12/19/2015two body problem fundamentalsslide 1 fundamentals of the two-body problem of celestial...

04/21/23 Two Body Problem Fundamentals Slide 1

Fundamentals of theTwo-Body Problem ofCelestial Mechanics

John L. Junkins


Estimation of Earth’s Spin Axis From Star Motion

• All star circles theoretically have the same center.

• Least square circle fits (leaving radius and center as free parameters to be estimated) provide an estimate of the Earth’sspin vector direction …

• This vector is normal to a plane that is (to high precision) the instantaneous Equatorial plane.


Vernal Equinox

Equinoxes And Solstices 2006-2010 Source: U.S. Naval Observatory


Vernal Equinox Mar 20 2006 12:26 PM EST Mar 20 2009 7:44 AM EDT Summer Solstice Jun 21 2006 7:26 PM EDT Jun 21 2009 1:45 AM EDT Autumnal Equinox Sep 23 2006 12:03 AM EDT Sep 22 2009 5:18 PM EDT Winter Solstice Dec 21 2006 8:22 PM EST Dec 21 2009 12:47 PM EST

Vernal Equinox Mar 20 2007 8:07 PM EDT Mar 20 2010 1:32 PM EDT Summer Solstice Jun 21 2007 2:06 PM EDT Jun 21 2010 7:28 AM EDT Autumnal Equinox Sep 23 2007 5:51 AM EDT Sep 22 2010 11:09 PM EDT Winter Solstice Dec 22 2007 1:08 AM EST Dec 21 2010 6:38 PM EST

Vernal Equinox Mar 20 2008 1:48 AM EDT Summer Solstice Jun 20 2008 3:59 PM EDT Autumnal Equinox Sep 22 2008 7:44 AM EDT Winter Solstice Dec 21 2008 7:04 AM EST

http://aa.usno.navy.mil/data/docs/EarthSeasons.php


The Two Basic Initial Value Problems of Astrodynamics: Conceptually

1 1

2 2

33

ˆ ˆ1 0 0 0 0ˆ ˆ0 0 1 0 0 ,

ˆ ˆ0 0 0 0 1

c s c s

c s s c

s c s c

b n

b n

nb

1n2n

3n 1b

2b

3b

( ) cos( ), ( ) sin( ) c s

Orbital Motion Attitude Motion{ ( ), ( )} { ( ), ( )}o ot t t tr r r r& & {( , , , , , ) at time } {( , , , , , ) at time }ot t && &&& &

These two problems approximately decouple & except for drag computation and for the case of very large spacecraft, the orbit can usually be computed without considering attitude motion. Attitude torques, on the other hand, depend more strongly upon knowledge of the orbital motion.

r v&

rm

f

m


History of Celestial Mechanics: A Sketch

Pre – 1700s 1700s

Copernicus

Brahe

Kepler

Galileo

et al

dt

&

calculus differential equations

law of universal gravitationNewton

laws of dynamical motion

celestial mechanics

&

variational calculus PDEs

rigid body dynamicsEuler

fluid mechanics

celestial mechanics

&

probability theory

rigid body fluid dynGauss

systems of eqns

celestial mechanics


History of Celestial Mechanics: the 1800s

&

variational calculus

rigid body dynamicsJacobi

special fcts PDEs

celestial mechanics


variational mechanicsLagrange

generalized mechanics

celestial mechanics

&


canonical eqs of mechanicsHamilton

quaternions rotational dyn

celestial mechanics

&

special fcts PDEs

Laplace transformLaplace

potential theory

celestial mechanics


History of Celestial Mechanics:Late 1800s-early 1900s

&

vector analysis

matrix analysisGibbs

fluid mech thermodynamics

celestial mechanics

Laplace transforms

vector analysisHeaviside

differential equations

circuit analysis

quantum mechanics

general relativityEinstein

special relativity

modern physics

matrix analysis

differential equationsCayley

linear algebra

celestial mechanics

Since ~1900, it is harder to point to comparable “giants”, but rather we can point to theintegration of 1000s of smaller contributions leading to the state of the art. Many advanceshave been elsewhere: Electronics, computers, power, propulsion, guidance and control,communications, and lastly, integrated systems to realize modern aircraft, missiles, satellites, …


During the 1900s, and continuing …• Substantial Progress along Many Directions:

– Relativity Theory/Non-Newtonial Effects (Einstein, et al)– Integration of Astrodynamics and Control Theory– Numerical methods

• Estimation Theory & Algorithms, e.g., the Kalman Filter• Optimization Methods• Computational Linear Algebra (e.g., the SVD and QR algorithms)

– Orbit Transfer Methods/Optimization– Orbit Estimation/Navigation– Spacecraft Formations/Constellations

• Much of the progress has been associated with exploiting new sensing, computing, and propulsion systems, and generally to accommodate man-made spacecraft, including forces from man-made devices, as opposed to “dealing only with natural forces” & Earth-based measurements, prior to the space age.

• There have been many individuals contributing (un-named here for brevity), rather than the “double handful” of giant contributors who laid the foundations of Celestial Mechanics prior to 1900.


Keplers Laws2a

2b f

r 5

1

43

2

6

7

89 10 11

12

21

is the semi-major axis

is the semi-minor axis

is the orbit eccentricity

b a e

a

b

e

First Law Second Law Third LawThe possible paths are The radius vector r sweeps Keplers original law:Conic Sections (Newton) area at a constant rate, or • Ellipses (Kepler) equal areas in equal time. … actually (Newton):• Parabolas, or This is a geometrical • Hyperbolas interpretation of depending upon the Conservation of where P is the energy level Angular Momentum orbital period

2 3/ P a a constant

22 3

1 2

4

( )P a

G m m

(Newton)


Said Kepler on his discovery that the orbit of Mars is an ellipse:

"Why should I mince my words ? The truth of Nature, which I had rejected and chased away, returned through the back door, disguising itself to be accepted.... I thought and searched, until I went nearly mad, for why does the planet prefer to move in an elliptical orbit...."


