Astrodynamics_tudelft

7/27/2019 Astrodynamics_tudelft

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Chapter 1

Some Basic Concepts

1.1 Newtons law and inertial frames

The three laws of newton

First law: Every particle continues in its state of rest or uniform motion in a straight linerelative to an inertial reference frame, unless it is comepelled to change that state byforces acting upon it

Second law: The time rate of change of linear momentum of a particle relative to an inertialreference frame is proportional to the resultant of all forces acting upon thatparticle and is collinear with and in the direction of the resultant force

Third law: If two particles exert forces on each other, these forces are equal in magnitude andopposite in direction (action - reaction)

The rst law denes an inertial reference frame. If one inertial reference frame is know a whole classof inertial reference frames is known. These are non rotating frames which are in rest or in uniformmotion in a straight line with respect to the rst inertial reference frame. This can be shown with theGalilei transformations eq [1.1]. Assume that the rst frame moves with a constant speed of V and thatthe second frame moves with a constant velocity W with respect to the rst frame. The time in frameXYZ differs with a constant T and for simplicity sakes it is assumed that at t0 the origins coincided.The following holds: (also see g 1.1)

r = r W (t t0 ) ; t = t + T (1.1)V =

drdt

; V =drdt

(1.2)

differentiating eq [1.1] with respect to dt and substituting eq: [1.2]

drdt

=drdt W ;

dtdt

= 1 (1.3)

drdt

dtdt

= V W (1.4)drdt

= V = V W (1.5)We dened that V and W were constant so also V has to be constant so XYZ has to be a inertialreference frame.Newtons second law is:

F =ddt

(mV ) (1.6)

From wikipedia: In mathematics and theoretical physics, an invariant is a property of a system which remains unchanged under some transformation . It would be handy if this was also the case for Newtonslaws. We examine newtons second law in reference frame XYZ and substitute eq: [1.5]

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F =ddt

(mV ) =ddt

(m(V W )) = F =ddt

(mV ) W dmdt

(1.7)

So Newtons second law is only invariant if dmdt = 0 i.e. for a body of constant mass. For a rocket, whichhas a varying mass m is considered to be the instantaneous mass and extra apparent forces needs to beadded to compensate. The solidication principle can be used to describe the rocket as a body with aninstantaneous mass m and the thrust as an external force.

1.2 Gravity force and potential

The gravity force of body m 2 can be described as follows

F 2 = m 2 g2 = Gm 1 m 2

r 312r 12 (1.8)

g2 = Gm 1r 312

r 12 (1.9)

g2 is called the eld strength and can simply be found by dividing both sides of eq [1.8] with m 2 . The

eld strength can also be found by taking the (negative) gradient of the potential eld. (In astrophysicsit is assumed that the potential at innity is zero so that U 2 0 = 0). Furthermore we assume for the scalerU 2 eq [1.10]

U 2 = Gm 1r 12

+ U 2 0 = Gm 1r 12

+ 0 (1.10)

g2 = U 2 = Gm 1 1

r 12= Gm 1

ddx

1r 12

,d

dy1

r 12,

ddz

1r 12

= Gm 11

X 2,

1Y 2

,1

Z 2= G

m 1r 212

= Gm 1r 312

r 12

(1.11)

The end of equation [1.11] is the same as equation [1.9] so apparently the potential at an arbitrarydistance r can be written as:

U = Gm 1r (1.12)

eq [1.12] can be differentiated wrt the mass. To nd the gravity eld of a body this equation needs tobe integrated over the body. Only for spherical shells of constant mass density and spheres with a radialsymmetric mass density distribution there is a closed form analytical solution.

dU = Gdmr

(1.13)

For a spherical shell the following holds (see also g 1.3)

dm = (2 R sin )(Rd )t (1.14)

dU P =

G

2R 2 t sin d

r(1.15)

M = 4 R 2 t (1.16)

r 2 = ( R sin )2 + ( l R cos )2

= R 2 sin2 + l2 2Rl cos + R 2 cos2 = R 2 (sin 2 + cos 2 ) + l2 2Rl cos = R 2 + l2 2Rl cos (1.17)

2rdr = 2 RL sin d (1.18)drd

=Rl sin

r=

Rl sin R 2 + l2 2Rl cos

(1.19)

d = R 2 + l2 2Rl cos

Rl sin dr (1.20)

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substituting [1.14] into [1.13] gives [1.15]. Substituting [1.16] and [1.17] into [1.15] gives [1.21]. Substi-tuting [1.17] into [1.18] gives [1.19].

