problem for a hyperbolictarget with an upper-bounded tangent impulse Based onfour special initial true anomalies the solution-existencerange of the target true anomaly for any conic transfer isobtained Then the problem is solved by the secant methodfor a time-of-flight equation For a given impulse pointthere may be zero one many or infinite solutions howeverwhen considering the sphere of influence of the Earth thereare always finite solutions For different impulse points allfeasible solutions and the global minimum-time solution are

obtained This proposed method is valid for an assignedimpulse point and the time optimization with the initialcoasting time The interception with a tangent-impulse tech-nique can be used in orbital interception task for potentiallyhazardous objects during the Earth flyby with simple control

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work is supported in part by the China Postdoctoral Sci-ence Foundation Funded Project (Grant no 2012M520753)theHeilongjiang Postdoctoral Science Foundation (Grant noLBH-Z12125) and theNational Natural Scientific Foundationof China (Grant no 61273096)

References

[1] J C Solem ldquoInterception of comets and asteroids on collisioncourse with Earthrdquo Journal of Spacecraft and Rockets vol 30 no2 pp 222ndash228 1993

[2] IMRoss S-Y Park and SDV Porter ldquoGravitational effects ofearth in optimizing ΔV for deflecting earth-crossing asteroidsrdquoJournal of Spacecraft and Rockets vol 38 no 5 pp 759ndash7642001

[3] B Dachwald and B Wie ldquoSolar sail trajectory optimization forintercepting impacting and deflecting near-earth asteroidsrdquoin Proceedings of the AIAA Guidance Navigation and ControlConference and Exhibit pp 3331ndash3348 San Francisco CalifUSA August 2005

[4] D Izzo ldquoOptimization of interplanetary trajectories for impul-sive and continuous asteroid deflectionrdquo Journal of GuidanceControl and Dynamics vol 30 no 2 pp 401ndash408 2007

[5] J Park R G Rhinehart and P T Kabamba ldquoMiss analysisin Lambert interceptions with application to a new guidancelawrdquo Mathematical Problems in Engineering Theory Methodsand Applications vol 6 no 2-3 pp 225ndash265 2000

[6] Y Ulybyshev ldquoDirect high-speed interception analytic solu-tions qualitative analysis and applicationsrdquo Journal of Space-craft and Rockets vol 38 no 3 pp 351ndash359 2001

[7] S P Burns and J J Scherock ldquoLambert guidance routinedesigned to match position and velocity of ballistic targetrdquoJournal of Guidance Control and Dynamics vol 27 no 6 pp989ndash996 2004

[8] R H Battin An Introduction to the Mathematics and Methodsof Astrodynamics American Institute of Aeronautics and Astro-nautics (AIAA) Reston Va USA Revised edition 1999

[9] S L Nelson and P Zarchan ldquoAlternative approach to thesolution of Lambertrsquos problemrdquo Journal of Guidance Controland Dynamics vol 15 no 4 pp 1003ndash1009 1992

[10] P V Arlulkar and S D Naik ldquoSolution based on dynamicalapproach for multiple-revolution lambert problemrdquo Journal ofGuidance Control and Dynamics vol 34 no 3 pp 920ndash9232011

[11] G Zhang D Mortari and D Zhou ldquoConstrained multiple-revolution Lambertrsquos problemrdquo Journal of Guidance Controland Dynamics vol 33 no 6 pp 1779ndash1786 2010

10 Mathematical Problems in Engineering

[12] C A Lembeck and J E Prussing ldquoOptimal impulsive interceptwith low-thrust rendezvous returnrdquo Journal of Guidance Con-trol and Dynamics vol 16 no 3 pp 426ndash433 1993

[13] D-R Taur V Coverstone-Carroll and J E Prussing ldquoOptimalimpulsive time-fixed orbital rendezvous and interception withpath constraintsrdquo Journal of Guidance Control and Dynamicsvol 18 no 1 pp 54ndash60 1995

[14] H Leeghim ldquoSpacecraft intercept using minimum controlenergy andwait timerdquoCelestialMechanics ampDynamical Astron-omy vol 115 no 1 pp 1ndash19 2013

[15] D A Vallado Fundamentals of Astrodynamics and ApplicationsMicrocosm Press El Segundo Calif USA 2nd edition 2001

[16] B F Thompson K K Choi S W Piggott and S R BeaverldquoOrbital targeting based on hodograph theory for improvedrendezvous safetyrdquo Journal of Guidance Control and Dynamicsvol 33 no 5 pp 1566ndash1576 2010

[17] V G Adamyan L V Adamyan G M Zaimtsyan and LT Manandyan ldquoDouble-pulse cotangential transfers betweencoplanar elliptic orbitsrdquo Journal of Applied Mathematics andMechanics vol 73 no 6 pp 664ndash672 2009

[18] G Zhang D Zhou D Mortari and T A Henderson ldquoAna-lytical study of tangent orbit and conditions for its solutionexistencerdquo Journal of Guidance Control and Dynamics vol 35no 1 pp 186ndash194 2012

[19] G Zhang D Zhou Z Sun and X Cao ldquoOptimal two-impulsecotangent rendezvous between coplanar elliptical orbitsrdquo Jour-nal of Guidance Control and Dynamics vol 36 no 3 pp 677ndash685 2013

[20] G Zhang X Cao and D Zhou ldquoTwo-impulse cotangent ren-dezvous between coplanar elliptic and hyperbolic orbitsrdquo Jour-nal of Guidance Control and Dynamics vol 37 no 3 pp 964ndash969 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 5

Parallel line of asymptotes Target hyperbolic orbit

Initial elliptic orbit

F1

Ps1

us1

rs1 u2max

120574s1

(a) 11989111199041

Parallel line of asymptotesTarget hyperbolic orbit

Initial elliptic orbit

F1

Ps2

us2rs2

u2min120574s2

(b) 11989111199042

Parallel line of asymptotes

Target hyperbolic orbit

Initial elliptic orbit

F1

Ps3

us3

rs3u2max

120574s3

(c) 11989111199043

Parallel line of asymptotes

Target hyperbolic orbit

Initial elliptic orbit

F1

Ps4

us4

rs4u2min

120574s4

(d) 11989111199044

Figure 2 Geometry interpretation for special initial true anomalies

and the function only of 11989111199044

is

11989111199044

+ 1205961minus3120587

2minus arctan(

1198901sin11989111199044

1 + 1198901cos11989111199044

)

= arccos( 1

1198902

) + 1205962

(30)

All the values of 1198911119904119895

(119895 = 1 2 3 4) are obtained by Newton-Raphson iterative algorithm

Once the special initial true anomalies are obtained thesolution-existence range of target true anomaly 119891

2can be

discussed and obtained as follows

(1) If 1198911isin [11989111199041

11989111199042

] there is only one intersectionpoint for the velocity-vector line and the targethyperbolic orbit Assume that the true anomaly of

the intersection point on the target hyperbolic orbitis 1198912119902 which can be obtained by (19) then the

target true anomaly is in the range (minus1198912max 1198912119902)

The solution-existence range for any conic transferorbit is obtained as follows (a) If there is no rangefor elliptic transfer orbits there is also no range forany conic transfer orbit (b) If there exists a range(1198912119886 1198912119887) sub (minus119891

2max 1198912119902) for elliptic transfer orbitswhich is obtained by (21) then the range for any conictransfer orbit can be (119891

2119886 1198912119887) or (minus119891

2max 1198912119887) When1199053by (11)ndash(13) is positive for an arbitrarily selected

1198912isin (minus119891

2max 1198912119886) for example 1198912= (minus119891

2max +1198912119886)2 the solution-existence range is (minus119891

2max 1198912119887)otherwise the range is (119891

2119886 1198912119887)

(2) If 1198911

isin (11989111199042

11989111199043

) there are two intersectionpoints for the velocity-vector line and the target

6 Mathematical Problems in Engineering

hyperbolic orbit Assume that the true anomalies ofthe intersection points on the target hyperbolic orbitare 1198912119901

and 1198912119902 then the solution-existence range for

any conic transfer orbit is in the range (1198912119901 1198912119902) The

solution-existence range for elliptic transfer orbits isobtained as (119891

2119886 1198912119887) sub (119891

2119901 1198912119902) by (21) Then the

solution-existence range for any conic transfer orbitis (1198912119901 1198912119887)

(3) If 1198911isin [11989111199043

11989111199044

] there is only one intersectionpoint for the velocity-vector line and the targethyperbolic orbit Assume that the true anomaly ofthe intersection point on the target hyperbolic orbitis 1198912119901 which is obtained by (19) then the target true

anomaly is in the range (1198912119901 1198912max) The solution-

existence range for any conic transfer orbit is obtainedas follows (a) If there is no range for elliptic transferorbits then the solution-existence range for any conictransfer orbit is (119891

2119901 1198912max) (b) If there exists a range

(1198912119886 1198912119887) sub (119891

2119901 1198912max) for elliptic transfer orbits

then the range for any conic transfer orbit can be(1198912119901 1198912119887) or (119891

2119901 1198912max) When 119905

3by ((11)ndash(13)) is

positive for an arbitrarily selected 1198912isin (1198912119887 1198912max)

for example 1198912

= (1198912119887

+ 1198912max)2 the solution-

existence range is (1198912119901 1198912max) otherwise the range is

(1198912119901 1198912119887)

(4) If 1198911

isin (11989111199044

11989111199041

) there are two intersectionpoints for the velocity-vector line and the targethyperbolic orbit Assume that the true anomalies ofthe intersection points on the target hyperbolic orbitare 1198912119901

and 1198912119902 then the target true anomaly must be

in the range (1198912119902 1198912max) There is no range for elliptic

transfer orbits and the solution-existence range forany conic (hyperbolic) transfer orbit is (119891

2119902 1198912max)

From the above discussions it is known that there maybe no solution-existence range for any conic transfer orbitonly when 119891

1isin [119891

11199041 11989111199042

] and there is certainly nosolution-existence range for elliptic transfer orbits when 119891

1isin

[11989111199044

11989111199041

]Let the solution-existence range for any conic transfer

by the above method be (1198912119898 1198912119899) The target true anomaly

1198912119894at the impulse time is obtained by the hyperbolic orbit

propagationThen there are three cases (1) If 1198912119894lt 1198912119898 then

the final solution should be obtained in the range (1198912119898 1198912119899)

(2) If 1198912119894

isin [1198912119898 1198912119899] then the final solution should be

obtained in the range (1198912119894 1198912119899) (3) If 119891

2119894gt 1198912119899 then there

is no solution

6 Solving the Problem

For a given impulse point the solution-existence range forany conic transfer and that for elliptic transfers are obtainedin Section 5 Assume that the final solution-existence rangefor any conic transfer considering the target true anomalyat the impulse time is (119891

21198921 11989121198922

) The interception problemcan be solved by numerical methods

Thevalue of 1205781015840 can be calculated by (15) then the solutionsof 1198912for 1205781015840(119891

2) = 0 can be obtained by a numerical iterative

algorithm for example the golden section search If thereis no solution for 1205781015840(119891