Celestial Mechanics

m r&&F

Most of the “fun” arises

due to nonlinear terms contained in the

force model

( , ), , , , , ...t gravity atmospheric density attitude &F F r r

This vector eqn can be written as 3 scalar 2nd

order differential eqns.These eqns are nonlinear,but depending on the force model, may still beanalytically solvable.

A frustrating truth about orbital mechanics: The more we learn about the earth’s gravitational field, the more it hurts! (i.e., greater mathematical and computational effort is required to compute model toobtain gravitational acceleration with higher accuracy). Two pieces of good news are:

(i) Near-earth gravitational acceleration can be modeled with errors in the 7th or 8th figure and (ii) We are able to make efficient use of analytical insights to construct accurate solutions.

Newton’s 2nd Law:

( ) ( )Nd

dt g


CLASSICAL SOLUTIONOF THE

TWO-BODY PROBLEM


Newton’s Law ofUniversal Gravitation

12r1 212 21 2

12

Gm mf f

r

21f

1 212 12 12 213

12r

Gm mf

r f i r f

21f

1m

2m12

12r r

ri

Newton conjectured this force law to be consistent with Kepler’s laws,his calculus, differential equations, and to make the Earth-Moondynamics ( ) become consistent with Newton’s corrected version of Kepler’s Laws.

1 280 80earth moonm M M m


Geometry of Conic Sections: A Terse Review

p

p/e

q/e

r

majoraxis F

f

The directrix definitionof a conic section:

The most general conic sectionis the locus of all points whosedistance (r) from a fixed point (the occupied focus F) have aconstant ratio (the eccentricity e)to the perpendicular distance froma typical point on the curve to afixed line (the directrix).

q

r/e

directrix


Geometry of Elliptic Orbits

a

b

ae

a p

q

y

xE f

a(1+e)

p/e

r r/e

reference circle

orbit

directrix


Some Conic Geometrical Relationships

Universal Elliptic Orbit Special Case

2

2

(1 cos ), (1 ) 1 cos

cos (cos ), 1

1sin sin , tan

2 1 2

pr r a e E p a e

e f

x r f x a E e b a e

f e Ey r f y b E tan

e

This is a small but important subset of an incredible number ofelegant geometrical relationships. You will require skill in doinggeometry of conic sections. These skills themselves are rather“universal”, so the investment is worth the effort.


Integration of theTwo-Body Problem

• Fundamental Integrals– Angular Momentum– Eccentricity Integral– Energy Integral– Orientation of the Orbit Plane– Kepler’s Equation– …

• Classical Solution for Elliptic Orbits• Universal Solutions


Two Body Equations of Motion

Inertial EOM for m1: Inertial EOM for m2:

SinceThen the relative EOM for m2 relative to m1 is

1 2 1 21 1 1 2 2 23 3

, d d

Gm m Gm mm m

r r r r + f r r + f&& &&

12 2 1 2 1 r r = r r r = r r&& && &&

2 11 23

2 1

where ( ), d dd dG m m

r m m

f f

r r + a a&&This vector differential equation is the most important equationin Celestial Mechanics!

1r

r1 22

Gm m

r

2mr r

ri1 2

2

Gm m

r

O

N

2r1m


Conservation of Angular Momentum

{ 3

Define angular momentum/unit mass as:

Take inertial time derivative of as: d dzero r

r

h r r

h h r r + r r = r r + a r a

&&

&

& & & &&1444442444443

So, we conclude, if 0, then angular momentum is a constant vector.

It is immediately obvious that whose

ˆnormal is .

d

h

the motion occurs in an inertially fixed plane

h

a = h r r

h r r = i

&

&

2 2

Since the motion occurs in a plane, we can introduce

polar coordinates and use the radial and transverse components of , to obtain:

ˆ ˆ ˆ ˆ 2(r r hr r r r h r r

r r

h r r = i i i i

&

& & && & )

Thus Kepler's second law has been proven analytically and is simply a geometrical

property of conservation of angular momemtum.

ate area swept by r

rrr = i

1m

&r

r&

ˆ ˆrr r && &v r = i i

2 2 2 2

, :

,

r

Kinematics Notation note

v r and v r

v r r

& &&&&& &

r

r r =

Path of m2 relative to m1.

iri


The Eccentricity Vector Integral

3

3

Investigate the vector This vector obviously lies in the orbit plane!

making use of , and

, ( )

d

dt r

r

&

& && && &

&

r h

r h r h r = r h r r

r r r a b c a c b a b c

3

23

2

ˆ ˆ ˆ, ,

,

,

r rr r rr

r r rr

r rr

&& & & &

&&

& &

r r r r r r r i r i i

r r

r r

, so ... both sides are perfect differentials:

, and we can integrate to obtain the exact integral:

d

dt r

d d

dt dt r

&

r

rr h

ˆ = a constant = ... can be shown... =

And so we conclude that , .

eer r

is a constant vector directed toward perigee

& & r h = r + c c = r h r i

c


The Eccentricity Vector IntegralKepler’s 1st LawWe just established: = a constant

To gain more insight, investigate: cos( ) (Eqn #)

We can also take the dot product this way:

( ) , (

r

rc

r

r

&

&

&

c = r h r

r c = r,c

r c = r r h r

= r r h a

2

2

) ( )

( ) ,

, (Eqn ##)

/Equate (Eqn #) to (Eqn ##), solve for :

1 cos1 cos( )

the paths are conic s

From which we draw several insights:

r

h r

h pr r r

c e f

& &b c = c a b

= h r r h = r r

r c =

r,c

22

ections

, the true anomaly

ˆ = is toward perigeee

h p

f

e

&r r

r,c

c i c


Two Body Problem: Conservation of Energy2

3

3

3

2

Consider the kinetic energy/unit mass: 2

Take the time derivative: ,

From which: ( )

( )

Integrate the

relative T v

dT

dt r

dT

dt r

r rr

r

r

d

dt r

&&

&&& &&

&

&

&

r r =

r r r r

r r

=

=

=

2 2

2

perfect differential to obtain: constant (energy equation)

2Making use of /2, we have: , the "vis-viva" equation

2 1Evaluate the energy constant at perigee:

Tr

T v vr

vr a

4 2 2 22 2 2

2 2 2 2 2

22

(1 ) (1 )(1 )Evaluation of

(1 ) (1 )the energy const

2 2 2 (1 ) (1 ) 1at perigee:(1 ) (1 ) (1 )

p pp p p p

p p p

p

p

r h p a e er q a e v r

r r r a e a e

vv e e

r r a e a e a e a

&&


2 2 2 22 1 2 2 , or , or

Three obvious paths:

2 1 1. Given ( , ), compute

2. G

v v v vr a r r

r a vr a

The energy (vis - viva) equation is so simple, it is subtle :

-12 2

12

1 2 2iven ( , ), compute

1 1 3. Given ( , ), compute

2

Also notice the important special cases:

1. Circular orbit: (

v vr v a

a r r

vv a r

a

r a circular orbit spee

2 2

2 2

2 2

) =

2 2. Parabolic orbit: , so ( ) =

3. Hyperbolic orbit speed at infinity: gives

d vr a

a escape speed vr

r v va


Vary Launch Speed From Zero to Infinity: “What Happens?”

0v3

2

1

4

5

6

7

0r

0 0

0 00

0 00

00 0

1: 0, rectilinear ellipse ( 1, /2)

2 : 0 , ellipse, apogee launch (0 1, )

3 : , circular orbit ( 0, )

2 4 : , ellipse, perigee launch (0 1,

case v e a r

case v e a rr

case v e a rr

case v e ar r

0

00

00

0

)

2 5 : , parabola ( 1, , 0)

2 6 : , hyperbola (1 , 0, 0)

7 : , straight line hyperbola ( , )

r

case v e a vr

case v e a vr

case v e a


Two Body Solution as a Function of (h, e, i, , f )

ˆ(cos cos sin sin cos )

ˆ (sin cos cos sin cos ) "all" we need is (

ˆ (sin sin )

and

ˆ[cos (sin sin ) sin cos cos cos ]

[si

)

n (sin s

x

y

z

x

t

r i

r i

r i

e e ih

eh

r i

i

i

r v i&

2

2 2

ˆin ) cos cos cos cos ]

ˆ cos cos sin

where

/ +

1 cos

constant /

y

z

e i

e ih

hf r

e f

r h h r

i

i

& &

X

Y

Z

iz

ie

iy

ix

f

i

r


Projection of Orbital Unit Vectors Onto Inertial Axes: Orientation of the Orbit Plane

T

ˆ ˆ ˆ ˆ

ˆ ˆ ˆ ˆ

ˆ ˆ ˆ ˆ

where the direction cosine matrix is

e x x e

m y y m

h z z h

C C

i i i i

i i i i

i i i i

11 12 13

21 22 23

31 31 33

0 1 0 0 0

0 0 0 ,

0 0 1 0 0 0 1

C C C c s c s where

C C C C s c ci si s c c cos

C C C si ci s sin

=

The inverse relationships are

s

c c s ci s c s s ci c s si

s c c ci s s s c ci c c i

si s si c ci

-131 1333

32 23

= , = ,

Also, the same direction cosine matrix accomplishes the coordinate transformation:

0

-1 -1C Ctan tan i = cos C

C C

x

y

X X

Y Y

Z Z 0

T

x

C C y


Motion as a Function of Time: A Frontal Assault

2 2

2

Begin with conservation of angular mementum:

= =

= =

r r f h constant

df r h constant

dt

&&

0

2

23/ 2

0 23/ 2

= , = p,1 cos

1

!!

1

Is there a way to "duck" thi

f

f

p hdt r d f substitute h r

e f

d fdt

p e cos f

dt t not much fun

p e cos

Question : s non-standard elliptic integral?

Yes! We need to look at the analogous integral using eccentric

anomaly instead of true anomaly as the angle variable.

Answer :


Motion as a Function of Time: Kepler’s Equation

Go another route: Use as the angle variable

constant + y , + y

cos

E

x x

x ax y y x h

& & & &

& &

e m e m

h h

h r r r i i r i i

h i i

2

2

2 2 2 2 2

3/2

, 1 sin

sin , 1 cos

1 cos sin cos 1

1 cos Integrate this (much easier than for true a

E e y a e E

dE dEx a E y a e E

dt dth x y y x

dEh a e E E e E h p a e

dt

dt e E dEa

& &

& &

0

0

03/2

0 0 03/2

'

nomaly) equation

( sin ) This is "Kepler's Equation"...

( sin ) sin Re-arranged classical form of "Kepler's Equation"...

