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Transcript of 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
04/21/23 Two Body Problem Fundamentals Slide 2
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.
04/21/23 Two Body Problem Fundamentals Slide 3
Vernal Equinox
Equinoxes And Solstices 2006-2010 Source: U.S. Naval Observatory
04/21/23 Two Body Problem Fundamentals Slide 4
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
04/21/23 Two Body Problem Fundamentals Slide 5
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
04/21/23 Two Body Problem Fundamentals Slide 6
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
04/21/23 Two Body Problem Fundamentals Slide 7
History of Celestial Mechanics: the 1800s
&
variational calculus
rigid body dynamicsJacobi
special fcts PDEs
celestial mechanics
variational calculus
variational mechanicsLagrange
generalized mechanics
celestial mechanics
&
variational calculus
canonical eqs of mechanicsHamilton
quaternions rotational dyn
celestial mechanics
&
special fcts PDEs
Laplace transformLaplace
potential theory
celestial mechanics
04/21/23 Two Body Problem Fundamentals Slide 8
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, …
04/21/23 Two Body Problem Fundamentals Slide 9
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.
04/21/23 Two Body Problem Fundamentals Slide 10
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)
04/21/23 Two Body Problem Fundamentals Slide 11
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...."
04/21/23 Two Body Problem Fundamentals Slide 12
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
04/21/23 Two Body Problem Fundamentals Slide 13
CLASSICAL SOLUTIONOF THE
TWO-BODY PROBLEM
04/21/23 Two Body Problem Fundamentals Slide 14
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
04/21/23 Two Body Problem Fundamentals Slide 15
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
04/21/23 Two Body Problem Fundamentals Slide 16
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
04/21/23 Two Body Problem Fundamentals Slide 17
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.
04/21/23 Two Body Problem Fundamentals Slide 18
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
04/21/23 Two Body Problem Fundamentals Slide 19
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
04/21/23 Two Body Problem Fundamentals Slide 20
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
04/21/23 Two Body Problem Fundamentals Slide 21
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
04/21/23 Two Body Problem Fundamentals Slide 22
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
04/21/23 Two Body Problem Fundamentals Slide 23
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
&&
04/21/23 Two Body Problem Fundamentals Slide 24
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
04/21/23 Two Body Problem Fundamentals Slide 25
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
04/21/23 Two Body Problem Fundamentals Slide 26
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
04/21/23 Two Body Problem Fundamentals Slide 27
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
04/21/23 Two Body Problem Fundamentals Slide 28
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 :
04/21/23 Two Body Problem Fundamentals Slide 29
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
04/21/23 Two Body Problem Fundamentals Slide 30
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
04/21/23 Two Body Problem Fundamentals Slide 31
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
04/21/23 Two Body Problem Fundamentals Slide 32
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
04/21/23 Two Body Problem Fundamentals Slide 33
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
04/21/23 Two Body Problem Fundamentals Slide 34
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&
04/21/23 Two Body Problem Fundamentals Slide 35
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)
04/21/23 Two Body Problem Fundamentals Slide 36
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
04/21/23 Two Body Problem Fundamentals Slide 37
-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)
04/21/23 Two Body Problem Fundamentals Slide 38
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)
04/21/23 Two Body Problem Fundamentals Slide 39
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)
04/21/23 Two Body Problem Fundamentals Slide 40
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)
04/21/23 Two Body Problem Fundamentals Slide 41
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
04/21/23 Two Body Problem Fundamentals Slide 42
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
04/21/23 Two Body Problem Fundamentals Slide 43
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)
04/21/23 Two Body Problem Fundamentals Slide 44
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 ...
04/21/23 Two Body Problem Fundamentals Slide 45
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)
04/21/23 Two Body Problem Fundamentals Slide 46
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)
04/21/23 Two Body Problem Fundamentals Slide 47
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)
04/21/23 Two Body Problem Fundamentals Slide 48
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)
04/21/23 Two Body Problem Fundamentals Slide 49
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)
04/21/23 Two Body Problem Fundamentals Slide 50
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)
04/21/23 Two Body Problem Fundamentals Slide 51
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)
04/21/23 Two Body Problem Fundamentals Slide 52
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
04/21/23 Two Body Problem Fundamentals Slide 53
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)
04/21/23 Two Body Problem Fundamentals Slide 54
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
04/21/23 Two Body Problem Fundamentals Slide 55
UNIVERSAL SOLUTIONOF THE
TWO-BODY PROBLEM
04/21/23 Two Body Problem Fundamentals Slide 56
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
04/21/23 Two Body Problem Fundamentals Slide 57
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
04/21/23 Two Body Problem Fundamentals Slide 58
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.
04/21/23 Two Body Problem Fundamentals Slide 59
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
04/21/23 Two Body Problem Fundamentals Slide 60
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
04/21/23 Two Body Problem Fundamentals Slide 61
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
04/21/23 Two Body Problem Fundamentals Slide 62
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
04/21/23 Two Body Problem Fundamentals Slide 63
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)
04/21/23 Two Body Problem Fundamentals Slide 64
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
04/21/23 Two Body Problem Fundamentals Slide 65
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
04/21/23 Two Body Problem Fundamentals Slide 66
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)
04/21/23 Two Body Problem Fundamentals Slide 67
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
04/21/23 Two Body Problem Fundamentals Slide 68
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
04/21/23 Two Body Problem Fundamentals Slide 69
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)
04/21/23 Two Body Problem Fundamentals Slide 70
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…
04/21/23 Two Body Problem Fundamentals Slide 71
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
04/21/23 Two Body Problem Fundamentals Slide 72
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
04/21/23 Two Body Problem Fundamentals Slide 73
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
04/21/23 Two Body Problem Fundamentals Slide 74
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!
04/21/23 Two Body Problem Fundamentals Slide 75
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
04/21/23 Two Body Problem Fundamentals Slide 76
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
04/21/23 Two Body Problem Fundamentals Slide 77
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
04/21/23 Two Body Problem Fundamentals Slide 78
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
04/21/23 Two Body Problem Fundamentals Slide 79
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
04/21/23 Two Body Problem Fundamentals Slide 80
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