U P = 12

GM

=0

sin d R 2 + l2 2Rl cos

(1.21)

Substituting [ ?? ] into [1.21] gives [1.22] for l< R (point inside sphere) and [1.23] for l > R (point outsidesphere).

U p inside = 12

GM Rl

r = l+ R

r = R ldr =

12

GM Rl

(R + l (R l)) = GM

R(1.22)

U poutside = 12

GM Rl

r = l R

r = R ldr =

12

GM Rl

(R + l (l R)) = GM

l(1.23)

If a body has a radially symmetric density distribution it can be modeled as a series of spherical shellswith the same center. Giving each shell an index i with a mass M i the following is valid

U P outside =

G

l iM i =

GM T

l(1.24)

Looking back one can say that a force in an arbitrary direction, lets say x can be expressed as [1.25]

F P x = m pU px

(1.25)

F P x inside = m p

xGM

R= 0 (1.26)

F P x outside = m p

xGM

l=

GM T m pl2

(1.27)

So if the point lies inside the sphere the neto force is 0 and if it lies outside the sphere the force is directedalong l and can be calculated with [1.27].

It is of course also possible to write out [1.13] for arbitrary bodies. To come to a solution one cando a series expension and truncated all higher order terms. Rearranging and simplifying gives [1.28]

U = GM

l 12

Gl3

(A + B + C 3D ) (1.28)A = (y2 + z2 )dm (1.29)B = (x2 + z2 )dm (1.30)C =

(x2 + y2 )dm (1.31)

D = r 2 sin2 dm (1.32)These terms might look familiar because A,B and C are the moment of inertia around the X,Y and Zaxis and D is the moment of inertia around the line OP (the line between the Origin of the referenceframe of the body and the particle P). For a sphere all the inertias are equal which gives equation [1.27]again.

One can also simplify equation [1.28] for rotational symmetric bodies. For such bodies A=B andD = A cos2 + C sin2 . This results in [1.34]

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C AMl 2 earth

= J 2 1.086 10 3 (1.33)

U = GM

l[1

C AMl 2

(3sin 2 1)] = GM

l[1J 2 (3sin

2 1)] (1.34)This is solved by me, and not in the reader. Please conrm yourself Per denition the force in the direction of l, so the nd it, you only have to differentiate with respect to l

F l = m 2dU dl

= m 2ddl

GMl 2 +12

J 2 R 2 (3sin 2 1)l3 (1.35)

= m 2 GM

l2 32

J 2 R 2GM

l4(3 sin 2 1) (1.36)

=GMm 2

l21 + J 2

32

Rl

2

(3 sin 2 1) (1.37)

hmm This is incorrect because the nal answer should be:

F l =GMm 2

l21 + J 2

32

Rl

2

(3cos2 1) (1.38)

1.3 rocket maneuvering

When using an impulsive shot the following is valid:

H = r 0 V 1 r 0 V 0 = r 0 V (1.39) E =

1

2(V 21

V 20 ) =

1

2(V 0 + v)

(V 0 + v)

V 20 =

1

2( V )2 + V 0

V (1.40)

From these equation the following can be determined:

For a given V the maximum change in orbital angular momentum (H) is achieved if the impulsiveshot is executed when the spacecraft is farthest away from Earth and of V is perpendicular tor 0 .