EE

M

Classical form of Kepler

t t E e Ea

M t t E e E E e Ea

@

@ 1444442444443

0 0 0 03/2

3/2

: "mean anomaly" where: sin

2 note that: 0 2 , "mean angular motion"

s Equation M M t t M n t ta

M E e EM n

P a


Mean vs. Eccentric Anomaly 0 0 sinM M n t t E e E

2

M

2E

00

0.0

0.2

0.4

0.6

0.8

1.0

ee


Classical Solution of the Two Body Problem

0 0, , , , , , , ,pt t a e i t t t r r r r& &K

& shape angles perigeesize orientation time of

“orbit elements”

X

Y

Z

iz

ie

iy

ix

f

i

y

a aexie

r&

r f


0 0 0 0, , , , , , , ,t a e i M r r& K

2 2 2 20 0 0 0 0 0

2 2 2 20 0 0 0 0 0

0 0 0 0 0

2 2 2 20 0 0

20

0

2

00

0

1 2

/

/

sin 1NOTE: ,

x x y y z z

x y z

x y z

r X Y Z

X Y Z

h h h

h h h h h h

a r

p h

c c cr

e

r a e E Er rE

ra

r r

r r

r r

rr

r r

& & && &

& K

& K

&& & &

x y z

h i i i

c h i i i

c

1

sin sin / tan

1 /1 cos cos 1 /

a e E e E aEa

r ar a e E e E r a

Transformation from Rectangular Coordinates to Orbit Elements

31 32 33

11 12 13

21 22 23

133

1 31

32

1 13

23

0 0 0 0 0 0 0 00

1 00

0

cos ,0

tan ,0 2

tan ,0 2

/tan

1 /

h C C C

e C C C

C C C

i C i

C

C

C

C

X X Y Y Z Z

aE

r a

r r & & &&

h x x z

e x y z

m h e x y z

i h / i i i

i c / i i i

i i i i i i

0

00 0 0 0 3/2

, 0 2

, /

p

E

MM E e sin E t t

a

ALSO

Program 1


Classical Solution for Position and Velocity

0 0, , , , , ' . ,a e i M t t Kepler s Eqn t t r r&

0 03/ 2

2

2

,

' . ( . ., ' )

sin

cos , 1 sin

1 sin , cos

1 cos

n M M n t ta

Solve Kepler s Eqn for E e g via Newton s Method

M E e E E

x a E e y a e E

a eax E y E

r rr a e E

X

Y

Z

& &

11 12 13 11 12 13

21 22 23 21 22 23

31 32 33 31 32 33

,

0 0

T TC C C x X C C C x

C C C y Y C C C y

C C C Z C C C

& && &&

Program 2


Lagrange/Gibbs F and G Solutions

0 0

0 0

-The motion is planar

- & lie in the orbit plane.

- Any other vector in the plane can be written as

a linear combination of &

t t

t t

&

&

r r

r r

0 0

0 0

t tIn particular

t t

t F G

t F G

&&&& &

r r r

r r r

3

3 3

Eqns of motion: , substitution of the above projection...

This immediately gives ,

So if we can compute the functions and , we have an elegant

analytica

r

F F G Gr rF G

r&&

&&&&

r

0 0

0 0

l solution of the two body problem.

We also note the initial conditions 1, 0

0, 1

F t F t

G t G t

&

&

0tr& 0tr

tr

tr&


F and G solutions of the 2-body problem

3

0 00 0

0 00 0

0 0

initial cond

Eq. of Motion: =

Since the motion is planar, we seek solution in the form

1

0

Sinc

r

t F GF t G t

t F GF t G t

t F G

&&

& &&&& &

&&&&&&& &

r r

r r r

r r r

r r r

0 0

3 3

0 0

e Eqs. (2) hold for arbitrary , , 2 1 yields

,

& must depend upon time as well as ,

how can we find these functions?

F F G Gr r

F G

r r&

&&&&

&K

L

r r

(2)

(1)

(3)


Power Series Solutions

0

0

0

00

1

2

03 2 30

3 30

0 03 4 3 3 4 30 0

4

4

As a "brute force" approach to solving for , consider

!

Note by using Eq. (1)

33

r

12

n n

nn t

t

t

t

t t dt

n d t

d

r dt r

rd r d

dt r dt r r

d r

dt

&&

&&&&& & &

&

r

rr r

rr r r

r rr r r r r

r

2 4

5 4 4 3 4

3

n

0 0

0

0

3 6+

substitute

n

t

t

r r d

r r r r dt

r

d

dt

&& &&&

&&M L

&

rr r r r

r r

rr r


-1 -1

0 0 0 0 0 0 0 0 0-1 -1

0 0 0

Notice the structure

, , , , , , , , ,

Use Eqs of motion to eliminate acceleration & related terms, the

n n n

n n n

t t t

d d r d rfct r r r t t fct r r t t

d t d t d t

&&&& &L L

rr r

0 0 0

0 0 030

00 0 04 3

0 0

1st 3 are easy

0 + 1

+ 0

3 +

Thereafter, the "fun" begins! Notice no recursion is immediately obvi

r

r

r r

& &

&& &

&&&& &

r r r

r r r

r r r

0 0 0

0

ous.

