keplerian trajectories

1

KEPLERIAN TRAJECTORIES

SOLO HERMELIN

2

SOLO

Table of Contents

2. Newton’s Laws

3. Newton’s Law of Universal Gravitation4. Two Body Problem

4.1 Polar Representation4.2 Conservation of Angular Momentum


1. Kepler’s Laws

4.3 Conservation of Mechanical Energy

4.4 Integration of Equations of Motion

4.5 Velocity on the Trajectory

4.6 Specific Mechanic Energy

4.7 Eccentricity Vector

4.8 Orbit Determination from Initial Conditions

4.9 Flat Earth Approximation4.10 Orbital Trajectories

4.11 Time of Flight on an Elliptic Orbit

5. References

3

SOLO

Kepler’s Laws


Johannes Kepler1571 - 1630

Tycho Brache1546 - 1601

From 1601 to 1606 Kepler tried to fit various geometrical curves to Tycho Brache’sdata on Mars orbit.

4

SOLO

Kepler’s Laws


1. The orbit of each planet is an ellipse, with the sun at a focus.

2. The line joining the planet to the sun sweeps out equal area in equal times.

3. The square of the period of a planet orbit is equal to the cube of its mean distance to the sun.

b

a

a dAh

dt2

b

a 2/322a

GMGM

abTP

b

5

SOLO

Newton’s Laws of Motion


1. Every body continues in its state of rest or of uniform motion instraight line unless it is compelled to change that state by forcesimpressed upon it.

2. The rate of change of momentum of a given body is proportional to the force impressed on the body and is in the same direction as the force.

3. To every action there is always an equal reaction

“THE MATHEMATICAL PRINCIPLES OF NATURAL PHILOSOPHY”1687

Isaac Newton1643-1727

6

SOLO

Newton’s Law of Universal Gravitation


“THE MATHEMATICAL PRINCIPLES OF NATURAL PHILOSOPHY”1687

Isaac Newton1643-1727

Any two body attract one on other with a force proportional to theproduct of the masses and inversely proportional to the square of the distance between them.

EQPOISSON

G

Gr

GM

r

GMg

gmr

GMmr

r

mMGF

4&&

1

2

2

228 /1067.6 gmcmdyneG The Universal Gravitational Constant

Instantaneous propagation of the force along the direction between the masses (“Action at a Distance”)

GF

GF

M m

7

SOLO

Two Body Problem


Assumptions

1. The bodies are spherical symmetric.

2. The gravitational forces are the only forces acting on the bodies.

Mr

mr

r

M

m

Cr

Let

Mr

-the position vector of the mass M relative to an inertial point.

mr

-the position vector of the mass m relative to an inertial point.

Mm rrr

-the relative position vector of the mass m relative to M.

r

rr

:1 -unit vector in direction. r

Let apply the Newton’s Law of Universal Gravitation to describe the forces each mass applies on the other.

rr

mGr

r

r

r

MmGrM

rr

MGr

r

r

r

MmGrm

MM

mm

1

1

22

22

8

SOLO

Two Body Problem (continue – 1)


rr

mGr

r

r

r

MmGrM

rr

MGr

r

r

r

MmGrm

MM

mm

1

1

22

22

Mm

rMrmr Mm

C

The Center of Mass of those masses is

MmGrr

rr

MmGrrr Mm

:1122

From those equations we obtain:

and

0

Mm

rMrmr Mm

C The Center of Mass is not accelerating

9

SOLO



Polar Representation

Define

r

- Angular Velocity Vector

nrR 11

Decompose in and perpendicular to direction. r

r

r

M

m

r1

n1

R

t1

Since we can define a right handCartesian system using vector product:

011

rn

rnt 11:1

tnnrnnntd

d

nrtnrttttd

d

trnrrrrtd

d

RR

RR

R

1111111

11111111

1111111

Differentiating we obtain:

10

SOLO



Polar Representation (continue – 1)

or

n

t

r

n

t

r

td

d

R

R

1

1

1

00

0

00

1

1

1

rrr 1

Therefore

trrrrrrrr 1111

nrtrrrrr

nrrtrrtrrrr

R

R

1121

11111

2

rrr 1

vtrrrr

11

rr

nrtrrrrrr R 111212

2

r

M

m

r1

n1

R

t1

11

SOLO



Conservation of Angular Momentum

Vector Solution

Cross-multiply the equation of motion by

rr

r 12

r

03

rrr

rr

Define the specific angular momentum as: h

vrrrh

:

Differentiate h

0

rrrr

td

hd

Therefore is a constant vector in space that is perpendicular to thetrajectory vectors and , therefore those vectors (and the trajectory) remainin the plane defined by the initial conditions

vrh

r

v

000 tvtrth

12

SOLO



Conservation of Angular Momentum (continue – 1)

Polar Coordinate Solution

nrtrrrrrvrh 1111 2

The equation of motion in polar coordinates is

rr

nrtrrrrrr R 111212

2

from which

02 rr andr

rdd2

Integrated both sides

ConstrhConstrConstr 222 lnlnlnlnln

1

000 rr RR 2

constntn

td

dR 1011

also

constnrvrh 12 therefore

13

SOLO



Conservation of Mechanical Energy

Vector Solution

Dot-multiply the equation of motion by

rr

r 12

r

rrr

rr

3

Use vrvr &

03

rrr

vv

Use also the identity aaaaaaaaaaaa

011

111

022

022

3

r

v

td

d

rtd

dv

td

drr

rvv

const

r

vE

energypotentialspecific

energykineticspecific

2:

2

We can see that

E is the specific mechanical energy.

14

SOLO



Integration of Equations of Motion

Vector Solution

Cross-multiply the equation of motion with the specific angular momentum

rr

r 12

h

hrr

hr

3

rrtd

drhrh

td

drhrhr

td

d

The left side is

hrrr

rrhrrrrrrrhr

0

3

0

The right side can be written

r

r

td

dr

r

rv

rrrrrv

rvrrrrv

rvrr

rhr

r

2

2

3333

Equaling both sides gives

r

r

td

dhr

td

d

Integrating both sides Br

rhr

where is the constant of integration B

15

SOLO



Integration of Equations of Motion (continue – 1)

Vector Solution (continue – 1)

Dot-multiply this equation by :

Br

rhr

r Br

r

rrhrr

Use 2hhhhrrhrr

rr

rr

pBrBr cos

to obtain pBrrh cos2 or

pp e

p

B

hr

cos1cos/1

/2

where/: 2hp

/: Be

p

parameter (semi-latus rectum)eccentricityperiapsis angle

16

SOLO




Polar Coordinate Solution

The equations of motion in polar coordinates are:

22

rrr

Consthr 2

Let define a new variable: r

u1

:

from which urrrr

u 22

1

If , we can write 00 h

hr 2 and 0

h

d

udhu

hr

Differentiating again we obtain

02

222

22

2

2

2

hd

uduh

r

h

d

udh

d

udhr

We also have 32

4

22 uh

r

hrr

The first equation of motion becomes 02322

222

huuh

d

uduh

17

SOLO




Polar Coordinate Solution (continue – 1)

Therefore

32

4

22 uh

r

hrr

The solution of this equation is

02322

222

huuh

d

uduh

02

32

2

hh

ud

ud

221 cos

1

hCC

ru

where C1 and C2 are constant of integration

pCp

eC

hp 212

,:,:Define

to obtain pe

pr

cos1 This is the equation of a conic section in polar coordinates.

18

ConeApex C

ConeAxis

Generators

BaseCircle

SOLO

A right circular cone is a cone obtained by generators (straight lines) passing througha circle, and the apex C that is situated on the normal to the circle plane and passingtrough the center of the circle. β is the angle between the cone axis and the generators.

CONIC SECTIONS

CuttingPlane

generating a"hyperbola"

RightCircular

Cone

ConeApex

ConicalSection

C

ConeAxis

CuttingPlane

generating a"parabola"

CuttingPlane

generating a"ellipse"

CuttingPlane

generating a"circle"

CuttingPlane

generatingtwo

"lines"

2

2

2

0

lines

line

po

22

12

int2

P

F

F*

CuttingPlane

(Hyperbola)

RightCircular

Cone

Hyperbola2

Branches

C

Ellipse

Parabola

CuttingPlane

(Ellipse)

CuttingPlane

(Circle)

CuttingPlane

(Parabola)

By cutting the right circular conic by a plane we obtain different conic sections, as afunction of the inclination angle α of the plane relative to the base of the conic sectionand the angle β between the generators and the base.

The discovery of theConical Sections isattributed to the greekMenachmus who livedaround 350 B.C..