If the direction of the orbital angular momentum vector should bot be changed, V should bedirected in the initial orbital plane and vice versa The maximum change in (total) orbital energy is achieved if the impulsive shot is executed at thepoint in the orbit where the velocity reaches a maximum value, and if V is directed along the

velocity vector

V 0Most rockets burns are not impulse shot so Tsiolkovskis law has an extra term for the gravity losses(second right hand term)

V = v e lnM 0M e

t e

t 0g sin dt (1.41)

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Chapter 2

Many-body problem

Second law of Newton : F = m d2 rdt 2

(2.1)

Gravity law of Newton : F ij = Gm i m j

r 3ij r ij (2.2)

EoM of body i : m id2 r idt 2

= Gj = i

m i m jr 3ij

r ij (2.3)

Potential eld

g = U = j = i

Gm i m j

r ij+ U i 0 (2.4)

= Gj = i

m jj = i

r 1ij = Gj = i

m jj = i

r ij 2 = Gj = i

m jr 3ij

r ij (2.5)

This forceeld is not conservative because when looking to a point xed in an inertial frame the potentialeld will differ because the bodies j move in time. The sum of the potential and kinetic and potentialenergy of body m i is not constant. (Part its energy will ow into the bodies j and vice versa)

With the use of a auxiliary variable and differentiation the next equation can be found:

i

12

m i V 2i 12

Gi j = i

m i m jr ij

= C (2.6)

Or in short (watch the sign change and denition of E p)

E k + E P = C (2.7)

When taking a closer look one can see that as two bodies approach each other very closely , r ij , thepotential energy goes to innity and the only solution is that the speed of one of the bodies also goes toinnity ( V i ). Or in other words, at least one body reaches escape velocity ( r i ).By using steiners rule for the moment of inertia, differentiating twice using an auxiliary variable andsubstituting previous equations the following can be found:

dI dt

= 2i

m i r i V i (2.8)d2 I dt 2

= 4 E k + 2 E P (2.9)

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To have a stable system it is necessary that d2 I

dt 2 < 0 or else the moment of inertia at the end will growto inty, meaning that at least one body is a innity distance which is not of course not a stable system.Substituting 2.7 in 2.9 gives:

d2 I dt 2

= 4 E k + 2( C E k ) = 2 C + 2 E k (2.10)So to have a stable system C < 0 but one also has to look further. To determine if a system is reallystable we integrate 2.9 over a long period of time, te , and take the average.

1te

t e

0

d2 I dt 2

dt =4te

t e

0

E k dt +2te

t e

0

E P dt (2.11)

1te

dI dt

t e

0= 4 E k + 2 E P (2.12)

substitute 2.8 gives:

2

te[

i

m i r i

V i ]t e0 = 4 E k + 2 E P (2.13)

1te

[i

m i r i V i ]t e0 = 2 E k + E P (2.14)

Because in a stable system no collision or escapes happen the term between square brackets stays nite.In the limit case (lim t e 0 ) the left hand side of the equation will become zero. Giving:

2E k + E P = 0 (2.15)

E k = 12

E P (2.16)

Also the following holds when substituting 2.7 in 2.15

2E k + ( C E k ) = 0 (2.17)E k = C (2.18)

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Chapter 3

Three body problem

3.1 Equations of motion

When describing the motion of a body while there are only two other bodies the EoM reduces to:

d2 r idt 2

= Gm jr ij

r ij + Gmkr ik

r ik (3.1)

When this set is written as an set of rst order differential equations we have three bodies with each sixdifferential equations (dx dy and dz with respect to two other bodies) giving us a total of 18 unknowns,or in other words of the order 18. By using the ten integrals of motion, the invariable plane of Laplace asreference plane and a angular coordinate to replace time, the order can reduced with a factor 12. Thisstill leaves an order six differential equation to be solved. An other approach is by using center-of-massintegrals, rewriting the equations in such a way that they are with respect to the center of mass of thesystem and with respect to the center of mass of body P 1 and P 2 combined, Jacobi reduced this set toan order 12. (which can be solved with the above mentioned tricks) The Jacobi set is:

d2 Rdt 2

= GM r13

r 313+ (1 ) r

23

r 323(3.2)

d2 r 12dt 2

= G (m 1 + m 2 )r 12

r 312+ m 3

r 13r 313

r 23r 323

(3.3)

=m 1

m 1 + m 2(3.4)

R is the vector between the CoM of body P 1 and P 2 combined and P 3 . For the Lunar case were P 1 isthe Earth, P 2 the Moon and P 3 is the sun. And for the planetary case were P 1 is the Sun, P 2 is theEarth and P 3 and planet, the following holds:

lunar case: 1 ; r 13 r 23 R (3.5)planetary case: 1 ;

m 3m 1 + m 2 1 ; r 13 R (3.6)