2) = 0 then 120578 decreases or increases

monotonically Then for the range (11989121198921

11989121198922

) if lfloor120578(1198911198921

+

120575)rfloor = lfloor120578(1198911198922minus 120575)rfloor or max(lfloor120578(119891

1198921+ 120575)rfloor lfloor120578(119891

1198922minus 120575)rfloor) lt 0

where the function lfloor119910rfloor denotes the nearest integer less thanor equal to 119910 and 120575 is a small positive number then there isno solution for (2) Otherwise there are two cases

(1) If min(lfloor120578(1198911198921

+ 120575)rfloor lfloor120578(1198911198922

minus 120575)rfloor) ge 0 there are|lfloor120578(1198911198922minus 120575)rfloor minus lfloor120578(119891

1198921+ 120575)rfloor| solutions which can be

obtained by the secant method as

1198912119899+2

= 1198912119899+1

minus (1198912119899+1

minus 1198912119899)

120578 (1198912119899+1

) minus 1198731

120578 (1198912119899+1

) minus 120578 (1198912119899)

119899 isin N

(31)

where the initial guesses are11989120

= 1198911198921+12057511989121

= 1198911198922minus

1205751198731= lfloor120578(119891

1198921+120575)rfloor+1 lfloor120578(119891

1198921+120575)rfloor+2 lfloor120578(119891

1198922minus120575)rfloor

if lfloor120578(1198911198921+120575)rfloor lt lfloor120578(119891

1198922minus120575)rfloor and119873

1= lfloor120578(119891

1198922minus120575)rfloor+1

lfloor120578(1198911198922minus 120575)rfloor + 2 lfloor120578(119891

1198921+ 120575)rfloor if lfloor120578(119891

1198921+ 120575)rfloor gt

lfloor120578(1198911198922minus 120575)rfloor

(2) If min(lfloor120578(1198911198921

+ 120575)rfloor lfloor120578(1198911198922

minus 120575)rfloor) lt 0 there are[max(lfloor120578(119891

1198921+ 120575)rfloor lfloor120578(119891

1198922minus 120575)rfloor) + 1] solutions

which can be obtained by (31) with 1198731= 0 1

max(lfloor120578(1198911198921

+ 120575)rfloor lfloor120578(1198911198922

minus 120575)rfloor) Note that 120578(1198911198921)

instead of 120578(1198911198921

+ 120575) should be used for the range[1198911198921 1198911198922)

If there exist 119896119899ge 1 solutions for 1205781015840(119891

2) = 0 and these

119896119899extreme points for 120578(119891

2) are obtained as 119891

21198961lt 11989121198962

lt

sdot sdot sdot lt 1198912119896119899

then there are 119896119899+ 1 piecewise ranges that is

(11989121198921

11989121198961

) [11989121198961

11989121198962

) [1198912119896119899

11989121198922

) For each piecewiserange the function of 120578 decreases or increasesmonotonicallyThen the solution for each piecewise range can be obtainedby (31) with different initial guesses If lim

(1198912rarr11989121198922)120578 = +infin

(see Figure 3(d)) there are infinite solutions in the range[1198912119896119899

11989121198922

) and the minimum-time solution (or with theminimum 119891

2) will be obtained

In the patched-conic approximation method the radiusof the Earthrsquos sphere of influence is 119877SOI = 925 000 km thenthe true anomaly range of the solution is 119891

2isin (minus119891SOI 119891SOI)

where 119891SOI = arccos([1199012119877SOI minus 1]119890

2) Thus the solutions

beyond this range should be removedThe above section gives a method to obtain all solutions

for a given initial impulse point 1198911 For each solution the

corresponding fuel consumption is

ΔV = 1003816100381610038161003816V1119905 minus V1

1003816100381610038161003816

=

1003816100381610038161003816100381610038161003816100381610038161003816

radic120583

1199011

120582 (1 + 1198901cos1198911) minus radic

120583

1199011

(1 + 1198902

1+ 21198901cos1198911)

1003816100381610038161003816100381610038161003816100381610038161003816

(32)

The solution whose cost is greater than the given maxi-mal value should be removed Thus the feasible solutionneeds to satisfy two conditions including the radius of theEarthrsquos sphere of influence and the given maximal valueof fuel When considering the initial coasting time before

Mathematical Problems in Engineering 7

minus120 minus100 minus80 minus60 minus40 minus20 0 20 40 60minus1

minus08

minus06

minus04

minus02

0

02

True anomaly f2 (∘)

120578

(a) 1198911 = 160∘

minus150 minus100 minus50 0 50 100minus1

minus05

0

05

True anomaly f2 (∘)

120578

(b) 1198911 = 170∘

minus40 minus20 0 20 40 60 80 100 120 140minus5

minus4

minus3

minus2

minus1

0

1

True anomaly f2 (∘)

120578

(c) 1198911 = 240∘

60 70 80 90 100 110 120 1300

1

2

3

4

5

6

7

8

9

10

True anomaly f2 (∘)

120578

(d) 1198911 = 330∘

Figure 3 The curves of 120578 versus 1198912for different 119891

1

the impulse the impulse point is free then for each initialimpulse point all feasible solutions can be obtained and theminimum-time one is with the minimum value of 119891

2

Reference [20] dealt with the cotangent rendezvousbetween two coplanar elliptic and hyperbolic orbits (Problem1) whereas this paper is for the tangent-impulse interceptionproblem from an interceptor on elliptic parking orbits toa target on hyperbolic orbits (Problem 2) Three maindifferences between these two problems are listed as follows

(1) For Problem 1 there is a single closed-form solutionfor an arbitrary point on the target hyperbolic orbitthat is 119891

2isin (minus119891

2max 1198912max) where 1198912max =

arccos(minus11198902) thus there is only one degree of

freedom in (2) for zero revolution however in aspecified (eg counterclockwise) motion directionthe solution-existence range is not the whole rangeof target true anomaly For Problem 2 there isno certain relationship between the tangent-impulse

point1198911and the interception point119891

2 In other words

there are two degrees of freedom in (2) with zerorevolutions

(2) For Problem 1 with nonintersecting case (counter-clockwise) the solution-existence range for cotangentrendezvous is 119891

2isin (1198912119886 1198912max) where (1198912119886 1198912119887) is

obtained by the solution-existence range for elliptictransfers [20] But for Problem 2with nonintersectingcase (counterclockwise) the solution-existence rangeof 1198912can be obtained by the proposed method for

different1198911 which is dependent on four special initial

true anomalies whose velocity vectors are parallel tothe lines of asymptotes for the target hyperbolic orbitMoreover there may be no solution-existence rangefor any conic transfer orbit when 119891

1isin [11989111199041

11989111199042

](3) For Problem 1 with a given revolution number 119873

3

of the transfer orbit usually there is only one orbitsolution (see the examples in [20]) However for

8 Mathematical Problems in Engineering

Problem 2 with an upper-bounded impulse theminimum-time solution is interesting then 119873

3is set

to be 0 For a given impulse point 1198911 there may exist

multiple solutions of 1198912(see Figures 3(a) and 3(b))

With the initial coasting time all feasible solutionsfor different impulse points are obtained and theminimum-time one is with the minimum value of 119891

2

7 Numerical Examples

Assume that the interceptorrsquos orbit elements are 1198861= 2119877119864+

4000 km 1198901= 06 120596

1= 10∘ and 119891

10= 60∘ and the targetrsquos

orbit elements are 1198862= minus2119877

119864minus 10 000 km 119890

2= 16 120596

2= 0∘

and 11989120

= minus120∘ where 119877

119864= 637813 km is the Earthrsquos

radius The gravitational parameter used for the example is120583 = 3986004415 km3s2 The upper bound of the tangentimpulse is set to be 5 kms Four special initial true anomaliesare obtained as 119891

11199041= 454187

∘ 11989111199042

= 1581023∘ 11989111199043

=

1919457∘ and 119891

11199044= 2845334

∘ via (26) (28) (29) and(30) To obtain elliptic transfers the range of the initial trueanomaly of the impulse point is119891

1isin (799244

∘ 2499268

∘) via

(24) In addition considering the radius of the Earthrsquos sphereof influence that is 925000 km the range of true anomaly is1198912isin (minus1269425

∘ 1269425

∘)

If the impulse is imposed at the initial time that is 1198911=

11989110

= 60∘ then 119891

1isin [11989111199041

11989111199042

] is satisfied and there is norange for elliptic transfers thus no solution-existence rangeexists for 119891

2

If the impulse is imposed at 1198911

= 160∘ the target

true anomaly at the impulse time is 1198912

= minus1183547∘

by solving Keplerrsquos equation for hyperbolic orbits andthe plots of 120578 versus 119891

2can be obtained in Figure 3(a)

The final solution-existence range for any conic transferconsidering the target true anomaly at the impulse timeis (11989121198921

11989121198922

) = (minus1183547∘ 724802

∘) There are two

solutions that is minus928589∘ and 207296∘ the correspondingrevolution numbers are both 0 and the corresponding costsare 13182 kms and 10211 kms respectively The minimum-time solution is minus928589∘ the corresponding transfer time is219724 s and the total interception time 119905

2is 283104 s via (3)

The value of 120582 is obtained as 11178 by (9) thus the trajectoryis elliptic and it is plotted in Figure 4

If the impulse is imposed at 1198911= 170

∘ the target trueanomaly at the impulse time is 119891

2= minus1177828

∘ the plots of120578 versus 119891

2can be obtained in Figure 3(b)The final solution-

existence range is (11989121198921

11989121198922

) = (minus1163047∘ 883112

∘)

There are three solutions minus1158579∘ minus1092932

∘ and532432

∘ all the corresponding revolution numbers are 0and the corresponding costs are 211998 kms 39088 kmsand 11770 kms respectivelyThe energy cost ofminus1158579∘ ismuch greater than 5 kms thus it cannot be usedThe feasibleminimum-time solution is minus1092932∘ the correspondingtransfer time is 131883 s and the total interception time is212588 s The value of 120582 is 27636 thus the trajectory ishyperbolic and it is plotted in Figure 5

If the impulse is imposed at 1198911= 240

∘ the target trueanomaly at the impulse time is 119891

2= minus1125985

∘ the plots of120578 versus 119891

2can be obtained in Figure 3(c) There is a single

minus15 minus10 minus5 0 5

minus16

minus14

minus12

minus10

minus8

minus6

minus4

minus2

0

2

X (km)

Y (k

m)

InitialTargetTransfer

P1

P20

P10

P2

F1

times104

times104

Figure 4 Tangent-impulse interception with elliptic transfer trajec-tory

minus15 minus1 minus05 0 05 1minus16

minus14

minus12

minus10

minus8

minus6

minus4

minus2

0

2

4

Y (k

m)

X (km)

times104

times105

P1

P20

P10

P2

F1

InitialTargetTransfer

Figure 5 Tangent-impulse interception with hyperbolic transfertrajectory

solution 1152336∘ the corresponding revolution number is0 the cost is 17743 kms the transfer time is 363182 s andthe total interception time is 542081 sThe value of 120582 is 19316thus the trajectory is elliptic