19

RightCircular

Cone

ConeApex

ConicalSection

C

ConeAxis

CuttingPlane


0

RightCircular

Cone

C

CuttingPlane

(Circle)

2

SOLO

The conical sections are:

CONIC SECTIONS

RightCircular

Cone

ConeApex

ConicalSection

C

ConeAxis

CuttingPlane


2

C

Ellipse

CuttingPlane

(Ellipse)

1. Circle if the cutting plane is normal to the cone axis (α=0) and is above or bellow the apex.

2. Ellipse if the cutting plane is inclined to the basis at an angle that falls short of the angle between generators to the base (α<π/2-β) (in greek word elleipsis means falls, short or leaves out.

3. Hyperbola if the cutting plane is inclined to the basis at an angle that exceeds of the angle between generators to the base (α>π/2-β)(in greek word hyperbole means excess.

4. Parabola if the cutting plane is parallel to a generator of the right circular cone (α=π/2-β) (in greek word parabole is the origin of the words parabola and parallel.

5. A point- apex (α<π/2-β), one straight line (α=π/2-β), two straight lines (α>π/2-β), if the cutting plane passes through the apex and intersects the cone basis.

CuttingPlane


RightCircular

Cone

ConeApex

ConicalSection

C

ConeAxis

2

P

F

F*

CuttingPlane

(Hyperbola)

RightCircular

Cone

Hyperbola2

Branches

C

RightCircular

Cone

ConeApex

ConicalSection

C

ConeAxis

CuttingPlane


2

C

Parabola

CuttingPlane

(Parabola)

RightCircular

Cone

ConeApex

ConicalSection

C

ConeAxis

CuttingPlane

generatingtwo

"lines"

lines

line

po

22

12

int2

C

CuttingPlane

generatingtwo

"lines"

CuttingPlane


RightCircular

Cone

ConeApex

ConicalSection

C

ConeAxis

CuttingPlane


CuttingPlane


CuttingPlane


CuttingPlane

generatingtwo

"lines"

2

2

2

0

lines

line

po

22

12

int2

P

F

F*

CuttingPlane

(Hyperbola)

RightCircular

Cone

Hyperbola2

Branches

C

Ellipse

Parabola

CuttingPlane

(Ellipse)

CuttingPlane

(Circle)

CuttingPlane

(Parabola)

20

SOLO



Integration of Equations of Motion (continue – 6) pe

pr

cos1

Conic Section (continue – 2)

pp

r d

directrixfocus conicsection

x

y

P1

Q1

r1

t1

periapsis

Let definee

pd :

and rewrite pe

edr

cos1 as follows prd

re

cos

From which we obtain the following definition of conic sections:

Conic Section A conic section is a planar curve, such that the ratio of distances, for any point on the curve, to a fixed point F* (focus) and to aline directrix is constant and equal to e.

21

SOLO





Let write the conic section equation in cartesian coordinates using

QP 1,1

QPt

QPr

pp

pp

1cos1sin1

1sin1cos1

and p

p

ry

rx

sin

cos

pe

pr

cos1Substitute those in the equation rewritten as

prer p cos

xepyx 22to obtain

Squaring both sides gives 22222 2 xexpepyx

or 2222 21 pyxpexe

22

SOLO





For e ≠ 1 can we can write

2222 21 pyxpexe

2

2

2

2222

2

22

1111

e

p

e

eppy

e

epxe

We see that for p ≠ 0 ( h ≠ 0 )

1 0e222 pyx circle

2 10 e 1

11

1

2

2

2

2

2

e

py

e

pe

epx

ellipse

3 1epe

ypx

2

22 parabola

4 1e hyperbola1

11

1

2

2

2

2

2

e

py

e

pe

epx

F*

directrix2

directrix1

F

P

x

yx=cx=-c

e

ax

e

ax

Hyperbola

a

Circle

y

x

FF*

P M

x

yr

0

a

b

directrix2directrix

1

e

ax

e

ax

x=-c x=c

Ellipse

r

Parabola

px

y

directrix

M

F

23

SOLO



Velocity on the Trajectory

Let find the velocity on the trajectory in and coordinates

tr 1,1

QP 1,1

QyPxtrrrrv 1111

QrrPrr pppp 1cossin1sincos

From the equation of motion

pe

pr

cos1 hr 2 /: 2hp

we obtain

p

p

p

pr

ep

p

eh

r

h

e

ep

td

d

d

drrv

sin

sin

cos1

sin22

pt epr

p

r

hrrv cos1

2

24

SOLO



Velocity on the Trajectory (continue – 1)

Let substitute those results in the velocity equation to obtain the componentsin coordinates.