Both set of approximations give the same simplied results:

d2 Rdt 2

= GM RR 3

(3.7)

d2 r 12dt 2

= G(m 1 + m 2 )r 12

r 312(3.8)

The rst equation describes the motion of CoM of body P 1 and P 2 with respect to P 3 and the secondequation describes the relative motion of body P 1 with respect to P 2 . In these special cases the motionof the bodies may be approximated by a superposition of two two-body trajectories. (The earth moon

system circles around the sun, and the moon circles around the earth)

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3.2 Circular restricted three-body problem

By assuming that the mass of body P 1 and P 2 are much larger than the mass of P 3 and by assumingthat the two massive bodies move in circular orbits around the CoM of the system equation 3.1 reduces.When also a new (pseudo-)inertial reference frame is used this gives: (see also g 3.4)

d2

rdt 2 = G m1

r 1r 1 + G m 2r 2

r 2 (3.9)

By rewriting this equation for the movement of body P 3 in a rotating reference frame with the sameorigin and its X-axis alligned with the line r 12 , assuming that the velocity body P 3 in this referenceframe is 0 gives the equation for the surface of Hill . In this equation x and y are the coordinates in therotating reference frame and r 1 and r 2 are the distances between body P 1 and P 3 , and between P 2 andP 3 . is a mass ratio and C is an intergration constant which is determined by the position and velocityof body P 3 at time t =0.

x2 + y2 +2(1 )

r 1+

2r 2

= C (3.10)

=m 2

m 1 + m 2 (3.11)

By looking at cross section in the XY plane for z=0 simplies this equation to:

r 2 +2(1 )

r 1+

2r 2

= C (3.12)

If we look for solutions when C is very large there are three possibilities: r 2 is very large (so body P 3very far away from the CoM of the whole system), r 1 is very small (Body P 3 is very close to Body P 1 )and r 2 is very small (Body P 3 is very close to Body P 2 ). If C becomes smaller there will be more possiblesolutions, It is possible for body P 3 to be located further from P 1 and P 2 . So the areas will grow andthe places where to areas meet for the rst time a Lagrange point is located. (see g 3.6)

A Lagrange libration point is place where centripetal forces on body P 3 have the same magnitude andopposite direction as the combined gravitational force of body P 1 and P 2 of body P 3 . (Remember thewhole reference frame is rotating, Body P 1 and P 2 are still rotating around the CoM of the whole system,and with them body P 3 . Only the relative position of body P 3 does not change). The equilibrium inpoint L1 , L2 and L3 are not stable, point L4 and L5 are stable for certain values of and eccentricity.Sidenote, a lissajous orbit is a quasi-periodic orbital trajectory and does not have to be round or elliptical(think gure of eight and even more complex gures). However for realistic cases they can be consideredas slowly-changing elliptical paths. It can be shown that for the collinear libration points ( L1 , L2 andL3 ) the major axis is aligned with the Y axis (perpendicular to the line between P 1 and P 2 ). The orbitdirection is clockwise looking top down. For the equilateral libration points( L4 and L5 ) it can be shownthat major axis is rotated 30 degrees with respect to the X axis. Halo orbits are special lissajous orbitsaround collinear libration points which are very long elongated but have xed geometry and size and areclosed loop. (They are real orbits).

Halo orbits are very handy if you want the have a xed position with respect to the other body. TheEarth-Sun L1 point is multiple times used to study the sun. The Sun-Earth L2 point can be used if youalways want to be in the dark (cold), to study deep space. Furthermore the gravity gradient is very low(gravity eld is weak) so gravity distorsions are lower and higer pointing accuracy can be obtained (deepspace telescopes)

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Chapter 4

Relative motion in the many-bodyproblem

In real life there are no real two body problems. There are always other bodies like, the sun, other

planets and even galaxies which have an inuence on the trajectory. These forces are small compared tothe forces between the two main bodies and can be seen as disturbance forces. These forces are smallbecause the distance is very large or because the mass is relatively low. To derive a usefull equation westart again with

m id2 r idt 2

= Gj = i

m i m jr 3ij

r ij (4.1)

