Back of the Envelope Calculations for Mission Design

Table of Contents1 LAUNCH AZIMUTH ______________________________________________________3

2 ORBIT INCLINATION FROM VISUAL SIGHTING____________________________4

3 SERIALLY STAGED BOOSTERS____________________________________________5

4 PARALLEL STAGED BOOSTERS ___________________________________________6

5 HOHMANN TRANSFERS__________________________________________________8

6 VELOCITY INCREMENT REQUIRED TO CIRCULARIZE AN ORBIT____________9

7 TRANSFERS TO GEOSYNCHRONOUS EQUATORIAL ORBITS _______________10

8 GREAT CIRCLE DISTANCES _____________________________________________13

9 SINGLE IMPULSE MANEUVERS _________________________________________14

10 OMS/RCS BURNS _____________________________________________________18

11 ORBIT PERIOD _______________________________________________________19

12 Pointing At Geostationary Satellites_________________________________________20

1 LAUNCH AZIMUTH

A first approximation to the launch azimuth. AZ , for an orbit can be determined from thetarget orbit inclination, i, and the latitude of the launch site, LS . This approximation is given by therelations:

1. "North" launches (AZ north of an east-west axis)

AZ north( ) = arcsin cos i / cos LS( )

If this calculation results in a negative angle, add 360 degrees.

2. "South" launches (AZ south of an east-west axis)

AZ south( ) = 180 deg. - arcsin cos i / cos LS( )

If the arcsin calculation results in an angle less than zero. KEEP the sign. It willcancel with the other minus sign and result in a southwest launch.

The reason that the launch azimuths produced in azimuths produced in this manner are onlyapproximate is that the insertion of the spacecraft into orbit takes a finite amount of time and thetarget orbit plane moves during that time. Also, the launch vehicle starts from the pad with theeastward motion of the launch site. Thus, small changes in launch azimuth from that calculatedabove should result in launch vehicle performance improvements.

NOTE: When cos LS < cos i, the equations result in values for sin AZ > 1. In such a case, wecannot take the arcsin and must be able to recognize what this means. This situationoccurs when the orbit plane lies closer to the equator than the launch site does.Hence, it is impossible to launch directly into this orbit and no launch azimuth existsin the simple sense. A plane change will be necessary to reach the target plane inthis case.

NOTE: The calculations described above can be used to give the heading of a spacecraft at anypoint in the orbit. To do this, the current latitude is used in place of the latitude ofthe launch site.

CAUTION: ALWAYS CHECK YOUR RESULT USING LOGIC!

2 ORBIT INCLINATION FROM VISUAL SIGHTING

If you visually detect a satellite moving among the stars at night, you can quickly estimatethe inclination of its orbit by using the following procedure.

1. Estimate the satellite’s heading ( AZ ). ( AZ should lie between zero and 360 degrees)

2. Determine the local latitude ( ).

3. Calculate the approximate orbit inclination, i, from

i = arccos sin AZ( )cos ( )( )The resulting value for i will be in the range from zero to 180° and no quadrant check is

needed. Recall that orbits with inclinations between zero and 90°, are direct orbits (posigradeorbits) and orbits with inclinations between 90° and 180° are retrograde orbits. Orbits withinclinations of 90° are polar orbits. Orbits with inclinations i = 0° are direct equatorial orbits andorbits with i = 180° are retrograde polar orbits.

EXAMPLE: For sightings from the US Air Force Academy ( = 39.01° ), the following table isprovided.

HEADINGESTIMATE

AZ(deg)

APPROXIMATEINCLINATION

i(deg)

HEADINGESTIMATE

AZ(deg)

ApproximateINCLINATION

i(deg)

0 90 180 90

10,170 82 I90,350 98

20,160 75 200,340 105

30,150 67 210,330 113

40,140 60 220,320 120

50,130, 53 230,310 127

60,120 48 240,300 132

70,110 43 250,290 137

80,100 40 260,280 140

90 30.01 270 140.99

3 SERIALLY STAGED BOOSTERS

Most staged boosters in the past have been serially staged. This means: that only one stage(and more importantly, only one type of engine/fuel combination) is in operation at any given timeduring the boost. There can be more than one engine on a given stage, but all engines on that stagemust have the same exhaust velocity.