QP 1,1

Qep

Pp

QyPxv pp 1cos1sin11

2/12222/1222/122 cos1sin pptr eep

yxvvv

2/12 cos21 peep

v

pp

r d

directrixfocus conic

section

x

y

P1

Q1

r1

t1

v

rv tv

periapsis

The velocities at the periapsisand apoapsis are

p

p

ep

va 1 e

pv p 1

25

SOLO



Specific Mechanic Energy

The specific mechanic energy is

ap

e

p

e

p

e

r

v

r

vE

a

a

22

11

2

1

22:

2222

or

ar

vE

22:

2

where we used 21 eap

From the last equation we get a

pe 1

Substitute and to obtain/: 2hp Ea 2/

2

221

hE

e

26

SOLO



Eccentricity Vector

We defined the eccentricity as a scalar /: Be

Let define the eccentricity vector as /: Be

where is defined as B

Br

rhr

Therefore

Pevvr

rvr

r

rvvrvr

r

rvrv

r

rhve 1

22

The last equation is obtained from the definition of , and , and canbe checked by substituting the range and velocity vectors in coordinates,in previous equation.

B

e p

QP 1,1

27

SOLO



Orbit Determination from Initial Conditions

Given the initial position and velocity vectors , find the orbital parameters 00 ,vr

pr

d


section

x

y

P1

Q1

r1

t1

vrv

tv p

1 From the specific angular momentumof the orbit we can find 00 vrh

01 00

hh

vrR

2hp

2 From the specific mechanical energy on an elliptic orbit

ar

vvE

22 0

00

we obtain

000

2 vvr

a

28

SOLO



Orbit Determination from Initial Conditions (continue – 1)

pr

d


section

x

y

P1

Q1

r1

t1

vrv

tv p

3 The eccentrity vector of a Keplerian trajectoryis given by

from which

Pevvrr

rvve 1

10000

000

ee

01

ee

eP

PRQ 111

pe

pr

cos1

The position and velocity vectors are given by

QPrr pp 1sin1cos

QePp

QyPxv pp 1cos1sin11

29

SOLO




pr

d


section

x

y

P1

Q1

r1

t1

vrv

tv p

4 Initial trajectory angle , on the conicsection, from the periapsis is given by

p 01

1

111

0

01

0

01

cos

sintan

11sin

1cos

Rr

rP

r

rP

One other way to find is by using

RQP 1,1,1

QePvp

QPr

1cos1sin

1sin1cos1

110

110

or

01011

01011

cos1sin1cos1

sin1cos1cos1

vp

rQe

vp

rePe

1

1

sin

cos

e

1

1

cos

sin

30

SOLO




pr

d


section

x

y

P1

Q1

r1

t1

vrv

tv p

Using 1

0 cos1

e

pr

we finally obtain

010

01

010

01

cossin

1

sincos

1

vp

rr

pQ

vp

rr

p

eP

010

01

010

01

0110

011

sincos11

sincos1cos1

sincoscossinsinsincoscos

1sin1cos

vp

rrr

p

r

vp

rrre

p

r

vp

rrre

p

r

QPrr

Therefore

31

SOLO




pr

d


section

x

y

P1

Q1

r1

t1

vrv

tv p

where we used

0110

011

0110

011

coscossinsinsin

coscossinsinsincoscossin

1cos1sin

vep

rr

p

ee

p

vep

rr

p

ee

p

QePp

v

010

01

0

1

0

00

010

01111

0110

0111111

cos11sin1

cos1

cos11cos1sincos1sin

cos1cos1

sinsincoscossinsin

vp

rr

pre

pr

vr

vp

rr

p

eee

p

vep

r

rp

ee

p

10

1111000

sin

1cos1sin1sin1cos

ep

r

QePp

QPrvr

32

SOLO




pr

d


section

x

y

P1

Q1

r1

t1

vrv

tv p

Summarize

010

01 sincos11 vp

rrr

p

rr

010

010

10

00 cos11sin1

cos1 vp

rr

pre

pr

vrv

33

SOLO



Orbit Determination from Initial Conditions (Second Method)

Start from the solution for Keplerian orbit 221 cos

1

hCC

r

where C1 and C2 must be determined from initial conditions

0r - initial range from the center of mass M

0v - initial velocity magnitude

000002 cosvrvrrvrh

0 - initial trajectory angle (between velocity vector and the horizon plane)

Force Center

FiringPoint

0r

0v

pr

0

p

r

0

Trajectory

Periapsis

34

SOLO



Orbit Determination from Initial Conditions (Second Method - 1)

We have 0

20

21

022

02

0

21 cos

1cos

coscos

1

rCC

vrCC

r

where0

20

/:

r

v

Let differentiate 21212

212sinsinsin

1CChCCrrCC

r

r

rdt

d

or 21000 sincos CCvrr

Let use now the initial conditions to find C1, C2.