Lets assume that there is one main body k, a body i and a disturbing body j. First step is the writethe EoM of body i and k with respect to an inertial reference frame. By substrating the EoM of bodyk from body i, the equation of motion for body i in a non rotating (non inertial) with as center body kcan be found. (The EoM of body i with respect to body k)

m i r i = Gm i mk

r 3ikr ik + G

j = i = k

m i m jr ij

rij (4.2)

mk r k = Gmk m i

r 3kir ki + G

j = i = k

mk m jr kj

rkj (4.3)

r i = Gmkr 3ik

r ik + Gj = i = k

m jr ij

rij (4.4)

r k = Gm ir 3ki

r ki + Gj = i = k

m jr kj

rkj (4.5)

using the following vector rules

r ik = r ki ; r ki = r i r k ; r ij = r j r i = r kj r ki (4.6)It can be simplied to

r ki = r i r k = Gmkr 3ik

r ik + Gj = i = k

m jr ij

rij Gm ir 3ki

r ki Gj = i = k

m jr kj

rkj

= Gm i + mk

r 3kir ki + G

j = i = k

m jr kj r ki

r 3ij r kj

r kj(4.7)

The subscript k can now be dropped because everything is described with respect to k, and the above

mention reference frame is used.

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r ki = Gm i + mk

r 3ir i + G

j = i = k

m jr j r i

r 3ij r jr j

(4.8)

Assuming a satellite orbiting the earth the equation can be simplied because m s mE . The magni-

tude of the main force and the disturbing force can now be written as.

am = GmE r 2i

(4.9)

ad = Gm did

r 3id r jr j

(4.10)

= Gm d idr 3id r jr j idr 3id r jr j = Gm d idr 3id idr 3id 2 idr 3id dr 3d dr 3d dr 3d= Gm d r 2idr 6id + r

2d

r 6d 2r d r id cos

r 3d r3id

= Gm d 1r 4id + 1r 4d 2cos r 2d r 2id(4.11)

used geometry rules

a b = |a||b|cos ; a a = ||a||2 = |a2 | ; c2 = a2 + b2 2ab cos (4.12)Lets introduce = r i /r d and which is the angle between r i and r d ( is the angle between r d and r i d)

r 2id = r2d + r

2i 2r i r d cos (4.13)

cos =AS

=r d r i cos

rid

(4.14)

=r ir d

(4.15)

ad = Gm d 1(r 2d + r 2i 2r i r d cos )2 + 1r 4d 2(r d r i cos )r 2d r 3id= G

mdr 2d 1(1 + 2 2 cos )2 + 1 2(1 cos )r 5d (r 2d + r 2i + r d r i cos )3 / 2

= Gmdr 2d 1 + 1(1 2 cos + 2 )2

2(1 cos )(1 2 cos + 2 )3 / 2

(4.16)

THERE IS AN ERROR SOMEWHERE r 5d

SHOULD BE r 6d

In this equation k is the main body with mass mk m i . is the angle between the vector r kiand r kd and is the angle between the vector r kd and r di . (see g 4.2). With this equation it can beshown that bodies outside the solar system have a negligible inuence on the trajectories around thesun, and that the planets have a negligible inuence on satellites circling the earth.

When ying further away from a planet the sun will become more dominant in play of forces. Atsome point it can not be regarded any more as a disturbing force. The region were it can be regardedas an disturbing force is called the sphere of inuence of a planet. Beyond this sphere the planet willbecome a disturbing force. These sphere is are relatively small. An interplanetary trajectory can bemodeled in three separate parts, making the analysis easier.