For the jth stage in a serially staged vehicle, the "ideal" ∆V that can be produced is given by

∆Vj = c j lnMoj

M fj

where ∆Vj is the theoretical "ideal" velocity increment which can be imparted to the vehicle bythat stage,

c j is the exhaust velocity (relative to the vehicle) of the fuel/engine combination on the jthstage,

Moj is the mass of the vehicle at the start of the jth stage burn, and

M fj is the mass of the vehicle at the end of the jth stage burn BEFORE the empty jth stageis dropped.

Note that M fj ≠ Mo j +1( ). In fact, the mass dropped when the spent jth stage is given by

∆M j = M fj − Mo j +1( )

For an N stage vehicle, the "ideal" ∆V is given by ∆V = a c j lnMoj

M fj

j =1

N

∑ .

If all of the stages have the same exhaust velocity, c, then the "ideal" c, relation can be simplifiedto yield

∆V = c lnMo1

M f 1

×Mo2

M f 2

× ... ×MoN

M fN

NOTE: The mass ratios in the equations above can be replaced weight ratios with no change inthe end results. The units which result for the "ideal" ∆V will be the units in which theexhaust velocities are stated.

4 PARALLEL STAGED BOOSTERS

Some boost vehicles (Titan III, Space Shuttle) use two (or more) types of fuel/enginecombinations simultaneously (usually solid fueled strap-on boosters firing simultaneously withliquid fueled main engines). For such a combination (a Parallel Staged Booster), the "ideal" ∆Vequation developed for serial staged boost vehicles does not work. However, a slight modificationin the derivation of the equation results in a relation appropriate for parallel staged boosters.

A parallel staged booster is considered to have "staged" at any time that the ratio of themass flows out of the different types of engines changes. Usually this happens when the solidrocket boosters stop thrusting and are dropped. The liquid fueled main engines usually keep onthrusting through the staging point. The "ideal" ∆V calculation must be divided at the stagingpoint even though the main engines continue to thrust through that point.

The "ideal" ∆V for a parallel staged booster stage is given by

∆V = c* lnMo

M f

where

Mo is the total vehicle mass at the start of the for that stage,

M f is the total vehicle mass at the end of burn (at the stage point) PRIOR TO ANY MASSJETTISON,

c* = c1 f1 + c2 f2 (for the two engine type case)

ci is the exhaust velocity for fuel/engine type i,

fi =˙ m i

˙ m 1 + ˙ m 2 (i =1,2 for two engine types), and

fi is, the mass flow fraction due to fuel/engine type i (this is the mass flow for that type offuel/engine divided by the total mass flow for stage. Note that the sum of the mass flowfractions for a stage is one.

EXAMPLE:

For the Space Shuttle, the staging point is Solid Rocket Booster Separation (SRB SEP), and

∆V1 = c* lnM01

M f 1

∆V2 = cSSME lnM02

M f 2

∆V = ∆V1 +∆V2

where c∗ = cSRB fSRB + cSSME fssme

and cSRB is the Solid Rocket Booster (SRB) exhaust velocity,

cSSME is the Space Shuttle Main Engine (SSME) exhaust velocity,

fSRB =˙ m SRB

˙ m SRB + ˙ m SSME

fSSME =˙ m SSME

˙ m SRB + ˙ m SSME

˙ m SRB is the mass flow per second from BOTH SRB’S

˙ m SSME is the mass flow per second from all three SSME’s

M01 is the total mass prior to liftoff,

M f 1 is the total vehicle mass just prior to SRB SEP,

M02 is the total vehicle mass just after SRB SEP,

M f 2 is the total vehicle mass at Main Engine CutOff (MECO), including theempty External Tank (ET)

5 HOHMANN TRANSFERS

For most Earth orbits which will be serviced by the Space Shuttle, the distances from theearth will be small in terms of Earth radii. For such orbits (even out to and beyond geosynchro-nous distance), the Hohmann transfer is the best transfer to use when transferring between circularcoplanar orbits.