Question: What can be done to eliminate , , , , ? There are

several elegant answers -- see Battin 3.2.

n

n

t

d rr r r

d t& &&&&& L

(5)


3

32 3 2

2 3 3 4 3 5 3

One route introduce

You can verify that

(looks familar!), Also

Now, observe that

so

r r rr

rr r

d d r

d t r d t r r r r

r r =

r rr = r, r - r r r

& & &

&& &

&&& & &

3 3 34 2 2 2

4 6 5 5 4 3

3 3 34 22 2 2

4 6 5 6 5 3

5

5

5 3

5 3 used ,

Note: will show up in substi

dr r

d t r r r r r

dr

d t r r r r r r

d

d t

r r + r + r + r r

r r + r r = r

r

& & &&& & &

& &&& &

&&3

tute . . . .

r

&&

(6)

(7)


0 0 0

0

3

23

Following the above pattern, all derivatives will be functions of , , . . .

how do we evaluate ?

Notice

so

r

r

r

r r

r r + r r , r = r

r r

2

0 00

0 0

2

2

and finally

1 1 2 ,

So all derivatives can be "easily" ex

r

r r

r r r

, =

r r

0 0

th

pressed as functions of , , . . .

but still, we don't have convenient recursions for n term.

r

(8)


2

3 2 2 2

A much more elegant power series expansion results if we introduce

"LaGrange's Fundamental Invariants":

r

r r r r r

r r r r

4 2 2 33 2 3

2

3 2

Notice the derivatives of , , :

3 2 2 2

2 23

3 r

r rr r r r r r

r r r

r

r

r

r

r

r

r r r r r r r r r r

5

2

2

2

2 2 2

2 :

3 22

Bottom Lines

So , , are "closed" wrt differentiation... notice these expressions are

all polynomials... no fractions...

(10)


2

2

3

3

42 2

4

53 2

5

Battin shows that the successive derivatives of can easily be

established in terms of , , :

3

15 3 2 6

105 45 30

d

dt

d

dt

d

dt

d

dt

r

rr

rr r = r r

rr + r

rr

2 2 + 45 9 8

polynomial in , , polynomial in , ,

Polynomials are easy to differentiate, but what about recursions

for the coefficients??

n

n n n

d

dt

r

rr + r


3Notice the differential equation for & can be written as:

0, 0

Since this type shows up often -- Battin lets be a generic variable and

considers:

0

and seek

F Gr

F F G G

Q

Q Q

0 0

00

00

s power series solutions of the form

1 1

note... ,

, ,

Subst u

! !

it

n

nn

n

n

n n

n nn n

t t

n

d Q dQ

n dt nQ Q t t

d

t t

t

0te and equate like powers of to identify recursion ,s for . ., .,n

n n n nt t Q


n 2 0 n 1 n-1 n 0

n 1 0 n 1 n-1 n 0

n 1 n n 0 n 1 n-1 n 0

n 1 0 n n 1 n

To produce the recursions on page 113.

1 2

1 3

1 2

1 2

n n Q Q Q Q

n

n

n

1 n 1 n 0 0

0 0

1 1

2 0

start with, for example

1 0

0 1

1 :

2

F G

F G

F

(11)


0 3 0 0

2 21 4 0 0 0 0 0

32 0 5 0 0

In this way, the series coefficients for the Lagrange & functions found to be

11

25 1 1

0 8 8 12

1 7 3

2 8 8

F G

F F

F F

F F

20 0 0 0 0

0 3 0

1 4 0 0

2 22 5 0 0 0 0 0

1 etc.

4and

10

61

1 43 3 1

0 etc.8 40 15

The higher-order coefficients are considerably more com

G G

G G

G G

0 0

3 2 20 0 0 0 0 0 0 0

plex.

NOTE: , ,

For high order solutions, the recursions (3.10), eqn (11) above are more attractive carrying

thru algebra to obtain explicit , . The problem with pn n

r r r

F G

r

ower series is slow convergence ...


Well, even though we've been successful in establishing recursions

to generate as many terms as we wish -- the above is not satisfactory

as a basis for solving the 2-body problem. Power series don't co

0

0

nverge

fast enough unless ( - ) is a small fraction of an orbit period . . . we

need exact analytic expressions for & which we can compute

accurately, to arbitrary precision, for arbitrary ( - )

t t

F G

t t

0 0

0 0

0 0

0 0

. . . this can be done!

Consider:

Write the orbit plane components of

F G

F G

x Fx Gx

y Fy Gy

r r r

r r r

r

(12)

(13)


0 0

0 0

0 0

0 00 0 0 0

0 0 0 0 0 0

0 0

0 0

and solved for & as

1

notice , so

1

1

These don't look

x xx F

y yy G

F G

y xF x

y xG yx y y x

x y y x h p

F x y y xp

G x y y xp

r r

2

2

like much help, until you recognize & are functions of

from geometry

cos , sin

1

1 sin , cos

x y

ax a E e x E

r

a ey a e E y E

r

(14)

(15)

(16)

(17)

(18)


000

2

20 0

20 0

0 0 00

00

cos cos 1

Substitution of (18) into (17) gives

11 [ cos cos 1 sin sin ]

1

[ cos cos sin sin cos ]

1 1 cos a

rE E e Ea

a e aF a E e E a e E E

r ra e

aE E E E e E

r

aF E E

r

0

0

n exact equation for

ˆintroduce change in eccentric anomaly

ˆ 1 1 cos

A very nice result, when you see we can develop a similar equation for

ˆ and a "Kepler" Eqn. to determine from

F

E E E

aF E

r

G E t

0.t

(19)

(20)


0

0 02

2 20

2

32

0 0 0

sin

3

Substituting Eqn. (18) into Eqn. (17), I get

1 +

1

1 cos 1 sin 1 sin cos

1

cos sin sin cos sin sin

or

E E

G x y y xa e

a E e a e E a e E a E ea e

aE E E E e E E

aG

2

0

*

ˆsin sin sin

ˆOne way to get rid of * as a fct ( ) is to use Kepler's Eqn. next slide.