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Chapter 5

Two-body problem

Remember these formulas:

r =p

(1 + e cos )(5.1)

p =H 2

(5.2)

H = r 2 = r 2 (5.3)

In the pericenter = 0 and in the apocenter = . By differentiating 5.5 en substitute 5.5 back into itsderivative and substituting 5.3 a simple function for r can be found

2a = r pericenter + r apocenter = r =0 + r = =p

(1 + e cos 0)+

p(1 + e cos )

=p

(1 + e)+

p(1 e)

=p + ep + p ep

12 + ep ep e2=

2 p1 e2

p = a(1

e2 ) (5.4)

r =p

(1 + e cos )=

a(1 e2 )(1 + e cos )

=H 2

(1 + e cos )(5.5)

r =drdt

=H 2

((1 + e cos ) 2 )(e sin ) =

r 2H 2

e sin = r 2

H 2e sin =

H

e sin (5.6)

Combining 5.5 and 5.3 gives:

r =H 2

(1 + e cos )=

r 2 H (1 + e cos )

(5.7)

r =

H

(1 + e cos ) (5.8)

One can also write the speed in a component which is perpendicular to the radius vector V n and acomponent which is perpendicular to the axis of symmetry of the conic section. V l . (see g 5.8)

V n = r +r

tan ( 2 )= r +

rtan ( 2 ( 2 ))

= r +r

tan ( )= r +

rtan

(5.9)

=H

(1 + e cos ) H

e sin cos sin

=H

+H

e cos H

e cos =H

(5.10)

V l =r

cos( 2 )=

rsin

=H

e sin 1

sin = e

H

(5.11)

Formula 5.6 and 5.8 can be used to make velocity hedographs. For elliptical (e < 1), parabolic (e=1) andhyperbolic (e > 1) trajectories different speeds are calculated. The pericenter is located at the top, where

the tangential velocity is the largest, the apogee at the bottom. See also g 5.7

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Ellipse parabola Hyperbola r r 2 r r 2 r r 2 0 0 H (1 + e) 0 2

H 0

H (1 + e)

2 e

H

H

H

H e

H

H

0 H (1 e) 0 0 0 H (1 e) < 03 2

e H

H

H

H

e H

H

The Theorem of Whittaker states that at the pericenter V n and V l have to same direction and cantherefore be added to nd the maximum speed. In the apocenter V n is directed in the the opposite way,and the minimum speed can be obtained by subtracting V n from V l .

V p = V l + V n = eH

+H

=H

(1 + e) (5.12)

V a = V l V n = eH

H

=H

(1 e) (5.13)

5.1 Relativistic effects

With the use of the auxiliary variable u =1/r and some rearranging equation 5.14 can be writtin as eq5.16

r r 2 =

r 2

(5.14)

u =1r

(5.15)

d2 ud2

+ u =

H 2(5.16)

According to general theory of relativity a good approximation for the relativistic motion of a body canbe given by:

d2 u

d2 + u =

H 2 + 3

c2 u

2 (5.17)

In these equation is the gravitation parameter of the large body, c is the speed of light, u = 1/r and His the classical angular momentum. Comparing the two equations it becomes clear that the most rightpart of the right hand side of the equations is due to relativity. The right hand side can be rewritten

H = r 2 = 1

u2(5.18)

H 2

+ 3c2

u2 =

H 21 + 3

H 2 u2

c2=

H 2

1 + 3

2

u2 c2=

H 2

1 + 3( r )2

c2=

H 2

1 + 3V 2c2

(5.19)

The relativistic part depends on the ration between the velocity of the body and the speed of light. Incelestical mechanics the speed of the body is much smaller than the speed of light ( V c). Thereforethe ration will be small and therefore the right term will be much smaller than the left term. A rstorder approximation is:

u =

H 2[1 + e cos()] +

2

H 41 +

12

e2 + e sin() 16

e2 cos 2() (5.20) = 3

c2

(5.21)

The relativistic part of this equation consist of three parts:

The constant part (1 + 1 / 2e2 ) which increase the value of u with the constant value 2 (1 +e2 / 2)/H 2 . Time has no inuence on this part. e sin is a uctuating term, of which the amplitude continuously increases with increasing

values of . Because is the total angle traveled, and this term will become larger in time.