If the impulse is imposed at 1198911= 330

∘ the target trueanomaly at the impulse time is119891

2= minus1102890

∘ the plots of 120578versus119891

2can be obtained in Figure 3(d) and there are infinite

solutions However when considering the sphere of influenceof the Earth there are three solutions The minimum-timesolution is 1186911∘ the corresponding revolution number

Mathematical Problems in Engineering 9

Table 1 Feasible solutions of1198912for different119891

1and the correspond-

ing interception time 1199052and ΔV

1198911(∘) Solutions of

(1198912(∘)119873

1) 119905

2(s) ΔV (kms)

150 (minus566897 0) 321084 09035(minus399134 0) 328443 08897

170 (minus1092932 0) 212588 39088(532432 1) 358345 11770

190 (891630 0) 390801 14596210 (1043983 0) 436651 16333230 (1123591 0) 499326 17338250 (1177130 0) 598472 18148270 (1219959 0) 802495 19082290 (1259555 0) 1602413 20322310 (1244383 1) 1114899 22976330 (1186911 1) 628954 29698

(1257512 2) 1506675 26872(1269266 3) 2368738 26464

350 (1142511 1) 525410 44692(1242882 2) 1085322 35407(1260145 3) 1632862 34192(1267542 4) 2175443 33698

10 (1261114 4) 1685896 49766(1266609 5) 2084768 48911

is 1 the corresponding costs is 29698 kms the correspond-ing transfer time is 209279 s and the total interception timeis 628954 s The value of 120582 is 27280 thus the trajectory ishyperbolic

For different impulse points all feasible solutions satisfy-ing 1198912isin (minus1269425

∘ 1269425

∘) and the fuel constraint and

the corresponding value of 1198732= 120578 for (2) can be obtained

by (31) and are listed in Table 1 The total interception time is1199052= 1199051+ 1199053 the corresponding costs are also listed in Table 1

It should be notified that there is no feasible solution when1198911= 30∘+ 119896119895sdot 20∘ 119896119895= 1 2 3 4 5 By using the numerical

optimization algorithm the minimum-time impulse point isat 1198911= 1705235

∘ the solution is 1198912= minus1108795

∘ the cost is49999 kms and the total interception time is 198106 s Theminimum-time interception trajectory is hyperbolic and it isnear to that in Figure 5

8 Conclusions

This paper studies the interception problem for a hyperbolictarget with an upper-bounded tangent impulse Based onfour special initial true anomalies the solution-existencerange of the target true anomaly for any conic transfer isobtained Then the problem is solved by the secant methodfor a time-of-flight equation For a given impulse pointthere may be zero one many or infinite solutions howeverwhen considering the sphere of influence of the Earth thereare always finite solutions For different impulse points allfeasible solutions and the global minimum-time solution are

obtained This proposed method is valid for an assignedimpulse point and the time optimization with the initialcoasting time The interception with a tangent-impulse tech-nique can be used in orbital interception task for potentiallyhazardous objects during the Earth flyby with simple control

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work is supported in part by the China Postdoctoral Sci-ence Foundation Funded Project (Grant no 2012M520753)theHeilongjiang Postdoctoral Science Foundation (Grant noLBH-Z12125) and theNational Natural Scientific Foundationof China (Grant no 61273096)

References

[1] J C Solem ldquoInterception of comets and asteroids on collisioncourse with Earthrdquo Journal of Spacecraft and Rockets vol 30 no2 pp 222ndash228 1993

[2] IMRoss S-Y Park and SDV Porter ldquoGravitational effects ofearth in optimizing ΔV for deflecting earth-crossing asteroidsrdquoJournal of Spacecraft and Rockets vol 38 no 5 pp 759ndash7642001

[3] B Dachwald and B Wie ldquoSolar sail trajectory optimization forintercepting impacting and deflecting near-earth asteroidsrdquoin Proceedings of the AIAA Guidance Navigation and ControlConference and Exhibit pp 3331ndash3348 San Francisco CalifUSA August 2005

[4] D Izzo ldquoOptimization of interplanetary trajectories for impul-sive and continuous asteroid deflectionrdquo Journal of GuidanceControl and Dynamics vol 30 no 2 pp 401ndash408 2007

[5] J Park R G Rhinehart and P T Kabamba ldquoMiss analysisin Lambert interceptions with application to a new guidancelawrdquo Mathematical Problems in Engineering Theory Methodsand Applications vol 6 no 2-3 pp 225ndash265 2000

[6] Y Ulybyshev ldquoDirect high-speed interception analytic solu-tions qualitative analysis and applicationsrdquo Journal of Space-craft and Rockets vol 38 no 3 pp 351ndash359 2001

[7] S P Burns and J J Scherock ldquoLambert guidance routinedesigned to match position and velocity of ballistic targetrdquoJournal of Guidance Control and Dynamics vol 27 no 6 pp989ndash996 2004

[8] R H Battin An Introduction to the Mathematics and Methodsof Astrodynamics American Institute of Aeronautics and Astro-nautics (AIAA) Reston Va USA Revised edition 1999

[9] S L Nelson and P Zarchan ldquoAlternative approach to thesolution of Lambertrsquos problemrdquo Journal of Guidance Controland Dynamics vol 15 no 4 pp 1003ndash1009 1992

[10] P V Arlulkar and S D Naik ldquoSolution based on dynamicalapproach for multiple-revolution lambert problemrdquo Journal ofGuidance Control and Dynamics vol 34 no 3 pp 920ndash9232011

[11] G Zhang D Mortari and D Zhou ldquoConstrained multiple-revolution Lambertrsquos problemrdquo Journal of Guidance Controland Dynamics vol 33 no 6 pp 1779ndash1786 2010

10 Mathematical Problems in Engineering

[12] C A Lembeck and J E Prussing ldquoOptimal impulsive interceptwith low-thrust rendezvous returnrdquo Journal of Guidance Con-trol and Dynamics vol 16 no 3 pp 426ndash433 1993

[13] D-R Taur V Coverstone-Carroll and J E Prussing ldquoOptimalimpulsive time-fixed orbital rendezvous and interception withpath constraintsrdquo Journal of Guidance Control and Dynamicsvol 18 no 1 pp 54ndash60 1995

[14] H Leeghim ldquoSpacecraft intercept using minimum controlenergy andwait timerdquoCelestialMechanics ampDynamical Astron-omy vol 115 no 1 pp 1ndash19 2013

[15] D A Vallado Fundamentals of Astrodynamics and ApplicationsMicrocosm Press El Segundo Calif USA 2nd edition 2001

[16] B F Thompson K K Choi S W Piggott and S R BeaverldquoOrbital targeting based on hodograph theory for improvedrendezvous safetyrdquo Journal of Guidance Control and Dynamicsvol 33 no 5 pp 1566ndash1576 2010

[17] V G Adamyan L V Adamyan G M Zaimtsyan and LT Manandyan ldquoDouble-pulse cotangential transfers betweencoplanar elliptic orbitsrdquo Journal of Applied Mathematics andMechanics vol 73 no 6 pp 664ndash672 2009

[18] G Zhang D Zhou D Mortari and T A Henderson ldquoAna-lytical study of tangent orbit and conditions for its solutionexistencerdquo Journal of Guidance Control and Dynamics vol 35no 1 pp 186ndash194 2012

[19] G Zhang D Zhou Z Sun and X Cao ldquoOptimal two-impulsecotangent rendezvous between coplanar elliptical orbitsrdquo Jour-nal of Guidance Control and Dynamics vol 36 no 3 pp 677ndash685 2013

[20] G Zhang X Cao and D Zhou ldquoTwo-impulse cotangent ren-dezvous between coplanar elliptic and hyperbolic orbitsrdquo Jour-nal of Guidance Control and Dynamics vol 37 no 3 pp 964ndash969 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Mathematical Problems in Engineering

hyperbolic orbit Assume that the true anomalies ofthe intersection points on the target hyperbolic orbitare 1198912119901

and 1198912119902 then the solution-existence range for

any conic transfer orbit is in the range (1198912119901 1198912119902) The

solution-existence range for elliptic transfer orbits isobtained as (119891

2119886 1198912119887) sub (119891

2119901 1198912119902) by (21) Then the

solution-existence range for any conic transfer orbitis (1198912119901 1198912119887)

(3) If 1198911isin [11989111199043

11989111199044

] there is only one intersectionpoint for the velocity-vector line and the targethyperbolic orbit Assume that the true anomaly ofthe intersection point on the target hyperbolic orbitis 1198912119901 which is obtained by (19) then the target true

anomaly is in the range (1198912119901 1198912max) The solution-

existence range for any conic transfer orbit is obtainedas follows (a) If there is no range for elliptic transferorbits then the solution-existence range for any conictransfer orbit is (119891

2119901 1198912max) (b) If there exists a range

(1198912119886 1198912119887) sub (119891

2119901 1198912max) for elliptic transfer orbits

then the range for any conic transfer orbit can be(1198912119901 1198912119887) or (119891

2119901 1198912max) When 119905

3by ((11)ndash(13)) is

positive for an arbitrarily selected 1198912isin (1198912119887 1198912max)

for example 1198912

= (1198912119887

+ 1198912max)2 the solution-

existence range is (1198912119901 1198912max) otherwise the range is

(1198912119901 1198912119887)

(4) If 1198911

isin (11989111199044

11989111199041

) there are two intersectionpoints for the velocity-vector line and the targethyperbolic orbit Assume that the true anomalies ofthe intersection points on the target hyperbolic orbitare 1198912119901

and 1198912119902 then the target true anomaly must be

in the range (1198912119902 1198912max) There is no range for elliptic

transfer orbits and the solution-existence range forany conic (hyperbolic) transfer orbit is (119891

2119902 1198912max)

From the above discussions it is known that there maybe no solution-existence range for any conic transfer orbitonly when 119891

1isin [119891

11199041 11989111199042

] and there is certainly nosolution-existence range for elliptic transfer orbits when 119891

1isin

[11989111199044

11989111199041

]Let the solution-existence range for any conic transfer

by the above method be (1198912119898 1198912119899) The target true anomaly

1198912119894at the impulse time is obtained by the hyperbolic orbit

propagationThen there are three cases (1) If 1198912119894lt 1198912119898 then

the final solution should be obtained in the range (1198912119898 1198912119899)

(2) If 1198912119894

isin [1198912119898 1198912119899] then the final solution should be

obtained in the range (1198912119894 1198912119899) (3) If 119891

2119894gt 1198912119899 then there

is no solution

6 Solving the Problem

For a given impulse point the solution-existence range forany conic transfer and that for elliptic transfers are obtainedin Section 5 Assume that the final solution-existence rangefor any conic transfer considering the target true anomalyat the impulse time is (119891

21198921 11989121198922

) The interception problemcan be solved by numerical methods

Thevalue of 1205781015840 can be calculated by (15) then the solutionsof 1198912for 1205781015840(119891