0

20

2010 cos

1cos

1

rCC

r 002010000 sinsincos vCCvrr

or

02

00201 cos

11cos

rrCC

00

201 tan1

sin r

CC

35

SOLO




We have

0

200

201 cos

11cos

rrCC

00

201 tan1

sin r

CC

0

20

000

00

200

02

002010201

02

02001

cos

1sintan

1cos

cos

11

cos

1sinsincoscos

cos

1cos

1

rrrr

rCCCC

rCC

r

0

00

02

0

0

0000

02

00

cos

cos

cos

cos1

cos

sinsincoscos

cos

cos1

r

r

which, when developed further, gives

36

SOLO




0

00

0

2

00

cos

cos

cos

cos1

r

r

Finally

0000

02

0

coscoscos1

cos

rr

37

SOLO




0000

02

0

coscoscos1

cos

rr

20

20

20

22

002

0002

02

02

022

002

000

0000

1coscossin

cos1cossincossin1coscossin1

cos1cossincossin1

coscoscos1

The denominator can be expressed as

02

0 cos: rp

1cos21cos2cos1coscossin: 02

02

0222

02

02

022 e

We define

1cos

cossintan:

02

0010

p

pp ee

cos1cos1

coscoscos1

00

0000 to obtain

38

SOLO




0000

02

0

coscoscos1

cos

rr

The denominator can be expressed as

pp ee cos1cos1coscoscos1 000000

We obtain again the fact that the trajectory is a conic section given by

pe

pr

cos1

We can also write

21cos21

cos

1: 0

020

20

2

rr

e

pa

and

02

0 cos: rp 1cos2: 0

2 e

39

SOLO



Flat Earth Approximation

We assume that , so we can use10

20000 2

11cos&sin

20

0200

0

0002

0022

0

02

0

0000

02

0

cos

11

2

1tan1

cossin2

11cos

2

1cos

coscoscos1

cos

r

r

rr

and

where gr

v

rr

v

r

v

0

20

20

0

20

0

20

The order of magnitude of λ is 64

000,1

10000,400,6122

0

0

v

gr

40

SOLO



Flat Earth approximation (continue – 1)

Therefore if γ is not close to 90° , we can assume that , and then1cos

1

02

0

220

20

200000

022

0

020000

022

0

02000

0

cos2

1tan

cos2

1tan1

cos2

1tan1

v

grrr

v

grr

v

grr

r

Use

to obtain

0000 && rxzRrzRr ee

022

0

200

cos2

1tan

v

gxxzz

41

SOLO



Flat Earth approximation (continue – 2)

We obtained the Flat Earth approximation of the trajectory, that is a parabola.

022

0

200

cos2

1tan

v

gxxzz

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5 Vacuum Ballistic Trajectories

g

V

2

20

g

V

2

20

022

0

200 cos2

V

gxtgxzz

g

V 20

0V

00z

42

SOLO



Orbital Trajectories

Let discuss in more detail the Keplerian trajectories.

Ellipse (e < 1) ( Kepler’s First Law ) and Circle ( e = 0 )

x

y

21 eap

eac

a a

2/121 eab

eara 1 earp 1

rFOCUS

EMPTYFOCUS

Apoapsis RadiusPeriapsis Radius

ApoapsisPeriapsisGeometric Center

DirectrixDirectrix

c

epd /Semimajor

axis

Semiminor axis

v

nv

rv

P1

Q1

1

11

1

2

2

2

2

2

e

py

e

pe

epx

We found that the ellipse equation in cartesian coordinates is given by:

43

SOLO



Orbital Trajectories (continue – 1) Ellipse (e < 1) ( Kepler’s First Law ) and Circle ( e = 0 )

x

y

21 eap

eac

a a

2/121 eab

eara 1 earp 1

rFOCUS

EMPTYFOCUS

Apoapsis RadiusPeriapsis Radius

ApoapsisPeriapsisGeometric Center

DirectrixDirectrix

c

epd /Semimajor

axis

Semiminor axis

v

nv

rv

P1

Q1Define:

21:

e

pa

semimajor axis

21:

e

pb

- semiminor axis

eae

epc

21: - distance of geometrical

center from focus

e

pd : - distance of directrix

from focus

eara 1: - apoapsis radius

earp 1: - periapsis radius

- true anomaly

- flight-path angle

2/pa rra

papa rrrre /

ererabeap ap 11/1 22

From those relations we get:

122

b

y

a

cx

and the ellipse equation will be:

44

SOLO



Time of Flight on an Elliptic Orbit

From the equation 2rh we can write h

Ad

h

drdt 2

2

where is the area defined by the radius vector as it moves through an angle

2

2

drAd

d

pp

r

focus conicsection

xy

P1

Q1

r1

t1

v

rv tv

d

drAd 2

2

1

periapsis

This proves the 2nd Kepler’s Law that equal area are swept out equal in equal timesby the radius vector.