For transfers between circular coplanar orbits, the information usually given consists of theradii of the initial and final orbits. The information desired consists of the semi-major axis of thetransfer orbit, the velocity increments at the ends of the transfer orbit, and the total velocityincrement required for the transfer. In order to calculate the required information. it is necessary tocalculate the following four velocities:

The velocity in the initial circular orbit ( Vc1 =r1

)

The velocity in the transfer orbit at initial orbit height ( VT 1 =2

r1

−1

aT

)

The velocity in the transfer orbit at final orbit height (

2

r2

−1

aT

)

The velocity in the final circular orbit ( r2

)

The initial velocity increment is then qiven by the equation

∆V1 =2

r1

−1

aT

−

r1

and the final velocity increment is given by

∆V2 =r2

−2

r2

−1

aT

r1 is the radius of the initial circular orbit

r2 is the radius of the final circular orbit

aT is the semi-mayor axis of the transfer orbit

aT = r1 + r2( ) / 2( ) , and

is the gravitational parameter of the central body.

6 VELOCITY INCREMENT REQUIRED TO CIRCULARIZE AN ORBIT

Assume that a spacecraft is in a non-circular orbit with semi-major axis, a , and eccentricity, e .Suppose that it is required that the orbit be circularized at a radius, r , such rp ≤ r ≤ ra using a singleburn (we will consider the planar case only). The magnitude of velocity in the non-circular orbit at

the radius where the orbit is to be circularized is given by v =2

r−

1

a

Recall that the orbit parameter (semi-latus rectum) of the orbit is given by the relation

p = a 1− e2( ) . The true anomaly, , is then given by = cos−1 p − r

e r

.

(For this calculation, do not worry about obtaining the proper quadrant for , use the valuebetween zero and 180 degrees which is output by your calculator. The velocity increment requiredis the same for both possible values of the angle, . If you need to know the proper value of ,recall that on the outbound half of the orbit, lies between zero and 180 degrees whereas on theinbound half,, lies between 180 degrees and 360 degrees.

At any point in the orbit, velocity is given by the vector equation

V =p

esin u R +p

r u

H

where uH

is a unit vector in the horizontal direction,

uR

is a unit vector in the radial (outward) direction

is the gravitational parameter of the central body.

At any radius, the required circular velocity is given by V c =r

uH

The ∆V required for circularization is given by ∆V = V c − V , or in vectorterminology, by

∆V =r

−p

r

uH

− p

esin u R

The magnitude of the required ∆V is then given by

∆V =r

−p

r

2

+p

e2 sin2

7 TRANSFERS TO GEOSYNCHRONOUS EQUATORIAL ORBITS

When a satellite is launched from a non-equatorial launch site. it enters an initial orbit which hasan inclination at least equal to the latitude of the launch site. In order for a satellite to be insertedinto a geosynchronous equatorial orbit, a plane change must be made. The total plane changenecessary equals the inclination of the initial orbit of the satellite. In addition to the plane change,the altitude of the orbit must be raised from that of the initial orbit to geosynchronous height.

The crudest "good" approximation to the required maneuver (assuming circular initial andfinal orbits) consists of:

1. A Hohmann transfer initial burn (velocity increment) at an ascending or descending node ofthe initial orbit.

2. A combined plane change and Hohmann transfer final burn at geosynchronous height.

This combination is a good approximation, but we can do better if we are willing to work a bit. Letus assume the following in order to make the analysis clear:

1. The initial and final orbits are circular, with the initial orbit being a low (shuttle type) orbitand the final orbit being geosynchronous.

2. The inclination of the initial orbit is equal to the latitude of the launch site while the finalorbit is equatorial.

3. All burns will combine plane changes with Hohmann transfer velocity increments. All orpart of the required plane change can be made at either end of the transfer orbit. The initialburn will take place at either the ascending node or the descending node of the initial orbit.