2) = 0 can be obtained by a numerical iterative

algorithm for example the golden section search If thereis no solution for 1205781015840(119891

2) = 0 then 120578 decreases or increases

monotonically Then for the range (11989121198921

11989121198922

) if lfloor120578(1198911198921

+

120575)rfloor = lfloor120578(1198911198922minus 120575)rfloor or max(lfloor120578(119891

1198921+ 120575)rfloor lfloor120578(119891

1198922minus 120575)rfloor) lt 0

where the function lfloor119910rfloor denotes the nearest integer less thanor equal to 119910 and 120575 is a small positive number then there isno solution for (2) Otherwise there are two cases

(1) If min(lfloor120578(1198911198921

+ 120575)rfloor lfloor120578(1198911198922

minus 120575)rfloor) ge 0 there are|lfloor120578(1198911198922minus 120575)rfloor minus lfloor120578(119891

1198921+ 120575)rfloor| solutions which can be

obtained by the secant method as

1198912119899+2

= 1198912119899+1

minus (1198912119899+1

minus 1198912119899)

120578 (1198912119899+1

) minus 1198731

120578 (1198912119899+1

) minus 120578 (1198912119899)

119899 isin N

(31)

where the initial guesses are11989120

= 1198911198921+12057511989121

= 1198911198922minus

1205751198731= lfloor120578(119891

1198921+120575)rfloor+1 lfloor120578(119891

1198921+120575)rfloor+2 lfloor120578(119891

1198922minus120575)rfloor

if lfloor120578(1198911198921+120575)rfloor lt lfloor120578(119891

1198922minus120575)rfloor and119873

1= lfloor120578(119891

1198922minus120575)rfloor+1

lfloor120578(1198911198922minus 120575)rfloor + 2 lfloor120578(119891

1198921+ 120575)rfloor if lfloor120578(119891

1198921+ 120575)rfloor gt

lfloor120578(1198911198922minus 120575)rfloor

(2) If min(lfloor120578(1198911198921

+ 120575)rfloor lfloor120578(1198911198922

minus 120575)rfloor) lt 0 there are[max(lfloor120578(119891

1198921+ 120575)rfloor lfloor120578(119891

1198922minus 120575)rfloor) + 1] solutions

which can be obtained by (31) with 1198731= 0 1

max(lfloor120578(1198911198921

+ 120575)rfloor lfloor120578(1198911198922

minus 120575)rfloor) Note that 120578(1198911198921)

instead of 120578(1198911198921

+ 120575) should be used for the range[1198911198921 1198911198922)

If there exist 119896119899ge 1 solutions for 1205781015840(119891

2) = 0 and these

119896119899extreme points for 120578(119891

2) are obtained as 119891

21198961lt 11989121198962

lt

sdot sdot sdot lt 1198912119896119899

then there are 119896119899+ 1 piecewise ranges that is

(11989121198921

11989121198961

) [11989121198961

11989121198962

) [1198912119896119899

11989121198922

) For each piecewiserange the function of 120578 decreases or increasesmonotonicallyThen the solution for each piecewise range can be obtainedby (31) with different initial guesses If lim

(1198912rarr11989121198922)120578 = +infin

(see Figure 3(d)) there are infinite solutions in the range[1198912119896119899

11989121198922

) and the minimum-time solution (or with theminimum 119891

2) will be obtained

In the patched-conic approximation method the radiusof the Earthrsquos sphere of influence is 119877SOI = 925 000 km thenthe true anomaly range of the solution is 119891

2isin (minus119891SOI 119891SOI)

where 119891SOI = arccos([1199012119877SOI minus 1]119890

2) Thus the solutions

beyond this range should be removedThe above section gives a method to obtain all solutions

for a given initial impulse point 1198911 For each solution the

corresponding fuel consumption is

ΔV = 1003816100381610038161003816V1119905 minus V1

1003816100381610038161003816

=

1003816100381610038161003816100381610038161003816100381610038161003816

radic120583

1199011

120582 (1 + 1198901cos1198911) minus radic

120583

1199011

(1 + 1198902

1+ 21198901cos1198911)

1003816100381610038161003816100381610038161003816100381610038161003816

(32)

The solution whose cost is greater than the given maxi-mal value should be removed Thus the feasible solutionneeds to satisfy two conditions including the radius of theEarthrsquos sphere of influence and the given maximal valueof fuel When considering the initial coasting time before

Mathematical Problems in Engineering 7

minus120 minus100 minus80 minus60 minus40 minus20 0 20 40 60minus1

minus08

minus06

minus04

minus02

0

02

True anomaly f2 (∘)

120578

(a) 1198911 = 160∘

minus150 minus100 minus50 0 50 100minus1

minus05

0

05

True anomaly f2 (∘)

120578

(b) 1198911 = 170∘

minus40 minus20 0 20 40 60 80 100 120 140minus5

minus4

minus3

minus2

minus1

0

1

True anomaly f2 (∘)

120578

(c) 1198911 = 240∘

60 70 80 90 100 110 120 1300

1

2

3

4

5

6

7

8

9

10

True anomaly f2 (∘)

120578

(d) 1198911 = 330∘

Figure 3 The curves of 120578 versus 1198912for different 119891

1

the impulse the impulse point is free then for each initialimpulse point all feasible solutions can be obtained and theminimum-time one is with the minimum value of 119891

2

Reference [20] dealt with the cotangent rendezvousbetween two coplanar elliptic and hyperbolic orbits (Problem1) whereas this paper is for the tangent-impulse interceptionproblem from an interceptor on elliptic parking orbits toa target on hyperbolic orbits (Problem 2) Three maindifferences between these two problems are listed as follows

(1) For Problem 1 there is a single closed-form solutionfor an arbitrary point on the target hyperbolic orbitthat is 119891

2isin (minus119891

2max 1198912max) where 1198912max =

arccos(minus11198902) thus there is only one degree of

freedom in (2) for zero revolution however in aspecified (eg counterclockwise) motion directionthe solution-existence range is not the whole rangeof target true anomaly For Problem 2 there isno certain relationship between the tangent-impulse

point1198911and the interception point119891

2 In other words

there are two degrees of freedom in (2) with zerorevolutions

(2) For Problem 1 with nonintersecting case (counter-clockwise) the solution-existence range for cotangentrendezvous is 119891

2isin (1198912119886 1198912max) where (1198912119886 1198912119887) is

obtained by the solution-existence range for elliptictransfers [20] But for Problem 2with nonintersectingcase (counterclockwise) the solution-existence rangeof 1198912can be obtained by the proposed method for

different1198911 which is dependent on four special initial

true anomalies whose velocity vectors are parallel tothe lines of asymptotes for the target hyperbolic orbitMoreover there may be no solution-existence rangefor any conic transfer orbit when 119891

1isin [11989111199041

11989111199042

](3) For Problem 1 with a given revolution number 119873

3

of the transfer orbit usually there is only one orbitsolution (see the examples in [20]) However for

8 Mathematical Problems in Engineering

Problem 2 with an upper-bounded impulse theminimum-time solution is interesting then 119873

3is set

to be 0 For a given impulse point 1198911 there may exist

multiple solutions of 1198912(see Figures 3(a) and 3(b))

With the initial coasting time all feasible solutionsfor different impulse points are obtained and theminimum-time one is with the minimum value of 119891

2

7 Numerical Examples

Assume that the interceptorrsquos orbit elements are 1198861= 2119877119864+

4000 km 1198901= 06 120596

1= 10∘ and 119891

10= 60∘ and the targetrsquos

orbit elements are 1198862= minus2119877

119864minus 10 000 km 119890

2= 16 120596

2= 0∘

and 11989120

= minus120∘ where 119877

119864= 637813 km is the Earthrsquos

radius The gravitational parameter used for the example is120583 = 3986004415 km3s2 The upper bound of the tangentimpulse is set to be 5 kms Four special initial true anomaliesare obtained as 119891

11199041= 454187

∘ 11989111199042

= 1581023∘ 11989111199043

=

1919457∘ and 119891

11199044= 2845334

∘ via (26) (28) (29) and(30) To obtain elliptic transfers the range of the initial trueanomaly of the impulse point is119891

1isin (799244

∘ 2499268

∘) via

(24) In addition considering the radius of the Earthrsquos sphereof influence that is 925000 km the range of true anomaly is1198912isin (minus1269425

∘ 1269425

∘)

If the impulse is imposed at the initial time that is 1198911=

11989110

= 60∘ then 119891

1isin [11989111199041

11989111199042

] is satisfied and there is norange for elliptic transfers thus no solution-existence rangeexists for 119891

2

If the impulse is imposed at 1198911

= 160∘ the target

true anomaly at the impulse time is 1198912

= minus1183547∘

by solving Keplerrsquos equation for hyperbolic orbits andthe plots of 120578 versus 119891

2can be obtained in Figure 3(a)

The final solution-existence range for any conic transferconsidering the target true anomaly at the impulse timeis (11989121198921

11989121198922

) = (minus1183547∘ 724802

∘) There are two

solutions that is minus928589∘ and 207296∘ the correspondingrevolution numbers are both 0 and the corresponding costsare 13182 kms and 10211 kms respectively The minimum-time solution is minus928589∘ the corresponding transfer time is219724 s and the total interception time 119905

2is 283104 s via (3)

The value of 120582 is obtained as 11178 by (9) thus the trajectoryis elliptic and it is plotted in Figure 4

If the impulse is imposed at 1198911= 170

∘ the target trueanomaly at the impulse time is 119891

2= minus1177828

∘ the plots of120578 versus 119891

2can be obtained in Figure 3(b)The final solution-

existence range is (11989121198921

11989121198922

) = (minus1163047∘ 883112

∘)

There are three solutions minus1158579∘ minus1092932

∘ and532432

∘ all the corresponding revolution numbers are 0and the corresponding costs are 211998 kms 39088 kmsand 11770 kms respectivelyThe energy cost ofminus1158579∘ ismuch greater than 5 kms thus it cannot be usedThe feasibleminimum-time solution is minus1092932∘ the correspondingtransfer time is 131883 s and the total interception time is212588 s The value of 120582 is 27636 thus the trajectory ishyperbolic and it is plotted in Figure 5

If the impulse is imposed at 1198911= 240

∘ the target trueanomaly at the impulse time is 119891

2= minus1125985

∘ the plots of120578 versus 119891

2can be obtained in Figure 3(c) There is a single

minus15 minus10 minus5 0 5

minus16

minus14

minus12

minus10

minus8

minus6

minus4

minus2

0

2

X (km)

Y (k

m)

InitialTargetTransfer

P1

P20

P10

P2

F1

times104

times104

Figure 4 Tangent-impulse interception with elliptic transfer trajec-tory

minus15 minus1 minus05 0 05 1minus16

minus14

minus12

minus10

minus8

minus6

minus4

minus2

0

2

4

Y (k

m)

X (km)

times104

times105

P1

P20

P10

P2

F1

InitialTargetTransfer

Figure 5 Tangent-impulse interception with hyperbolic transfertrajectory

solution 1152336∘ the corresponding revolution number is0 the cost is 17743 kms the transfer time is 363182 s andthe total interception time is 542081 sThe value of 120582 is 19316thus the trajectory is elliptic