E e E E

E

(21)


32

0

0 032

0 0

0 0 0 032

ˆ

0 032

ˆ sin sin sin

Kepler's Eqn.

sin

sin

Subtract to obtain

sin sin

so

ˆ sin sin

sub

E

aG E e E E

M M t t E e Ea

M E e E

t t M M E E e E Ea

e E E t t Ea

3

2

0

stitute (22) (21) to obtain

ˆ ˆ sin

aG t t E E

(21)

(22)

(23)


32

0

0

32

& are obtained by (surprise!) differentiations of & of Eqns. (20), (23)

ˆ ˆ sin , 1 cos 1

where

ˆ

is obtained by differentiation of Kepler's Eqn.

F G F G

a aF E E G E E

r

d dE E E E

dt dt

Ma

/

00 0

11 cos

(25) (24) gives

ˆsin sin

ˆ1 1 cos

r a

e E E Ea r

a aF E E E

r r r r

aG E

r

(24)

(25)

(26)


0

0 0 0

0 0 0

0 0

0 0

ˆA "cute trick" for eliminating functions of in favor of :

ˆ cos cos cos cos sin sin

ˆ sin sin sin cos cos sin

or

ˆ cos sincos cos

ˆ sin cos sinsin

E E E E

E E E E E E E

E E E E E E E

E EE E

E E EE

0 0

0 0

from which we have the (probably familar) identies

ˆ ˆcos cos cos sin sin

ˆ ˆsin sin cos cos sin

We can use these to write all of our previously developed functio

E E E E E

E E E E E

ns

ˆof as functions of . . . for example, 1 cos . . .E E r a e E

(27)


0 0

0 0

0 0

0 01

(1 cos )

ˆ [1 cos cos sin sin ]

ˆ ˆ ˆ [1 cos cos sin ]

ˆ ˆ cos sin

Now, since

ˆ 1

ra a

r a e E

r a e E E e E E

rr a E E E

a a

r a r a E a E

dEE

dt a r

(28)

(29)

0 0ˆcos cos sin sinE E E E


0 03

2

03

2

We can separate variables as

ˆ

and substitute Eqn. (28) into (30) to obtain

ˆ ˆ ˆ 1 1 cos sin

which integrates to a "modified" Kepler's equation

ˆ ˆ 1

dt r dEa

rdt E E dE

a aa

M t t E

a

0 0

0

1

ˆ ˆsin cos 1

ˆ

ˆ ˆNotice Newton's method has the recursion (start with )

ˆ ˆˆ ˆ ,

ˆ

ˆ

i

i i

i i

rE E

a a

f E

E M

M f E df rE E

adEdf

dE

(30)

(31)

(32)

(33)


SUMMARY: F&G Solution of the Elliptic 2-Body Problem (Singularity – Free!)

0 0 0 0

2 2 2 2 20 0 0 0 0 0 0 02 2 2 2 20 0 0 0 0 0 0 0

0 0 0

,

,

,

F t G t F t G t

r t t X Y Z r r

v t t X Y Z v v

t t

r r r r r r

Equations in Order of Solution

r r

r r

r r

0 003

0 0 0 0 0 0 0

20

0

0 0

3 2

0

2

0

0

+

1 2,

ˆ

ˆ ˆcos sin

ˆ ˆ ˆ1 1 cos , sin ,

ˆ ˆsin , 1 1 cos

ˆ ˆ ˆ1 sin 1 cos

,

0

rt t E E E

X X Y Y Z Z

aa r

E

r a r a E a E r

a aF E G t t E E F G

r

a aF E G E F G

r r r

X

Y

a a a

Z

0 0 0 0

0 0 0 0

0 0 0 0

, ,

X t X t X X t X t

F Y t G Y t Y F Y t G Y t t t

Z t Z t Z Z t Z t

r r

Newton’s Method

Program 4


UNIVERSAL SOLUTIONOF THE

TWO-BODY PROBLEM


Elliptic and Hyperbolic AnomaliesELLIPTIC ORBITS HYPERBOLIC ORBITS

b sinE

C F

Y

X

ae

r

a cosE

ref. circle(e = 0)

{

{referencehyperbola(e = )

- b sinhH r

2

-a coshH

Y

XF C

- ae

{

2 2

2 2

cos

sin

cos sin 1

gives the equation of an ellipse

1

X a E

Y b E

E E

X Y

a b

2 2

2 2

cosh , 0

sinh

cosh sinh 1

gives the equation of a hyperbola

1

X a H a

Y b H

H H

X Y

a b

0 1, 0e a

E

a

1 , 0e a

2 1b a e 21b a e


LaGrange Coefficients as Functions of the Change in Eccentric & Hyperbolic Anomalies

ELLIPTIC ORBITS (a>0) HYPERBOLIC ORBITS (a <0)

0

00

0

0 0

00 03

2

0

0

ˆ1 1 cos

ˆ ˆ1 cos sin

ˆsin

ˆ1 1 cos

ˆ ˆcos sin

ˆ ˆ1 cos

ˆ 1 sin

ˆ ˆ,

aF E

r

a aG E r E

aF E

r r

aG E

r

r a r a E a E

t t E Ea

ar

Ea

dtE E E a d E

r

2

0

00

0

0 0

00 03

2

0

0

ˆ1 1 cosh 4

ˆ ˆ1 cosh sinh

ˆsinh

ˆ1 1 cosh

ˆ ˆcosh sinh

ˆ ˆcosh 1

ˆ 1 sinh

ˆ ˆ,

aF H b ac

r

a aG H r H

aF H

r r

aG H

r

r a r a H a H

t t H Haa

rH

a

dtH H H a d H

r


KEY TO UNIVERSAL INTEGRATION:

0

0

Introduce new "time variable"

Notice can be interpreted as

Elliptic Case

Hyperbolic Case

We will find some new transcendental function

dt rdt rd

d

a E E

a H H

s of which are

"universal" -- they work for both elliptic and hyperbolic cases.