4. The objective of the calculations will be to find the combination of two burns whichminimizes the total velocity increment required to accomplish the transfer.

Assume that the satellite is launched from a launch site which has latitude . Thus, theinclination of the initial orbit is i = . Let the radius of the initial low Earth orbit be rLEO and theradius of the circular target orbit be rGEO (radius of a Geocentric Equatorial orbit.)

Assume that we wish to make a small bit of the inclination change at the first burn point. Call thissmall amount of inclination change ∆ iLow (DiL). Recall that the first burn must take place at theascending or descending node. This burn always leads to an elliptic transfer orbit with apogee atgeosynchronous height and perigee at initial orbit height. The velocity in the circular low Earth orbit

is given byVcLEO =rLEO

and the velocity at perigee in the transfer orbit is given by

VPT =2

rLEO

−1

aT

.

Using the diagram below and the law of cosines, the required velocity change at perigee of thetransfer orbit is given by

DiL

Vc

-

VPT

-

∆V1-

∆V1 = VPT2 + VcLeo

2 − 2VPTVcLeo cos ∆ iLow( )

Once we reach the apogee of the transfer orbit, a second combined Hohmann burn and planechange must be made. The speed in the transfer orbit at apogee is given by

VAT =2

rGeo

−1

aT

The required speed for the circular equatorial orbit at geosynchronous height is given by

VcGeo =rGeo

The required plane change at apogee is the part of the plane change not made at perigee. Since thetotal plane change required is degrees and the plane change made at perigee is ∆ iLow degrees, theremaining plane change, ∆ iHigh (DiH), is given by ∆ iHigh = - ∆ iLow . Using the diagram belowand the law of cosines, the required velocity change at apogee is given by

∆V2 = VAT2 + VcGeo

2 − 2VATVcGep cos ∆ iHigh( )

The total velocity change required is the sum of the change at perigee and the change at apogee.

∆VTotal = ∆V1 +∆V2

Once the process described above has, been worked out, it is a simple matter to compute the totalvelocity change necessary once a value for ∆ iLow has been chosen. A plot of total velocity changerequired versus DiL can be created and the minimum total velocity change found from the plot.The value of DiL which corresponds to the minimum total velocity change is the amount of theinclination change which should be made at the perigee of the transfer orbit.

PROBLEM:

DiH

VAT-

VCgeo

-

∆V2

Launch from KSC into a minimum inclination circular orbit with altitude of 160 nmi. Starting fromthis orbit, find the minimum total velocity change required and the corresponding value for ∆ iLow

for a transfer to a circular geostationary equatorial orbit.

DiL forMin. Total

V

TOTAL

LO DiL

Minimum Total

V

V

8 GREAT CIRCLE DISTANCES

The range angle, Λ , between two points on the surface of the earth, (LATl, LONG1) and (LAT2,LONG2) is given by

Λ = arccos sin[ LAT1( ) LAT2( )+ cos LAT1( ) cos LAT2( ) cos Long2 − LONG1( )]

The result of this calculation on a hand calculator or in the computer will be an angle Λ betweenzero and 180 degrees. This value for Λ is the range angle for the "short" great circle path betweenthe two points. If you want the "long" great circle path between the two points use 360 degreesminus the "short" range angle for the angle in the relations below.

Once the range angle between the two points is known, the relations below give the great circledistance, D, between the points in the units noted.

D = 60.108 Λ (nautical miles)

D = 111.32 Λ (kilometers)

D = 69.171 Λ (statute miles)

D = 0.1745 Λ (D.U.)