If the impulse is imposed at 1198911= 330

∘ the target trueanomaly at the impulse time is119891

2= minus1102890

∘ the plots of 120578versus119891

2can be obtained in Figure 3(d) and there are infinite

solutions However when considering the sphere of influenceof the Earth there are three solutions The minimum-timesolution is 1186911∘ the corresponding revolution number

Mathematical Problems in Engineering 9

Table 1 Feasible solutions of1198912for different119891

1and the correspond-

ing interception time 1199052and ΔV

1198911(∘) Solutions of

(1198912(∘)119873

1) 119905

2(s) ΔV (kms)

150 (minus566897 0) 321084 09035(minus399134 0) 328443 08897

170 (minus1092932 0) 212588 39088(532432 1) 358345 11770

190 (891630 0) 390801 14596210 (1043983 0) 436651 16333230 (1123591 0) 499326 17338250 (1177130 0) 598472 18148270 (1219959 0) 802495 19082290 (1259555 0) 1602413 20322310 (1244383 1) 1114899 22976330 (1186911 1) 628954 29698

(1257512 2) 1506675 26872(1269266 3) 2368738 26464

350 (1142511 1) 525410 44692(1242882 2) 1085322 35407(1260145 3) 1632862 34192(1267542 4) 2175443 33698

10 (1261114 4) 1685896 49766(1266609 5) 2084768 48911

is 1 the corresponding costs is 29698 kms the correspond-ing transfer time is 209279 s and the total interception timeis 628954 s The value of 120582 is 27280 thus the trajectory ishyperbolic

For different impulse points all feasible solutions satisfy-ing 1198912isin (minus1269425

∘ 1269425

∘) and the fuel constraint and

the corresponding value of 1198732= 120578 for (2) can be obtained

by (31) and are listed in Table 1 The total interception time is1199052= 1199051+ 1199053 the corresponding costs are also listed in Table 1

It should be notified that there is no feasible solution when1198911= 30∘+ 119896119895sdot 20∘ 119896119895= 1 2 3 4 5 By using the numerical

optimization algorithm the minimum-time impulse point isat 1198911= 1705235

∘ the solution is 1198912= minus1108795

∘ the cost is49999 kms and the total interception time is 198106 s Theminimum-time interception trajectory is hyperbolic and it isnear to that in Figure 5

8 Conclusions

This paper studies the interception problem for a hyperbolictarget with an upper-bounded tangent impulse Based onfour special initial true anomalies the solution-existencerange of the target true anomaly for any conic transfer isobtained Then the problem is solved by the secant methodfor a time-of-flight equation For a given impulse pointthere may be zero one many or infinite solutions howeverwhen considering the sphere of influence of the Earth thereare always finite solutions For different impulse points allfeasible solutions and the global minimum-time solution are

obtained This proposed method is valid for an assignedimpulse point and the time optimization with the initialcoasting time The interception with a tangent-impulse tech-nique can be used in orbital interception task for potentiallyhazardous objects during the Earth flyby with simple control

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work is supported in part by the China Postdoctoral Sci-ence Foundation Funded Project (Grant no 2012M520753)theHeilongjiang Postdoctoral Science Foundation (Grant noLBH-Z12125) and theNational Natural Scientific Foundationof China (Grant no 61273096)

References

[1] J C Solem ldquoInterception of comets and asteroids on collisioncourse with Earthrdquo Journal of Spacecraft and Rockets vol 30 no2 pp 222ndash228 1993

[2] IMRoss S-Y Park and SDV Porter ldquoGravitational effects ofearth in optimizing ΔV for deflecting earth-crossing asteroidsrdquoJournal of Spacecraft and Rockets vol 38 no 5 pp 759ndash7642001

[3] B Dachwald and B Wie ldquoSolar sail trajectory optimization forintercepting impacting and deflecting near-earth asteroidsrdquoin Proceedings of the AIAA Guidance Navigation and ControlConference and Exhibit pp 3331ndash3348 San Francisco CalifUSA August 2005

[4] D Izzo ldquoOptimization of interplanetary trajectories for impul-sive and continuous asteroid deflectionrdquo Journal of GuidanceControl and Dynamics vol 30 no 2 pp 401ndash408 2007

[5] J Park R G Rhinehart and P T Kabamba ldquoMiss analysisin Lambert interceptions with application to a new guidancelawrdquo Mathematical Problems in Engineering Theory Methodsand Applications vol 6 no 2-3 pp 225ndash265 2000

[6] Y Ulybyshev ldquoDirect high-speed interception analytic solu-tions qualitative analysis and applicationsrdquo Journal of Space-craft and Rockets vol 38 no 3 pp 351ndash359 2001

[7] S P Burns and J J Scherock ldquoLambert guidance routinedesigned to match position and velocity of ballistic targetrdquoJournal of Guidance Control and Dynamics vol 27 no 6 pp989ndash996 2004

[8] R H Battin An Introduction to the Mathematics and Methodsof Astrodynamics American Institute of Aeronautics and Astro-nautics (AIAA) Reston Va USA Revised edition 1999

[9] S L Nelson and P Zarchan ldquoAlternative approach to thesolution of Lambertrsquos problemrdquo Journal of Guidance Controland Dynamics vol 15 no 4 pp 1003ndash1009 1992

[10] P V Arlulkar and S D Naik ldquoSolution based on dynamicalapproach for multiple-revolution lambert problemrdquo Journal ofGuidance Control and Dynamics vol 34 no 3 pp 920ndash9232011

[11] G Zhang D Mortari and D Zhou ldquoConstrained multiple-revolution Lambertrsquos problemrdquo Journal of Guidance Controland Dynamics vol 33 no 6 pp 1779ndash1786 2010

10 Mathematical Problems in Engineering

[12] C A Lembeck and J E Prussing ldquoOptimal impulsive interceptwith low-thrust rendezvous returnrdquo Journal of Guidance Con-trol and Dynamics vol 16 no 3 pp 426ndash433 1993

[13] D-R Taur V Coverstone-Carroll and J E Prussing ldquoOptimalimpulsive time-fixed orbital rendezvous and interception withpath constraintsrdquo Journal of Guidance Control and Dynamicsvol 18 no 1 pp 54ndash60 1995

[14] H Leeghim ldquoSpacecraft intercept using minimum controlenergy andwait timerdquoCelestialMechanics ampDynamical Astron-omy vol 115 no 1 pp 1ndash19 2013

[15] D A Vallado Fundamentals of Astrodynamics and ApplicationsMicrocosm Press El Segundo Calif USA 2nd edition 2001

[16] B F Thompson K K Choi S W Piggott and S R BeaverldquoOrbital targeting based on hodograph theory for improvedrendezvous safetyrdquo Journal of Guidance Control and Dynamicsvol 33 no 5 pp 1566ndash1576 2010

[17] V G Adamyan L V Adamyan G M Zaimtsyan and LT Manandyan ldquoDouble-pulse cotangential transfers betweencoplanar elliptic orbitsrdquo Journal of Applied Mathematics andMechanics vol 73 no 6 pp 664ndash672 2009

[18] G Zhang D Zhou D Mortari and T A Henderson ldquoAna-lytical study of tangent orbit and conditions for its solutionexistencerdquo Journal of Guidance Control and Dynamics vol 35no 1 pp 186ndash194 2012

[19] G Zhang D Zhou Z Sun and X Cao ldquoOptimal two-impulsecotangent rendezvous between coplanar elliptical orbitsrdquo Jour-nal of Guidance Control and Dynamics vol 36 no 3 pp 677ndash685 2013

[20] G Zhang X Cao and D Zhou ldquoTwo-impulse cotangent ren-dezvous between coplanar elliptic and hyperbolic orbitsrdquo Jour-nal of Guidance Control and Dynamics vol 37 no 3 pp 964ndash969 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 7

minus120 minus100 minus80 minus60 minus40 minus20 0 20 40 60minus1

minus08

minus06

minus04

minus02

0

02

True anomaly f2 (∘)

120578

(a) 1198911 = 160∘

minus150 minus100 minus50 0 50 100minus1

minus05

0

05

True anomaly f2 (∘)

120578

(b) 1198911 = 170∘

minus40 minus20 0 20 40 60 80 100 120 140minus5

minus4

minus3

minus2

minus1

0

1

True anomaly f2 (∘)

120578

(c) 1198911 = 240∘

60 70 80 90 100 110 120 1300

1

2

3

4

5

6

7

8

9

10

True anomaly f2 (∘)

120578

(d) 1198911 = 330∘

Figure 3 The curves of 120578 versus 1198912for different 119891

1

the impulse the impulse point is free then for each initialimpulse point all feasible solutions can be obtained and theminimum-time one is with the minimum value of 119891

2

Reference [20] dealt with the cotangent rendezvousbetween two coplanar elliptic and hyperbolic orbits (Problem1) whereas this paper is for the tangent-impulse interceptionproblem from an interceptor on elliptic parking orbits toa target on hyperbolic orbits (Problem 2) Three maindifferences between these two problems are listed as follows

(1) For Problem 1 there is a single closed-form solutionfor an arbitrary point on the target hyperbolic orbitthat is 119891

2isin (minus119891

2max 1198912max) where 1198912max =

arccos(minus11198902) thus there is only one degree of

freedom in (2) for zero revolution however in aspecified (eg counterclockwise) motion directionthe solution-existence range is not the whole rangeof target true anomaly For Problem 2 there isno certain relationship between the tangent-impulse

point1198911and the interception point119891

2 In other words

there are two degrees of freedom in (2) with zerorevolutions

(2) For Problem 1 with nonintersecting case (counter-clockwise) the solution-existence range for cotangentrendezvous is 119891

2isin (1198912119886 1198912max) where (1198912119886 1198912119887) is

obtained by the solution-existence range for elliptictransfers [20] But for Problem 2with nonintersectingcase (counterclockwise) the solution-existence rangeof 1198912can be obtained by the proposed method for

different1198911 which is dependent on four special initial

true anomalies whose velocity vectors are parallel tothe lines of asymptotes for the target hyperbolic orbitMoreover there may be no solution-existence rangefor any conic transfer orbit when 119891

1isin [11989111199041

11989111199042

](3) For Problem 1 with a given revolution number 119873

3

of the transfer orbit usually there is only one orbitsolution (see the examples in [20]) However for

8 Mathematical Problems in Engineering

Problem 2 with an upper-bounded impulse theminimum-time solution is interesting then 119873

3is set

to be 0 For a given impulse point 1198911 there may exist

multiple solutions of 1198912(see Figures 3(a) and 3(b))

With the initial coasting time all feasible solutionsfor different impulse points are obtained and theminimum-time one is with the minimum value of 119891