ENERGY INTEGRAL & MORE

2

2

2

" derivatives"

take the first " derivative" of :

2 2 2 ,

r

vr

r

r

dr d d dt dt rr

d d dt d d

drr r

d

dr

d

r r

r rr r

r r

1

"we know"

elliptic motion

= parabolic motion

hyperbolic motion

a

so


2

2

3

" derivatives", continued from previous pages

, ,

one more derivative of

+

dt r dr

d d

r

dt r

dd r d d dt

d d dt d

r

r

r r

r r

r r r rr r

22

2 2

2 2

2

2

or

1 1

r

r r

r

r r

r

r

d r d rr r

d d

nice linear ode. for r


2 1

2 1

1

1

2

2

3

3

2

2

" derivatives", continued

1 1

we found

1

notice

also

n n

n n

n n

n n

dt r d t dr d t d r

d d d d d

dr d d r

d d d

d rr

d

d r dr

d d

d

d

4 2

4 2

linear ode for

linear ode for d t d t

td d


3

3

2

2

4 2

4 2

22 2

2 2

,

linear odes for , , as functions of ,

how about ?

dt r dr

d d

d r dr

d d

d rr t

d

d t d t

d d

d d dt d r

d dt d dt

d d r d

d dt dt

r, r

r r r

r r r

2

2 3 ,

1

d

dt r

d

r dt

r r

rr

3

3

you verify

linear ode for d d

d d

r rr


2

2

2 3

2 3

3 2 4 2

3 2 4 2

In summary, then

1

dr d t

d d

d r d d tr

d d d

d r d d t dr d t

d d d d d

2 3 4 2

2 3 4 2

2

2

so that , , and are solutions of the equations

0 0 0

The derivatives of the position vector r

1 1

t

r t

d d r dr d t d t

d d d d d

d r d d

d d r r d

r r rv v r r

3

3

hese lead to

0d d

d d

r r

NOTE: Using as indep. var. leads to regular,

linear differential equations, w/o approximation

(4.73)

(4.74)


2

2

0

To construct the family of special functions, we begin by determining the

power series solution of

0

by substituting

=

and equating coefficients of like powers of

kk

k

d

d

a

2

2 22 22 2

0 1

0 1

. We are led to

for 0, 1, 1 2

as a recursion formula for the coefficients. Hence

1 12! 4! 3! 5!

where and are tw

k ka a kk k

a a

a a

0 1

o arbitrary constants. We shall designate the two

series expansions by ; and ; so thatU U

;nU The Universal Functions

*

alternatively, I can simply

expand * in a Taylor series

and get this equation

0 0 1 1 0 0 1 = ; ; , , .d

a U a U a ad


The Universal Functions1 0

1 0 2 1 3 20 0 0

th

The function is simply the integral of so that we are motivated

to define a sequence of functions

etc.

The function of such a sequence is

U U

U U d U U d U U d

n

222

01 1

2

1

1

easily seen to be

1 ;

! 2 ! 4 !

Also, , except

other identities:

(can see this directly from above series)!

nn

nn

n

n n

mn

m

Un n n

dU dUU U

d d

U Un

d U

d

1

10 0,1,

mn

m

d Un m

d


2

2

222

0

0 0 1 1222

1

1 10

0

1

;

;

0

[1 ]

2! 4!

[1 ]3! 5!

1, 2, , 1, 2,

xn

n n n

U

U

d

d

aa U a

a

dUU U d n U n

d

U

222

222

2

2

1

! 2 ! 4 !

1 [ ]

! 2 ! 4 ! 6 !

! important identity

nn

nn

n

nn n

nU

Un n n

Un n n n

U U n

Universal Functions

We found

Generally

This gives

Re-arrange

So

(4.75)

(4.76)


2 21

0 2

1 3

1

1 0 2 1 1

Special cases of (4.76)

(a) 1

(b)

Derive a few other identities -- multiply (a) by

Integrate to establish

d U d Ud U

U U

U U

U d

U U d U U d U d

2 21 2 2

21 2 2 2 2 0

2 21 2 0 2 2 2 1 0 2

20 2 0 1 0 2

2

or

2 , from (a) 1

2 (c)

(c) (a) gives

1

U U U

U U U U U U

U U U U U U U U U

U U U U U U

20 1 0 2

continued...

U U U U


20 2 0 1 0 2

20 1 0 2

2 20 1 0 0

2 20 1

2 0 1

1

1

1

From which we conclude

Generalization of the more fam

1

U U

U U U U U U

U U U U

U U U U

U U

2 2

2 2

ilar

1 cos sin

1 cosh sinh


Identities for Universal Functions

0

2

2

1 0

; cos 0

cosh 0

02

1 cos; 0

cosh 1 0

U

U

1

3

3

0

sin; 0

sinh 0

06

sin; 0

sinh

U

U

0

2 20 1

0 0 0 1 1

1 1 0 0 1

2 2 2 20 0 1 0 1 1 0 1

1

2 2 1 1 2 ; 2 2

U U

U U U U U

U U U U U

U U U U U U U U

(4.89)

(4.88)

(4.87)


4.7 Continued Fractions for Universal Functions

12

02

2

0 1 2

/ 2;

/ 2; / 2

/ 2

/ 2

/ 2 "universal half-tangent"

1

3

57 ...