9 SINGLE IMPULSE MANEUVERS

Assume that a maneuverable spacecraft is in a circular orbit with a radius, RO. A numberof orbit changes can be made with a single impulse. Among them are:

1. SIMPLE PLANE CHANGE -- To execute a simple plane change (changeinclination an amount ∆i without changing the size or shape of the orbit), therequired ∆V is given by

∆V2 = 2 VC2 − VC

2 cos ∆2( )∆V 2 = VC 2 1 − cos∆i( )

Recall that the velocity in a circular orbit given by

VC =RO

and thus the required ∆V can be written as

∆V = VC 2 1 − cos∆i( ) =2

RO1 − cosi( )

2. MAKE ORBIT ELLIPTIC WITH PERIGEE AT CIRCULAR ORBIT

To make the new orbit elliptic with the perigee radius equal to th radius of theoriginal circular orbit (RO), an apogee radius (RA) must be specified. The semi-major axis (a ) of the new orbit is then qiven by

a = RA + RO( ) / 2

and the eccentricity e( )is given by

e =RA − RO

2a

The velocity at perigee Vp( ) in the elliptic orbit is given by

VP =2

RO−

1

a

Vc

∆V∆2'

Vc

and the required ∆V is then given by

∆V = VP − VC

The burn is posigrade, tangent to the circular orbit, and is at the perigee of the ellipticorbit.

3. MAKE ORBIT ELLIPTIC WITH APOGEE AT CIRCULAR ORBIT

This case is very similar to the case above. In order to make the new orbit ellipticwith the apogee radius equal to the radius of the original circular orbit, a desiredperigee radius (RP) must be specified. Then. the new semi-major axis is given by

a = RP + RO( ) / 2a

The eccentricity is given by

e =RO − RP

2a

and the velocity in the new orbit at the burn point (at apogee) is given by

VA =2

RO−

1

a

The required ∆V is then given by

∆V = Vc − VP

Burn is in direction

opposite to velocity

4. SIMPLE RADIAL BURNS -- Suppose that we want to see the effect of that a radial burnhas on the size and shape of an orbit. Assume that we start in a circular orbit and apply a radialvelocity impulse ∆V to the spacecraft. During the impulse, the radius, RO, will not change, but thenew velocity will be given by

VNEW = VC2 +∆V 2

The new semi-major axis will be given by

a =RO

2 −VNEW2 RO

and the new flight path angle will be

= sin− 1 ∆V

VNEW

VNEW

∆V

Vc

(if ∆V is outward. then > 0 )

(if ∆V is inward, then < 0 )

The new orbit parameter will be

P =RO VNEW cos( )2

and the eccentricity wil1 be given by

e = 1 − P a

The true anomaly at the burn point, , is given

= cos−1 P − RO

e RO

(when ∆V is outward, use 0 < < 180 deg.)

(when ∆V is inward, use 180 deg < < 360 deg.

10 OMS/RCS BURNS

Objective:To calculate a rough estimate of the amount (weight) of fuel which is required toproduce a given ∆V usirg the OMS or RCS engines on the Space Shuttle.

Given:

WO ~ the Space Shuttle weight just prior to the burn.

Isp ~ the Isp of the engine/fuel combination used

for the OMS, Isp = 313.2 sec

for the RCS, Isp = 280.0 sec

The basic equation is the "ideal ∆V ” equation, i.e.,

∆V = c / nMo

M f

= c / nWo

Wf

or, in terms of weights,

∆V = c nWO

WO − WFUEL

Solving the equation for the amount of fuel required to produce a given ∆V , we get

WFUEL = WO 1 − E− ∆V Isp g( ){ }

11 ORBIT PERIOD

The period of an elliptic (or circular) orbit can be calculated from the semi-major axis, a , byusing the equation

Τ = 2a3

.

In this equation, one must be careful to use consistent units in the values used for the semi-major axis and the gravitational parameter.

In many cases, we are not given the semi-major axis, but we are given means to calculate it.Common cases are:

Given perigee radius, rP and apogee radius, rA then

a = rP + rA( ) 2

Given perigee altitude, rP , and apogee altitude, rA then

a = rP + rA + 2Re( ) 2

Note: Re is the radius of the earth.

Given perigee radius, rP , F and orbit eccentricity, e , then

a = rP 1− e( )

In all cases, be very careful with the units.