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Because the body has a radial velocity component with respect to the sun, the frequency of the light isshifted because of the Doppler effect. The power density is linear with the frequency (a lower frequencymeans that the photons have less energy, so produce a smaller force) Therefore:

v = v(1

r

c) (5.32)

W = W (1 rc

) (5.33)

Furthermore because also light needs time to travel the distance between the sun and the object, theobject already has rotated and it appears that the light comes more from the front.

t =rc

(5.34)

= t =rc

(5.35)

A bit of geometry and because is very small (small angle approach)

sin = sin(2

+ ) = cos 1 ; cos = cos(2

+ ) = sin = (5.36)Substituting these answers back into 5.28 gives:

r r 2 =

r 2

+r 2

1 rc

(5.37)

ddt

(r 2 = rr 2

1 rc

r c

(5.38)

Rewriting and neglecting second order terms ( c2 )

r r 2 =

r 2 rcr 2

(5.39)

ddt

(r 2 =

c(5.40)

A couple of things can be noticed here. All particle will be slowed down because of equation 5.40. Dueto this drag force the object will spiral down in the direction of the sun. Equation 5.39 can be evenfurther simplied by neglected the second term.

r r 2 =

r 2

1

= r 2

(1 ). (5.41)For the right values of / ( > 1) particles will not spiral inwards but will be blown out of the solarsystem. When taking a closer look to denition of one can see that small, low density, high reectivityparticles will have larger forces applied to them. This is why our sun is bright and not hazy because alldust is blown away.

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Chapter 6

Parabolic and hyperbolic orbits

The motion of a satellite in an elliptical orbit can be described with:

r = a(1 e cos E ) ; tan2

=

1 + e1

e

tanE 2

; E e sin E = n(t ) = M (6.1)For a hyperbolic orbit:

r = a(1 e cosh F ) ; tan2

= 1 + e1 e tanhF 2

; e sinh F F = n(t ) = M (6.2)In these equations r is the radius, a is the semi-major axis, e is the eccentricity, is the true anomaly ,E is the eccentric anomaly , F is the hyperbolic anomaly , n is the mean angular motion , n is an angularvelocity, t is the time, is the time of (last) pericenter passage, M is the mean anomaly and M is alsoa kind of mean anomaly. see g 6.7 and 8.1

n = T 2 = a3 ; n = T 2 = a3 (6.3)To demonstrate how these equations work, the calculation of the orbital period T and the position ( and r) are calculated for a satellite orbiting the Earth in an orbit with e=0.05 and a perigee altitude of h p = 500 km after 60 minutes. Furthermore given is that = 398600.4 km 3 /s 2 and R = 6378.14 km.

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r =a(1 e2 )1 + e cos

(6.4)

a =r (1 + e cos )

1

e2

=(500 + 6378 .14)(1 + 0 .05cos0)

1 + 0 .052= 7204 .036908 Km (6.5)

T = 2 a3 = 2 72043398600.4 = 6085 s (6.6)n = a3 = 398600.472043 = 0 .0010325355 s 1 (6.7)t = 60 60 = 3600 < 6085 = 0 (6.8)

M = n(t ) = 0 .0010325355(3600 0) = 3 .717127 (6.9)E k =0 = M = 3 .68935344 (6.10)E k +1 = M + e sin E k (6.11)

E 1 = 3 .717127 + 0 .05sin3.717127 = 3 .689913439 (6.12)E 2 = 3 .717127 + 0 .05sin3.689913439 = 3 .691064876 (6.13)E 3 = 3 .717127 + 0 .05sin3.691064876 = 3 .691015761 (6.14)E 4 = 3 .717127 + 0 .05sin3.691015761 = 3 .691017855 (6.15)

r = a(1 e cos E ) = 7204 .036908(1 0.05cos3.691017855) = 7511 .225Km (6.16) = 2 arctan 1 + 0 .051 0.05 tan

3.6910178552

= 2.62 (6.17)

= + 2 .64

180 = 330 (6.18)

As can be seen in the above example, the eccentric anomaly is found with an iterative process (you vanalso plot two lines with your GRM and nd the intersection). For elliptical trajectories is possible to ndan analytical approximation by using a series expension. For the hyperbolic case this can not be donebecause these series do not converge because 1 > e > and because the function sinhF and M F arenon-periodic.Hyperbolic problems can be solved numerical or graphical. A graphical method is to plot the func-tion sinhF and the linear function ( F + M )/e . The intersection of the two is the solution. The linearfunction can be plotted by drawing a line through ( M ,0) and ( M + e),1), or use a GRM.

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