Now, using some of the basic identities for the universal functions, we can express , , and as

Uu

U

U U U

20 0

1 0 12

2 1

0 1 2

2 2

0 1 22 2 2

/ 2

/ 2 / 2

/ 2

2 1

2

2

Hence, , , and are determined from

1 2 2 ; ; ;

1 1 1As a consequence, values of the first three U-f

U U

U U U

U U

U U U

u u uU a U U

u u u

2

2 2

unctions can be calculated using only

a single continued-fraction evaluation. Analogous to familar trig identities such as

1 tan 2 tan2 2 cos , sin , ...

1+ tan 1+ tan2 2

x x

x xx x

(4.100)

(4.101)

(4.102)

We will subsequentlyderive this expression…


Summarize Key Results Developed Previously

2

2

3

3

1

d d dt d r

d dt d dt

d d

d r dt

d d

d d

r r r

r rr +

r r

dt r

dt r dd

1

1

2

2

3 2

3 2

4

4

1

1

n n

n n

dr d t d r

d d d

d rr

d

d r dr d

d d d

d t

d

3

3

d t

d

NOTE: “regularizes” the differential Eqs: gets rid of divisions by r singularities

makes all eqns. linear w.r.t. derivatives


3

3

2 2 2 2 22

2 2 2 2 2

3

3 2

3

Verification of

( ) ( )

1 ( ) Note: nonlinear fct.

1 1

rr

d d

d d

d d dt d r d d r

d dt d dt d dt

d d dt d d t d d d

d dt d dt d d r dt d

d d

d r d

r

r r

r r r r r

r r r r r rr +

r rr

2

2

2 2

2

3

( )

1 1 1

r dr r d

dd d dt d d

r d dt d dt

d dr

r r d r r d

r

rr

r r r

r rr r

1 d

r d

rr

2r

r

1 d

r d

r

3

3 Linear eqn. for ,

d

d

dd d

d d d

r

rr rr

2

21

d rr

d


2 3

2 3 3

22 2 2

2 2 2

2

2

Let's look at another (& more attractive?) solution path

Let's verify that

Notice

,

so

d F d F dFF

dt r d d

d d dt

d dt d

d d dt d d t

d dt d dt d

d F F

d r

2

3

2 3

2 3

or

nonlinear fct, what about ?

continued ...

r dF

dt

d F F dF d F

d r dt d


2

2

3 2

3 2 2

3

3 2 2

3

1

one more derivative

1

11

rF rr

d F F dF

d r dt

dd F dF F dr d F dt dF dd r d r d dt d dt

rd F dF F F dF

d r d r r dt

3

3

1

dF dF d dF

dt d dt d r

d F dF

d r d

2

F

r

2

F

r

1

dF

r dt

3 3

3 3

similarily

dF

d

d F dF d G dG

d d d d

LinearEquations!


3

3

0 0 1 1 2 2

0 0 1 2 0 0

0 0 1 1 2

Let's look a little at how to carry out solutions in terms of functions

0

at 0

1 0 0

U

d r dr

d d

c U c U c U

r c c c c r

r r U c c

r

U

2

0 1 1 0 2 1

1 0

0 0 2 0 1

take derivative

at 0

so

U

drr U c U c U

d

c

drU c r U

d

2

0

2

1

22

2

222

1 2!

13!

1

3! 4!

1;

! 2 ! 4 !

n

n

U

U

U

Un n n


2

0 1 2 0 02

0 2 0 2

0 0 0 1 2

0 0 0 1

one more derivative

1

evaluate at = 0

1 1

conclude

= 1

c

d rr U c r U

d

r c r c

r r U U U

drU r U

d

an make direct use of

to obtain the Universal Kepler's Eqn.

dt r d

Generalization of:

1 cosr a e E


0 0 0 1 2

0 0 0 1 2

0 0 1

2 31

,

integrates immediately to

UNIVERSAL KEPLER'S EQUATION:

dU dUdU

dt r d r r U U U

dt r U d U d U d

g t t r U

0 2 3 0

Notice Newton's Method:

U U

dgr

d

1

ii i

i

g

r


Solution for the F & G Functions

3

3

d F dF

d d

2

200 0

You verify I.C. are

0 1

10 ,

F

dF d F

d d r

0 0 1 1 2 2

0

1

0

2

220 00

at 0 0 1 1

at 0 0 0

1 1at 0

F c U c U c U

F c

dFc

d

d Fc

d r r

0 2 0 2 20 01

1 1F U U U U U

r r

20

11F U

r

so


20

0 0 31 2 0

10

2

Do the similar solution for & verify

1 1 ,

,

Battin's (4.84)

,

1 1

G

F Ur

r UG U U t t

F Ur r

G Ur


Summary of Universal Solution2 2

0 0 0 0 0 0 0 0 0

20

0

1 , ,

2

r v

v

r

r r r r r r

0 0 1 0 2 3

Note: ; , compute using Euler's "top down method"i i

t t r U U U

U U

0 0 1 2

0 0 32 1 2 0

0

1 2

0

0

0 0 0 0

11 ,

1, 1

,

r r U U

r UF U G U U t t

r

F U G Urr r

F G F G

U

r r r r r r

Solutionsfor stateAt time t:

Solve theUniversalKepler Equations

Constants:

0,1,2,3i

Program 5

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