Problem: Pluto + Charon orbiting the sun
This three body system is interesting because the orbital plane of PlutoCharon is nearly perpendicular to the orbital plane of the P-C center of mass
(COM) with the sun. Here to make the calculations simpler, we will assume
that the two orbital planes are exactly orthogonal.
Kinetic Energy
First, we will easily take care of the kinetic energy for this three body system.
We start by writing the kinetic energy as the sum of the kinetic energies of the
sun, Pluto, and Charon in some inertial frame external to the three bodies.
Then, following the same procedure used in Ch. 3, the kinetic energy of Pluto
+ Charon can be rewritten as the sum of the P-C COM kinetic energy and the
P-C kinetic energy relative to its local COM:
 =  + 
+ 


(1)
We then note that the sun and the P-C COM form a separate two-body system
whose kinetic energy can similarly be decomposed into a COM piece and a
relative piece. The contribution for the three-body COM can be neglected
because it is simply an additive constant (no net force acts on this COM), and
the physically relevant kinetic energy is, thus, just the sum of the two relative
kinetic energies, which we can write using plane polar coordinates for the two
separate two body systems. We use  and Θ for the sun-PC COM coordinate
system and  and  for the PC system. Here, as shown in the figure,  is
the radial separation of the sun and the PC COM and  is the radial distance
between P and C,  = |r02 − r01 |. Also in the figure,  is the mass of the sun,
1 is Pluto’s mass, and 2 is Charon’s mass.
.
m1
r1/
R1
.
M
R
r2/
.
R2
m2
Fig. 1. Position vectors and masses for the Pluto-Charon-sun system.
The kinetic energy is thus given as the sum,
·2
·2
M ·2
 ·2
( + 2 Θ ) + ( + 2  )
2
2
where M is the reduced mass of the sun-PC COM system,
 =
M=


 +
1
(2)
(3)
with  = 1 + 2 , and  is the reduced mass of the PC system,
=
1 2

1 + 2
(4)
Potential energy
The potential energy requires more work to put it into a useable form. We
start by writing the exact gravitational potential as
 =−
 1  2 1 2
−
−

1
2

(5)
where 1 and 2 are defined in the figure. Now, we note from the figure, the
following vector identity
R = R + r0 
(6)
from which it follows that
 =
q
2 + 02 
(7)
after noting that R · r0 = 0 because of the assumed orthogonality of the two
orbital planes. Next, we take advantage of the huge disparity in size between
 and 0 to expand −1 to first order, thereby obtaining the excellent approximation,
−1 = −1 (1 − 02 (22 ))
(8)
After substituting Eq.(8) into Eq.(5), the first two terms of that equation, which
we will temporarily refer to as  , reduce to
 = −
   (1 102 + 2 202 )

+

23
(9)
Next, we know from our analysis of the two body problem that
 0 = 
(10)
   2

+

23
(11)
which allows us to simplify  as
 = −
and the total potential then becomes
 =−
1 2
   2
−
+


23

(12)
It is interesting to note that the central term in this approximate form is repulsive for the  motion, but is attractive (Hooke’s law potential, ∼ 2 ) for
the relative PC motion. This will have important consequences for the orbital
motion of the PC system.
2
(I) EOM for the sun-PC COM system
Let us first analyze the relative motion of the sun-PC COM system. Skipping
the details, the equation of motion for the one-dimensional radial problem is easy
to find as
··

2
M =   () = −
(13)
+ M3 

M
·
where M is the constant angular momentum, M = M2 Θ. The perturbative
simplification of  is seen to give rise to a 1-3 potential is defined as
 () = −
1
2 2

+

23
(14)
where 1 and 2 are positive constants,
1 =  
(15)
2 =  
(16)
and
With 2 = 0, we recover the unperturbed Kepler problem. Given how small
 is, the repulsive −3 term is a very small perturbation to the pure Kepler
potential. Because  is so small, any variations in  for typical PC orbits will
also be small compared to , and we may safely regard  as a constant in this
part of the problem.
The radial equation of motion can be written explicitly as
··
M =   () = −
1