12 POINTING AT GEOSTATIONARY SATELLITES

As more of us become involved with satellite-relayed TV, it becomes important to know howto point toward a geostationary satellite. The general problem can be stated as follows:

An antenna is located at a known point on the surface of the earth (latitude, , andlongitude, , are known). It is desired to point the antenna at a geostationary satellitelocated 22828 nmi. from the center of the earth in the equatorial plane directly abovelongitude G . Find the azimuth (angle measured east from north in the horizontalplane) and elevation (angle between the line of sight to the satellite and the localhorizontal plane) of the line of sight (LOS) to the geostationary satellite.

There are several ways to attack this problem, and the simplest take advantage of the factthat the geometry is fixed relative to the earth (the antenna site and the satellite are both"geostationary").

The vector from the center of the earth to the antenna site, r A , is given by

r A = rA cos cos i + rA sin cos j + rA sin k

where

rA is the distance between the antenna site and the center of the earth,

is the longitude of the antenna site (measured positive toward the east

0o <= < 360o( )

is the latitude of the antenna site (measured from the equatorial plane, positive to thenorth).

−90o <= <= 90o( )and i , j , k , are unit vectors in earth-fixed coordinates with k out the north pole, i inthe equatorial plane pointing toward 0

o longitude point and j such that ( i , j , k ) isa right-handed coordinate triad ( j points to the 90 o east longitude point on theequator - in the Indian Ocean).

The vector from the center of the earth to the geostationary satellite is

r G = rG cos G i + rG sin G j

Note that G ≡ 0 o since all geostationary orbits must be equatorial i ≡ 0o( )The vector from the antenna site to the satellite, r LOS , can be calculated from

Thus,r LOS = r G − r A

Now that we haver LOS , all that is necessary is to express this vector in terms of localhorizontal coordinates and then determine the azimuth and elevation angles.

The coordinate transformation between the earth fixed i, j, k( ) axes and a local set

′ i , ′ j , ′ k ( ) such that ′ i points south, ′ j points east, and ′ k points up is given by (Let

= 90 o − )

′ x

′ y

′ z

=sin 0 − cos

0 1 0

cos 0 sin

cos sin 0

−sin cos 0

0 0 1

x

y

z

′ x

′ y

′ z

=cos sin sin sin − cos

−sin cos 0

cos cos sin cos sin

x

y

z

where x, y, z( ) are components of any vector in the earth fixed polar/equatorial coordinatesand ′ x , ′ y , ′ z ( ) are the components of the same vector in local coordinates. The vector wewant to transform is r LOS . Let the components of r LOS be denoted by

r LOS = ai + bj + ck

where

a = rG cos G − rA cos cos

b = rG cos G − rA sin cos

c = rA sin

The components of the transformed vector,

r LOS = ′ a ′ i + ′ b ′ j + ′ c ′ k are ′ a , ′ b , and ′ c , and can be calculated using the transformation matrix. Specifically,

′ a = a cos sin + b sin sin − c cos

′ b = −a sin + b cos

′ c = a cos cos + b sin cos + c sin

Now all that remains is to determine the azimuth and elevation angles given ′ a , ′ b , and ′ c .

The situation is as shown in the figure below:

r-LOS

a'

αε

North

Southb'

k-

-i

c' j-

The angle is the elevation and the angle is the azimuth, . An easy way to carry outthe calculation is to rewriter LOS as

r LOS = A + ′ c ′ k

where

A = ′ a ′ i + ′ b ′ j

The elevation, , can now be written as

cos = A ⋅ r LOS / A r LOS( )This uniquely determines since −90o <= <= 90o .

The azimuth, = 90o + , can be determined from

cos =− ′ i ⋅ A ( )

A

and

− ′ k sin a =− ′ i ∗ A ( )

A

Knowing both sin and cos , we can determine uniquely 0o <= < 360 o( ) . [In

FORTRAN use the ATAN2 function.]

SATELLITE POINTING PROBLEM

The latitude and longitude of Houston are 30.0N , 95.5W( ) . Determine pointing angles (azimuths

and elevations) for geostationary satellites located at longitudes between 80oW and 140 oW (from

220 oE through 280 oE ) at 1o intervals.