2

7 Numerical Examples

Assume that the interceptorrsquos orbit elements are 1198861= 2119877119864+

4000 km 1198901= 06 120596

1= 10∘ and 119891

10= 60∘ and the targetrsquos

orbit elements are 1198862= minus2119877

119864minus 10 000 km 119890

2= 16 120596

2= 0∘

and 11989120

= minus120∘ where 119877

119864= 637813 km is the Earthrsquos

radius The gravitational parameter used for the example is120583 = 3986004415 km3s2 The upper bound of the tangentimpulse is set to be 5 kms Four special initial true anomaliesare obtained as 119891

11199041= 454187

∘ 11989111199042

= 1581023∘ 11989111199043

=

1919457∘ and 119891

11199044= 2845334

∘ via (26) (28) (29) and(30) To obtain elliptic transfers the range of the initial trueanomaly of the impulse point is119891

1isin (799244

∘ 2499268

∘) via

(24) In addition considering the radius of the Earthrsquos sphereof influence that is 925000 km the range of true anomaly is1198912isin (minus1269425

∘ 1269425

∘)

If the impulse is imposed at the initial time that is 1198911=

11989110

= 60∘ then 119891

1isin [11989111199041

11989111199042

] is satisfied and there is norange for elliptic transfers thus no solution-existence rangeexists for 119891

2

If the impulse is imposed at 1198911

= 160∘ the target

true anomaly at the impulse time is 1198912

= minus1183547∘

by solving Keplerrsquos equation for hyperbolic orbits andthe plots of 120578 versus 119891

2can be obtained in Figure 3(a)

The final solution-existence range for any conic transferconsidering the target true anomaly at the impulse timeis (11989121198921

11989121198922

) = (minus1183547∘ 724802

∘) There are two

solutions that is minus928589∘ and 207296∘ the correspondingrevolution numbers are both 0 and the corresponding costsare 13182 kms and 10211 kms respectively The minimum-time solution is minus928589∘ the corresponding transfer time is219724 s and the total interception time 119905

2is 283104 s via (3)

The value of 120582 is obtained as 11178 by (9) thus the trajectoryis elliptic and it is plotted in Figure 4

If the impulse is imposed at 1198911= 170

∘ the target trueanomaly at the impulse time is 119891

2= minus1177828

∘ the plots of120578 versus 119891

2can be obtained in Figure 3(b)The final solution-

existence range is (11989121198921

11989121198922

) = (minus1163047∘ 883112

∘)

There are three solutions minus1158579∘ minus1092932

∘ and532432

∘ all the corresponding revolution numbers are 0and the corresponding costs are 211998 kms 39088 kmsand 11770 kms respectivelyThe energy cost ofminus1158579∘ ismuch greater than 5 kms thus it cannot be usedThe feasibleminimum-time solution is minus1092932∘ the correspondingtransfer time is 131883 s and the total interception time is212588 s The value of 120582 is 27636 thus the trajectory ishyperbolic and it is plotted in Figure 5

If the impulse is imposed at 1198911= 240

∘ the target trueanomaly at the impulse time is 119891

2= minus1125985

∘ the plots of120578 versus 119891

2can be obtained in Figure 3(c) There is a single

minus15 minus10 minus5 0 5

minus16

minus14

minus12

minus10

minus8

minus6

minus4

minus2

0

2

X (km)

Y (k

m)

InitialTargetTransfer

P1

P20

P10

P2

F1

times104

times104

Figure 4 Tangent-impulse interception with elliptic transfer trajec-tory

minus15 minus1 minus05 0 05 1minus16

minus14

minus12

minus10

minus8

minus6

minus4

minus2

0

2

4

Y (k

m)

X (km)

times104

times105

P1

P20

P10

P2

F1

InitialTargetTransfer

Figure 5 Tangent-impulse interception with hyperbolic transfertrajectory

solution 1152336∘ the corresponding revolution number is0 the cost is 17743 kms the transfer time is 363182 s andthe total interception time is 542081 sThe value of 120582 is 19316thus the trajectory is elliptic

If the impulse is imposed at 1198911= 330

∘ the target trueanomaly at the impulse time is119891

2= minus1102890

∘ the plots of 120578versus119891

2can be obtained in Figure 3(d) and there are infinite

solutions However when considering the sphere of influenceof the Earth there are three solutions The minimum-timesolution is 1186911∘ the corresponding revolution number

Mathematical Problems in Engineering 9

Table 1 Feasible solutions of1198912for different119891

1and the correspond-

ing interception time 1199052and ΔV

1198911(∘) Solutions of

(1198912(∘)119873

1) 119905

2(s) ΔV (kms)

150 (minus566897 0) 321084 09035(minus399134 0) 328443 08897

170 (minus1092932 0) 212588 39088(532432 1) 358345 11770

190 (891630 0) 390801 14596210 (1043983 0) 436651 16333230 (1123591 0) 499326 17338250 (1177130 0) 598472 18148270 (1219959 0) 802495 19082290 (1259555 0) 1602413 20322310 (1244383 1) 1114899 22976330 (1186911 1) 628954 29698

(1257512 2) 1506675 26872(1269266 3) 2368738 26464

350 (1142511 1) 525410 44692(1242882 2) 1085322 35407(1260145 3) 1632862 34192(1267542 4) 2175443 33698

10 (1261114 4) 1685896 49766(1266609 5) 2084768 48911

is 1 the corresponding costs is 29698 kms the correspond-ing transfer time is 209279 s and the total interception timeis 628954 s The value of 120582 is 27280 thus the trajectory ishyperbolic

For different impulse points all feasible solutions satisfy-ing 1198912isin (minus1269425

∘ 1269425

∘) and the fuel constraint and

the corresponding value of 1198732= 120578 for (2) can be obtained

by (31) and are listed in Table 1 The total interception time is1199052= 1199051+ 1199053 the corresponding costs are also listed in Table 1

It should be notified that there is no feasible solution when1198911= 30∘+ 119896119895sdot 20∘ 119896119895= 1 2 3 4 5 By using the numerical

optimization algorithm the minimum-time impulse point isat 1198911= 1705235

∘ the solution is 1198912= minus1108795

∘ the cost is49999 kms and the total interception time is 198106 s Theminimum-time interception trajectory is hyperbolic and it isnear to that in Figure 5

8 Conclusions

This paper studies the interception problem for a hyperbolictarget with an upper-bounded tangent impulse Based onfour special initial true anomalies the solution-existencerange of the target true anomaly for any conic transfer isobtained Then the problem is solved by the secant methodfor a time-of-flight equation For a given impulse pointthere may be zero one many or infinite solutions howeverwhen considering the sphere of influence of the Earth thereare always finite solutions For different impulse points allfeasible solutions and the global minimum-time solution are

obtained This proposed method is valid for an assignedimpulse point and the time optimization with the initialcoasting time The interception with a tangent-impulse tech-nique can be used in orbital interception task for potentiallyhazardous objects during the Earth flyby with simple control

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work is supported in part by the China Postdoctoral Sci-ence Foundation Funded Project (Grant no 2012M520753)theHeilongjiang Postdoctoral Science Foundation (Grant noLBH-Z12125) and theNational Natural Scientific Foundationof China (Grant no 61273096)

References

[1] J C Solem ldquoInterception of comets and asteroids on collisioncourse with Earthrdquo Journal of Spacecraft and Rockets vol 30 no2 pp 222ndash228 1993

[2] IMRoss S-Y Park and SDV Porter ldquoGravitational effects ofearth in optimizing ΔV for deflecting earth-crossing asteroidsrdquoJournal of Spacecraft and Rockets vol 38 no 5 pp 759ndash7642001

[3] B Dachwald and B Wie ldquoSolar sail trajectory optimization forintercepting impacting and deflecting near-earth asteroidsrdquoin Proceedings of the AIAA Guidance Navigation and ControlConference and Exhibit pp 3331ndash3348 San Francisco CalifUSA August 2005

[4] D Izzo ldquoOptimization of interplanetary trajectories for impul-sive and continuous asteroid deflectionrdquo Journal of GuidanceControl and Dynamics vol 30 no 2 pp 401ndash408 2007

[5] J Park R G Rhinehart and P T Kabamba ldquoMiss analysisin Lambert interceptions with application to a new guidancelawrdquo Mathematical Problems in Engineering Theory Methodsand Applications vol 6 no 2-3 pp 225ndash265 2000

[6] Y Ulybyshev ldquoDirect high-speed interception analytic solu-tions qualitative analysis and applicationsrdquo Journal of Space-craft and Rockets vol 38 no 3 pp 351ndash359 2001

[7] S P Burns and J J Scherock ldquoLambert guidance routinedesigned to match position and velocity of ballistic targetrdquoJournal of Guidance Control and Dynamics vol 27 no 6 pp989ndash996 2004

[8] R H Battin An Introduction to the Mathematics and Methodsof Astrodynamics American Institute of Aeronautics and Astro-nautics (AIAA) Reston Va USA Revised edition 1999

[9] S L Nelson and P Zarchan ldquoAlternative approach to thesolution of Lambertrsquos problemrdquo Journal of Guidance Controland Dynamics vol 15 no 4 pp 1003ndash1009 1992

[10] P V Arlulkar and S D Naik ldquoSolution based on dynamicalapproach for multiple-revolution lambert problemrdquo Journal ofGuidance Control and Dynamics vol 34 no 3 pp 920ndash9232011

[11] G Zhang D Mortari and D Zhou ldquoConstrained multiple-revolution Lambertrsquos problemrdquo Journal of Guidance Controland Dynamics vol 33 no 6 pp 1779ndash1786 2010

10 Mathematical Problems in Engineering

[12] C A Lembeck and J E Prussing ldquoOptimal impulsive interceptwith low-thrust rendezvous returnrdquo Journal of Guidance Con-trol and Dynamics vol 16 no 3 pp 426ndash433 1993

[13] D-R Taur V Coverstone-Carroll and J E Prussing ldquoOptimalimpulsive time-fixed orbital rendezvous and interception withpath constraintsrdquo Journal of Guidance Control and Dynamicsvol 18 no 1 pp 54ndash60 1995

[14] H Leeghim ldquoSpacecraft intercept using minimum controlenergy andwait timerdquoCelestialMechanics ampDynamical Astron-omy vol 115 no 1 pp 1ndash19 2013

[15] D A Vallado Fundamentals of Astrodynamics and ApplicationsMicrocosm Press El Segundo Calif USA 2nd edition 2001

[16] B F Thompson K K Choi S W Piggott and S R BeaverldquoOrbital targeting based on hodograph theory for improvedrendezvous safetyrdquo Journal of Guidance Control and Dynamicsvol 33 no 5 pp 1566ndash1576 2010

[17] V G Adamyan L V Adamyan G M Zaimtsyan and LT Manandyan ldquoDouble-pulse cotangential transfers betweencoplanar elliptic orbitsrdquo Journal of Applied Mathematics andMechanics vol 73 no 6 pp 664ndash672 2009

[18] G Zhang D Zhou D Mortari and T A Henderson ldquoAna-lytical study of tangent orbit and conditions for its solutionexistencerdquo Journal of Guidance Control and Dynamics vol 35no 1 pp 186ndash194 2012