2
 0
+ 4 + M3 = −  
2


M

(17)
where  is just another constant,
 = 32 2 2
(18)
0
and 
is the effective potential for this problem,
0
=  () +

2
M

2M2
(19)
(Ia) Circular Orbits: The circular orbit condition is that the effective
0
( ) ] = 0, where  is
force must vanish when  =  ,   ( ) = −[
the radius of the circular orbit,
  ( ) = −
2
1

M
+
+
= 0
2 4 M3
(20)
This equation can be simplified and solved as follows. First, introduce a scaled
radial coordinate
 =  0 
(21)
3
where 0 is the radius of the circular orbit when  = 0,
2
M1 
0 = M
(22)
−3 − −2 + −4 = 0
(23)
Then Eq.(20) reduces to
where  is the positive, dimensionless constant
 = (1 02 )
(24)
Then, after multiplying Eq.(23) by 4 , we obtain a quadratic equation for the
variable  ,
2 −  −  = 0
(25)
with a solution given by the quadratic formula
√
2 = 1 + 1 + 4
(26)
Because  ≥ 0, a positive root is possible only for by choosing the positve
sign for the square root.
(Ib) Period of Circular Motion: From conservation of angular momentum for the circular orbit, we have
M = M2   
(27)
where   is the frequency of circular motion. The definition of   is   = 2  ,
from which, with the help of Eqs.(20) and (27), we find
1
1
  = 2M2 M = 2M2 (M1  − M ) 2 =  0 [5 (2 − )] 2  (28)
where  0 is the period of the circular orbit with  = 0
1
 0 = 2(M1 ) 2 (0 )32 
(29)
Two other forms for the answer are found by using Eq.(25)) to replace 2 − 
with  to get
  =  0 2 =  0 ( + )
(30)
The latter form follows from the use of Eq. (25) to replace 2 with  +
.
(Ic) Period of Small Oscillations: The oscillation frequency  of the
slightly perturbed orbit is defined as
 2 = −(1M)(  ())|= 
(31)
¸
¸
∙ 2
∙
3M
1
21
4

1 1
− 3 + 5 =
+

 =
M M4


M 3 5
(32)
It follows that
2
4
2
after substituting for M
from the circular orbit condition, Eq.(20). For  , we
have
∙
¸ 12
5
2

(33)
=
= 0

2 + 
using Eqs.(32), (21), (24), and (29) . For stable circular orbits,  2  0, and this
is manifestly true here. Taking the ratio of Eqs.(33) and (28) yields
∙ 2
¸1
 −  2

=


2 + 
(34)
and using Eq.(25) to replace 2 we also have
∙
¸ 12


=


 + 2
(35)
(Id) Precession of the stable perturbed orbits: The form of Eq.(35)
implies that     for all stable circular orbits. Because of this, the precession
is in the opposite direction as the orbital motion.
(Ie)Plot of the effective radial potential for V(R) = V
0
As defined by Eqs.(14), (18), and (19), 
is the effective potential for this
problem,
1
2

0
() = − +
+ M 2 
(36)

3

3
2M
To discuss the qualitative nature of the motion, we need to make a plot of
0
M
versus . It will be convenient to introduce a scaled radial coordinate
 = 0 , where 0 is the radius of the circular orbit when  = 0,
2
0 = M
M1 
This allows us to write
0
1
1
M

=
−

e () =
+
2
1 0
2

3 3
(37)
where  is the positive constant  = 1 02 We saw earlier that all positive
values of  were allowed.
To illustrate these possibilities, the dimensionless potential of Eq.(37) is
shown for  = 1800. Note that even this tiny value is probably unrealistically
large for the PC-sun system. The -axis corresponds to e ().
5
y
2
1
0
1
2
3
x
-1
Fig. 2. Effective potential for the sun-PC COM system with  =1/800.
p
There is an attractive well with a minimum at  = (1 + 210200)2 ≈
10012, where a stable circular orbit is possible. When   0 , only unbound
scattering orbits are possible. When   0, bound, but generally unclosed
orbits are possible in the shallow potential well.
(II) EOM for the perturbed PC system
Now we analyze the relative motion for the PC system. Notice that the
relevant part of the potential for this problem consists of the second and third
terms of Eq.(12), so we are treating a Keplerian potential with a Hooke’s law
perturbation.
(IIa) Again skipping the details, the equation of motion for the one-dimensional
radial problem is