45

SOLO



Time of Flight on an Elliptic Orbit (continue – 1)

The period of the orbit depends only on the major axis of the ellipse a.

p

pa

h

eaa

h

ea

h

baT

eap

ph

2/3122/322

21

21

222

or 2/32 aT

The period of an elliptical orbit T is obtained by integrating from Θ= 0 to Θ=2π , and the radius vector sweeps the area of the ellipse A = π a b.

This proves the Kepler’s third law: “the square of the period of a planet orbit is equalTo the cube of its mean distance to the sun”.

46

SOLO




Let draw an auxiliary circle of radius a, and the same center O as the geometric centerof the ellipse.

x

y

eac

a a

2/121 eab

r

FOCUS

EMPTYFOCUS

c

P1

Q1

a

F

Q

O VS

E

P

Let take any point P on the ellipse withpolar coordinates r,Θ and define the point Q on the circle at the same coordinate x as P.

Eeary

ra

xea

a

xby

Eaa

xay

ellipse

ellipse

circle

sin1sin

sin111

sin12

2

22

2

2

2

2

The angle E of OQ with x axis is called theeccentric anomaly.

aeEarxellipse coscos

47

SOLO




Let compute

x

y

eac

a a

2/121 eab

r

FOCUS

EMPTYFOCUS

c

P1

Q1

a

F

Q

O VS

E

P

0cos11

sin1sincos1cos

1

2

22

2

EEEeea

EEeaEaEEeaaeEa

xyyxvreah ellipseellipseellipseellpse

We obtain n

aEEe :cos1

3

pttntEetE sin

Integrating this equation gives

Kepler’s Equation

where tp is the time of periapsis ( E (tp) = 0 )

48

SOLO




From

x

y

eac

a a

2/121 eab

r

FOCUS

EMPTYFOCUS

c

P1

Q1

a

F

Q

O VS

E

P

Eeary

aeEarx

ellipse

ellipse

sin1sin

coscos

2

we have Eea

EeEeaEeaaeEar

cos1

coscos21sin1cos2/1222/12222

Therefore

cos1

sin1sin

cos1

sin1sin

cos1

coscos

cos1

coscos

22

e

eE

Ee

Ee

e

eE

Ee

eE

Ee

Ee

Ee

eEEe

sin1

cos11

sin1

coscos1

sin

cos1

2tan

22

From

2tan

1

1

2tan

E

e

e

or

and are always in the same quadrant.2

2

E

49

SOLO

References

2. Battin, R.H., “An Introduction to the Mathematics and Methods of Astrodynamics”, AIAA Education Series, AIAA, Washington. DV., 1987


3. Battin, R.H., “Astronautical Guidance”, McGraw Hill, 1964

1. Bate, R.R., Mueller, D.D., White, J.E., “Fundamentals of Astrodynamics”, Dover Publications, New York, 1971

4. Chobotov, V.A., Editor, “Orbital Mechanics”, AIAA Education Series, , AIAA,2nd Edition, 1996

5. Reagan, F.J., “Re-Entry Vehicle Dynamics”, AIAA Education Series, , AIAA,1984

6. Reagan, F.J. and Anandakrishnan, S.M., “Dynamics of Atmospheric Re-Entry”, AIAA Education Series, , AIAA, 1993

7. Montenbruck, O. and Eberhard, G., “Satellite Orbits”, Springer-Verlag, 2000

8. Kaplan, M.H., “Modern Spacecraft Dynamics and Control”, John Wiley & Sons,1976

April 18, 2023 50

SOLO

TechnionIsraeli Institute of Technology

1964 – 1968 BSc EE1968 – 1971 MSc EE

Israeli Air Force1970 – 1974

RAFAELIsraeli Armament Development Authority

1974 – 2013

Stanford University1983 – 1986 PhD AA

keplerian trajectories

Science