[19] G Zhang D Zhou Z Sun and X Cao ldquoOptimal two-impulsecotangent rendezvous between coplanar elliptical orbitsrdquo Jour-nal of Guidance Control and Dynamics vol 36 no 3 pp 677ndash685 2013

[20] G Zhang X Cao and D Zhou ldquoTwo-impulse cotangent ren-dezvous between coplanar elliptic and hyperbolic orbitsrdquo Jour-nal of Guidance Control and Dynamics vol 37 no 3 pp 964ndash969 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Mathematical Problems in Engineering

Problem 2 with an upper-bounded impulse theminimum-time solution is interesting then 119873

3is set

to be 0 For a given impulse point 1198911 there may exist

multiple solutions of 1198912(see Figures 3(a) and 3(b))

With the initial coasting time all feasible solutionsfor different impulse points are obtained and theminimum-time one is with the minimum value of 119891

2

7 Numerical Examples

Assume that the interceptorrsquos orbit elements are 1198861= 2119877119864+

4000 km 1198901= 06 120596

1= 10∘ and 119891

10= 60∘ and the targetrsquos

orbit elements are 1198862= minus2119877

119864minus 10 000 km 119890

2= 16 120596

2= 0∘

and 11989120

= minus120∘ where 119877

119864= 637813 km is the Earthrsquos

radius The gravitational parameter used for the example is120583 = 3986004415 km3s2 The upper bound of the tangentimpulse is set to be 5 kms Four special initial true anomaliesare obtained as 119891

11199041= 454187

∘ 11989111199042

= 1581023∘ 11989111199043

=

1919457∘ and 119891

11199044= 2845334

∘ via (26) (28) (29) and(30) To obtain elliptic transfers the range of the initial trueanomaly of the impulse point is119891

1isin (799244

∘ 2499268

∘) via

(24) In addition considering the radius of the Earthrsquos sphereof influence that is 925000 km the range of true anomaly is1198912isin (minus1269425

∘ 1269425

∘)

If the impulse is imposed at the initial time that is 1198911=

11989110

= 60∘ then 119891

1isin [11989111199041

11989111199042

] is satisfied and there is norange for elliptic transfers thus no solution-existence rangeexists for 119891

2

If the impulse is imposed at 1198911

= 160∘ the target

true anomaly at the impulse time is 1198912

= minus1183547∘

by solving Keplerrsquos equation for hyperbolic orbits andthe plots of 120578 versus 119891

2can be obtained in Figure 3(a)

The final solution-existence range for any conic transferconsidering the target true anomaly at the impulse timeis (11989121198921

11989121198922

) = (minus1183547∘ 724802

∘) There are two

solutions that is minus928589∘ and 207296∘ the correspondingrevolution numbers are both 0 and the corresponding costsare 13182 kms and 10211 kms respectively The minimum-time solution is minus928589∘ the corresponding transfer time is219724 s and the total interception time 119905

2is 283104 s via (3)

The value of 120582 is obtained as 11178 by (9) thus the trajectoryis elliptic and it is plotted in Figure 4

If the impulse is imposed at 1198911= 170

∘ the target trueanomaly at the impulse time is 119891

2= minus1177828

∘ the plots of120578 versus 119891

2can be obtained in Figure 3(b)The final solution-

existence range is (11989121198921

11989121198922

) = (minus1163047∘ 883112

∘)

There are three solutions minus1158579∘ minus1092932

∘ and532432

∘ all the corresponding revolution numbers are 0and the corresponding costs are 211998 kms 39088 kmsand 11770 kms respectivelyThe energy cost ofminus1158579∘ ismuch greater than 5 kms thus it cannot be usedThe feasibleminimum-time solution is minus1092932∘ the correspondingtransfer time is 131883 s and the total interception time is212588 s The value of 120582 is 27636 thus the trajectory ishyperbolic and it is plotted in Figure 5

If the impulse is imposed at 1198911= 240

∘ the target trueanomaly at the impulse time is 119891

2= minus1125985

∘ the plots of120578 versus 119891

2can be obtained in Figure 3(c) There is a single

minus15 minus10 minus5 0 5

minus16

minus14

minus12

minus10

minus8

minus6

minus4

minus2

0

2

X (km)

Y (k

m)

InitialTargetTransfer

P1

P20

P10

P2

F1

times104

times104

Figure 4 Tangent-impulse interception with elliptic transfer trajec-tory

minus15 minus1 minus05 0 05 1minus16

minus14

minus12

minus10

minus8

minus6

minus4

minus2

0

2

4

Y (k

m)

X (km)

times104

times105

P1

P20

P10

P2

F1

InitialTargetTransfer

Figure 5 Tangent-impulse interception with hyperbolic transfertrajectory

solution 1152336∘ the corresponding revolution number is0 the cost is 17743 kms the transfer time is 363182 s andthe total interception time is 542081 sThe value of 120582 is 19316thus the trajectory is elliptic

If the impulse is imposed at 1198911= 330

∘ the target trueanomaly at the impulse time is119891

2= minus1102890

∘ the plots of 120578versus119891

2can be obtained in Figure 3(d) and there are infinite

solutions However when considering the sphere of influenceof the Earth there are three solutions The minimum-timesolution is 1186911∘ the corresponding revolution number

Mathematical Problems in Engineering 9

Table 1 Feasible solutions of1198912for different119891

1and the correspond-

ing interception time 1199052and ΔV

1198911(∘) Solutions of

(1198912(∘)119873

1) 119905

2(s) ΔV (kms)

150 (minus566897 0) 321084 09035(minus399134 0) 328443 08897

170 (minus1092932 0) 212588 39088(532432 1) 358345 11770

190 (891630 0) 390801 14596210 (1043983 0) 436651 16333230 (1123591 0) 499326 17338250 (1177130 0) 598472 18148270 (1219959 0) 802495 19082290 (1259555 0) 1602413 20322310 (1244383 1) 1114899 22976330 (1186911 1) 628954 29698

(1257512 2) 1506675 26872(1269266 3) 2368738 26464

350 (1142511 1) 525410 44692(1242882 2) 1085322 35407(1260145 3) 1632862 34192(1267542 4) 2175443 33698

10 (1261114 4) 1685896 49766(1266609 5) 2084768 48911

is 1 the corresponding costs is 29698 kms the correspond-ing transfer time is 209279 s and the total interception timeis 628954 s The value of 120582 is 27280 thus the trajectory ishyperbolic

For different impulse points all feasible solutions satisfy-ing 1198912isin (minus1269425

∘ 1269425

∘) and the fuel constraint and

the corresponding value of 1198732= 120578 for (2) can be obtained

by (31) and are listed in Table 1 The total interception time is1199052= 1199051+ 1199053 the corresponding costs are also listed in Table 1

It should be notified that there is no feasible solution when1198911= 30∘+ 119896119895sdot 20∘ 119896119895= 1 2 3 4 5 By using the numerical

optimization algorithm the minimum-time impulse point isat 1198911= 1705235

∘ the solution is 1198912= minus1108795

∘ the cost is49999 kms and the total interception time is 198106 s Theminimum-time interception trajectory is hyperbolic and it isnear to that in Figure 5

8 Conclusions

This paper studies the interception problem for a hyperbolictarget with an upper-bounded tangent impulse Based onfour special initial true anomalies the solution-existencerange of the target true anomaly for any conic transfer isobtained Then the problem is solved by the secant methodfor a time-of-flight equation For a given impulse pointthere may be zero one many or infinite solutions howeverwhen considering the sphere of influence of the Earth thereare always finite solutions For different impulse points allfeasible solutions and the global minimum-time solution are

obtained This proposed method is valid for an assignedimpulse point and the time optimization with the initialcoasting time The interception with a tangent-impulse tech-nique can be used in orbital interception task for potentiallyhazardous objects during the Earth flyby with simple control

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work is supported in part by the China Postdoctoral Sci-ence Foundation Funded Project (Grant no 2012M520753)theHeilongjiang Postdoctoral Science Foundation (Grant noLBH-Z12125) and theNational Natural Scientific Foundationof China (Grant no 61273096)

References

[1] J C Solem ldquoInterception of comets and asteroids on collisioncourse with Earthrdquo Journal of Spacecraft and Rockets vol 30 no2 pp 222ndash228 1993

[2] IMRoss S-Y Park and SDV Porter ldquoGravitational effects ofearth in optimizing ΔV for deflecting earth-crossing asteroidsrdquoJournal of Spacecraft and Rockets vol 38 no 5 pp 759ndash7642001

[3] B Dachwald and B Wie ldquoSolar sail trajectory optimization forintercepting impacting and deflecting near-earth asteroidsrdquoin Proceedings of the AIAA Guidance Navigation and ControlConference and Exhibit pp 3331ndash3348 San Francisco CalifUSA August 2005

[4] D Izzo ldquoOptimization of interplanetary trajectories for impul-sive and continuous asteroid deflectionrdquo Journal of GuidanceControl and Dynamics vol 30 no 2 pp 401ndash408 2007

[5] J Park R G Rhinehart and P T Kabamba ldquoMiss analysisin Lambert interceptions with application to a new guidancelawrdquo Mathematical Problems in Engineering Theory Methodsand Applications vol 6 no 2-3 pp 225ndash265 2000

[6] Y Ulybyshev ldquoDirect high-speed interception analytic solu-tions qualitative analysis and applicationsrdquo Journal of Space-craft and Rockets vol 38 no 3 pp 351ndash359 2001

[7] S P Burns and J J Scherock ldquoLambert guidance routinedesigned to match position and velocity of ballistic targetrdquoJournal of Guidance Control and Dynamics vol 27 no 6 pp989ndash996 2004

[8] R H Battin An Introduction to the Mathematics and Methodsof Astrodynamics American Institute of Aeronautics and Astro-nautics (AIAA) Reston Va USA Revised edition 1999

[9] S L Nelson and P Zarchan ldquoAlternative approach to thesolution of Lambertrsquos problemrdquo Journal of Guidance Controland Dynamics vol 15 no 4 pp 1003ndash1009 1992

[10] P V Arlulkar and S D Naik ldquoSolution based on dynamicalapproach for multiple-revolution lambert problemrdquo Journal ofGuidance Control and Dynamics vol 34 no 3 pp 920ndash9232011

[11] G Zhang D Mortari and D Zhou ldquoConstrained multiple-revolution Lambertrsquos problemrdquo Journal of Guidance Controland Dynamics vol 33 no 6 pp 1779ndash1786 2010

10 Mathematical Problems in Engineering

[12] C A Lembeck and J E Prussing ldquoOptimal impulsive interceptwith low-thrust rendezvous returnrdquo Journal of Guidance Con-trol and Dynamics vol 16 no 3 pp 426ndash433 1993

[13] D-R Taur V Coverstone-Carroll and J E Prussing ldquoOptimalimpulsive time-fixed orbital rendezvous and interception withpath constraintsrdquo Journal of Guidance Control and Dynamicsvol 18 no 1 pp 54ndash60 1995