2
··
 =   () = −
(38)
+ 3 


with the potential defined as
 () = −
3

+ 2 

2
(39)
where 3 is the gravitational force constant,  is the Hooke’s law force constant,
·
 is the constant angular momentum, 2 , and  is the reduced mass of the
two particle system. For the PC-sun system, 3 and  are defined as
3 = 1 2 
(40)
and
 

(41)
3
Although  varies inversely with 3 , we may regard  as a constant for this
part of problem because on the timescale of motion for the PC pair,  will be
changing very slowly. The radial equation can be written explicitly as
=
··
 =   () = −
3
2
0
−

+
=
−

2
3

6
(42)
where 0 is the effective potential for this problem,
0 =  () +
2

22
(43)
To make a sketch of 0 versus  it will be convenient to introduce a scaled
radial coordinate  = 0 , where 0 is the radius of the circular orbit when
=0
2
0 =

(44)

This allows us to write
2
 0 0
1
1
e () =  = 2 +
− 

2
2

where  is the small positive constant
=
(45)
03

3
(46)
To illustrate these possibilities, the dimensionless potential of Eq.(45) is shown
below for three values of , 0, 0.1, and 0.01. The -axis corresponds to e ().
All curves are nearly the same at small values of , but differences emerge at
larger values of . The bottom curve is for the pure Kepler problem with  = 0.
The curve that crosses the  axis at  ≈ 77 corresponds to  = 0002. The
curve will rise monotonically as  increases, so the effect of the perturbative
term is to make all orbits bound. The curve that crosses the  axis at  ≈ 25
corresponds to  = 01. This is probably too big a value of  to be considered
a small perturbation, but you can see how much the potential well is narrowed
compared to the original effective Kepler potential. Even the value  = 0002
is probably far too large for the PC system.
y
0.4
0.2
0.0
1
2
3
4
5
6
7
8
9
10
x
-0.2
-0.4
Fig. 3. Effective potential for the PC system with  = 0, 0.002, and 0.1
(b)To find the period for a circular orbit with nonzero values of , we first
consider the condition that the effective force should vanish,
  ( ) = −

2
−

+
=0 

2
3
7
(47)
where  is the radius of the circular orbit. It follows that
2
=  + 4 

(48)
The period follows from the conservation of angular momentum for the circular
orbit
·

(49)
 =  = 2 

so that with Eq(48) we have
 2 =
1
( + 3 )
3
(50)
s
(51)
The period is
2
= 2
 =

3

 + 3
(c) To find the period  of a slightly noncircular orbit, we substitute
 =  +  
(52)
into the EOM, Eq.(38) or (42), and expand the effective force in a Taylor series
in the small parameter . The result is
∙  ¸
 ()
··
 =   ( ) +
 + ··· 
(53)

=
where, of course,   ( ) = 0 and
∙  ¸
 ()
2
32

= 3 −  − 4 = −( 3 + 4)







=
(54)
after substituting for 2 using Eq.(48). To first order terms in  we can write
Eq.(53) as
··
 = − 2  
(55)
where the orbital frequency  can be put in the form
2 =
¢
1 ¡
3
 + 43 =  2 +

3

after using Eq.(50). Since    =   , we have
s
s
3

 + 43
=

= 1+
2

 
 + 3
(56)
(57)
The square root definitely exceeds 1, so that
   
8
(58)
This implies that the perturbed orbit will generally precess.
·
(d) To find the direction of precession, we integrate  = 2 . For a slightly
·
noncircular orbit, because  is very small,  can be well approximated as   ,
·
=


= 2 (1 − 2 + · · · ) '   
2


(59)
We may, thus, integrate Eq.(59) over the difference in the periods of the circular
and noncircular orbit to obtain the angular displacement of the noncircular orbit
per revolution as
µ
¶

∆ =   ( −   ) = 2
−1 
(60)

Because ∆ is negative, the precession will be opposite (or clockwise) to the
direction of the orbital motion.
9