[14] H Leeghim ldquoSpacecraft intercept using minimum controlenergy andwait timerdquoCelestialMechanics ampDynamical Astron-omy vol 115 no 1 pp 1ndash19 2013

[15] D A Vallado Fundamentals of Astrodynamics and ApplicationsMicrocosm Press El Segundo Calif USA 2nd edition 2001

[16] B F Thompson K K Choi S W Piggott and S R BeaverldquoOrbital targeting based on hodograph theory for improvedrendezvous safetyrdquo Journal of Guidance Control and Dynamicsvol 33 no 5 pp 1566ndash1576 2010

[17] V G Adamyan L V Adamyan G M Zaimtsyan and LT Manandyan ldquoDouble-pulse cotangential transfers betweencoplanar elliptic orbitsrdquo Journal of Applied Mathematics andMechanics vol 73 no 6 pp 664ndash672 2009

[18] G Zhang D Zhou D Mortari and T A Henderson ldquoAna-lytical study of tangent orbit and conditions for its solutionexistencerdquo Journal of Guidance Control and Dynamics vol 35no 1 pp 186ndash194 2012

[19] G Zhang D Zhou Z Sun and X Cao ldquoOptimal two-impulsecotangent rendezvous between coplanar elliptical orbitsrdquo Jour-nal of Guidance Control and Dynamics vol 36 no 3 pp 677ndash685 2013

[20] G Zhang X Cao and D Zhou ldquoTwo-impulse cotangent ren-dezvous between coplanar elliptic and hyperbolic orbitsrdquo Jour-nal of Guidance Control and Dynamics vol 37 no 3 pp 964ndash969 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 9

Table 1 Feasible solutions of1198912for different119891

1and the correspond-

ing interception time 1199052and ΔV

1198911(∘) Solutions of

(1198912(∘)119873

1) 119905

2(s) ΔV (kms)

150 (minus566897 0) 321084 09035(minus399134 0) 328443 08897

170 (minus1092932 0) 212588 39088(532432 1) 358345 11770

190 (891630 0) 390801 14596210 (1043983 0) 436651 16333230 (1123591 0) 499326 17338250 (1177130 0) 598472 18148270 (1219959 0) 802495 19082290 (1259555 0) 1602413 20322310 (1244383 1) 1114899 22976330 (1186911 1) 628954 29698

(1257512 2) 1506675 26872(1269266 3) 2368738 26464

350 (1142511 1) 525410 44692(1242882 2) 1085322 35407(1260145 3) 1632862 34192(1267542 4) 2175443 33698

10 (1261114 4) 1685896 49766(1266609 5) 2084768 48911

is 1 the corresponding costs is 29698 kms the correspond-ing transfer time is 209279 s and the total interception timeis 628954 s The value of 120582 is 27280 thus the trajectory ishyperbolic

For different impulse points all feasible solutions satisfy-ing 1198912isin (minus1269425

∘ 1269425

∘) and the fuel constraint and

the corresponding value of 1198732= 120578 for (2) can be obtained

by (31) and are listed in Table 1 The total interception time is1199052= 1199051+ 1199053 the corresponding costs are also listed in Table 1

It should be notified that there is no feasible solution when1198911= 30∘+ 119896119895sdot 20∘ 119896119895= 1 2 3 4 5 By using the numerical

optimization algorithm the minimum-time impulse point isat 1198911= 1705235

∘ the solution is 1198912= minus1108795

∘ the cost is49999 kms and the total interception time is 198106 s Theminimum-time interception trajectory is hyperbolic and it isnear to that in Figure 5

8 Conclusions

This paper studies the interception problem for a hyperbolictarget with an upper-bounded tangent impulse Based onfour special initial true anomalies the solution-existencerange of the target true anomaly for any conic transfer isobtained Then the problem is solved by the secant methodfor a time-of-flight equation For a given impulse pointthere may be zero one many or infinite solutions howeverwhen considering the sphere of influence of the Earth thereare always finite solutions For different impulse points allfeasible solutions and the global minimum-time solution are

obtained This proposed method is valid for an assignedimpulse point and the time optimization with the initialcoasting time The interception with a tangent-impulse tech-nique can be used in orbital interception task for potentiallyhazardous objects during the Earth flyby with simple control

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work is supported in part by the China Postdoctoral Sci-ence Foundation Funded Project (Grant no 2012M520753)theHeilongjiang Postdoctoral Science Foundation (Grant noLBH-Z12125) and theNational Natural Scientific Foundationof China (Grant no 61273096)

References

[1] J C Solem ldquoInterception of comets and asteroids on collisioncourse with Earthrdquo Journal of Spacecraft and Rockets vol 30 no2 pp 222ndash228 1993

[2] IMRoss S-Y Park and SDV Porter ldquoGravitational effects ofearth in optimizing ΔV for deflecting earth-crossing asteroidsrdquoJournal of Spacecraft and Rockets vol 38 no 5 pp 759ndash7642001

[3] B Dachwald and B Wie ldquoSolar sail trajectory optimization forintercepting impacting and deflecting near-earth asteroidsrdquoin Proceedings of the AIAA Guidance Navigation and ControlConference and Exhibit pp 3331ndash3348 San Francisco CalifUSA August 2005

[4] D Izzo ldquoOptimization of interplanetary trajectories for impul-sive and continuous asteroid deflectionrdquo Journal of GuidanceControl and Dynamics vol 30 no 2 pp 401ndash408 2007

[5] J Park R G Rhinehart and P T Kabamba ldquoMiss analysisin Lambert interceptions with application to a new guidancelawrdquo Mathematical Problems in Engineering Theory Methodsand Applications vol 6 no 2-3 pp 225ndash265 2000

[6] Y Ulybyshev ldquoDirect high-speed interception analytic solu-tions qualitative analysis and applicationsrdquo Journal of Space-craft and Rockets vol 38 no 3 pp 351ndash359 2001

[7] S P Burns and J J Scherock ldquoLambert guidance routinedesigned to match position and velocity of ballistic targetrdquoJournal of Guidance Control and Dynamics vol 27 no 6 pp989ndash996 2004

[8] R H Battin An Introduction to the Mathematics and Methodsof Astrodynamics American Institute of Aeronautics and Astro-nautics (AIAA) Reston Va USA Revised edition 1999

[9] S L Nelson and P Zarchan ldquoAlternative approach to thesolution of Lambertrsquos problemrdquo Journal of Guidance Controland Dynamics vol 15 no 4 pp 1003ndash1009 1992

[10] P V Arlulkar and S D Naik ldquoSolution based on dynamicalapproach for multiple-revolution lambert problemrdquo Journal ofGuidance Control and Dynamics vol 34 no 3 pp 920ndash9232011

[11] G Zhang D Mortari and D Zhou ldquoConstrained multiple-revolution Lambertrsquos problemrdquo Journal of Guidance Controland Dynamics vol 33 no 6 pp 1779ndash1786 2010

10 Mathematical Problems in Engineering

[12] C A Lembeck and J E Prussing ldquoOptimal impulsive interceptwith low-thrust rendezvous returnrdquo Journal of Guidance Con-trol and Dynamics vol 16 no 3 pp 426ndash433 1993

[13] D-R Taur V Coverstone-Carroll and J E Prussing ldquoOptimalimpulsive time-fixed orbital rendezvous and interception withpath constraintsrdquo Journal of Guidance Control and Dynamicsvol 18 no 1 pp 54ndash60 1995

[14] H Leeghim ldquoSpacecraft intercept using minimum controlenergy andwait timerdquoCelestialMechanics ampDynamical Astron-omy vol 115 no 1 pp 1ndash19 2013

[15] D A Vallado Fundamentals of Astrodynamics and ApplicationsMicrocosm Press El Segundo Calif USA 2nd edition 2001

[16] B F Thompson K K Choi S W Piggott and S R BeaverldquoOrbital targeting based on hodograph theory for improvedrendezvous safetyrdquo Journal of Guidance Control and Dynamicsvol 33 no 5 pp 1566ndash1576 2010

[17] V G Adamyan L V Adamyan G M Zaimtsyan and LT Manandyan ldquoDouble-pulse cotangential transfers betweencoplanar elliptic orbitsrdquo Journal of Applied Mathematics andMechanics vol 73 no 6 pp 664ndash672 2009

[18] G Zhang D Zhou D Mortari and T A Henderson ldquoAna-lytical study of tangent orbit and conditions for its solutionexistencerdquo Journal of Guidance Control and Dynamics vol 35no 1 pp 186ndash194 2012

[19] G Zhang D Zhou Z Sun and X Cao ldquoOptimal two-impulsecotangent rendezvous between coplanar elliptical orbitsrdquo Jour-nal of Guidance Control and Dynamics vol 36 no 3 pp 677ndash685 2013

[20] G Zhang X Cao and D Zhou ldquoTwo-impulse cotangent ren-dezvous between coplanar elliptic and hyperbolic orbitsrdquo Jour-nal of Guidance Control and Dynamics vol 37 no 3 pp 964ndash969 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

10 Mathematical Problems in Engineering

[12] C A Lembeck and J E Prussing ldquoOptimal impulsive interceptwith low-thrust rendezvous returnrdquo Journal of Guidance Con-trol and Dynamics vol 16 no 3 pp 426ndash433 1993

[13] D-R Taur V Coverstone-Carroll and J E Prussing ldquoOptimalimpulsive time-fixed orbital rendezvous and interception withpath constraintsrdquo Journal of Guidance Control and Dynamicsvol 18 no 1 pp 54ndash60 1995

[14] H Leeghim ldquoSpacecraft intercept using minimum controlenergy andwait timerdquoCelestialMechanics ampDynamical Astron-omy vol 115 no 1 pp 1ndash19 2013

[15] D A Vallado Fundamentals of Astrodynamics and ApplicationsMicrocosm Press El Segundo Calif USA 2nd edition 2001

[16] B F Thompson K K Choi S W Piggott and S R BeaverldquoOrbital targeting based on hodograph theory for improvedrendezvous safetyrdquo Journal of Guidance Control and Dynamicsvol 33 no 5 pp 1566ndash1576 2010

[17] V G Adamyan L V Adamyan G M Zaimtsyan and LT Manandyan ldquoDouble-pulse cotangential transfers betweencoplanar elliptic orbitsrdquo Journal of Applied Mathematics andMechanics vol 73 no 6 pp 664ndash672 2009

[18] G Zhang D Zhou D Mortari and T A Henderson ldquoAna-lytical study of tangent orbit and conditions for its solutionexistencerdquo Journal of Guidance Control and Dynamics vol 35no 1 pp 186ndash194 2012

[19] G Zhang D Zhou Z Sun and X Cao ldquoOptimal two-impulsecotangent rendezvous between coplanar elliptical orbitsrdquo Jour-nal of Guidance Control and Dynamics vol 36 no 3 pp 677ndash685 2013

[20] G Zhang X Cao and D Zhou ldquoTwo-impulse cotangent ren-dezvous between coplanar elliptic and hyperbolic orbitsrdquo Jour-nal of Guidance Control and Dynamics vol 37 no 3 pp 964ndash969 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of