The Two Body Problems with Central Forces and Kepler’s Orbits
Concepts of primary interest:
Central Force
Angular Momentum
Reduced Mass
Examples:

Tools of the Trade
A central force is one that is directed along the line joining the particle acted upon to a fixed point or
center. Further the magnitude of that force depends only on the distance from that center.
F (r )  f (r )rˆ
This radial character ensures that the torque due to this force computed about the force center vanishes
so the angular momentum of the particle is conserved.
L  r  mv conserved
The two body problem with a central force interaction decomposes into the center of mass motion
responding to the net external force and the dynamics of a mythical particle with the reduced mass of the
two particles executing the relative motion of the particles. Extended spherical mass are treated as point
particles with the total mass considered to be located at the respective centers. Tidal effects occur due to
the gradient of the force across each extended mass and the associated distortions of the mass
distribution from spherical can lead to the transfer of angular momentum from one mass to the other as
has been the case of the earth-moon system. The center to center separation of the earth and moon is
about 240,000 miles, and the center of mass of the system lies about 1000 miles below the earth’s
surface. We will study the motion of a particle with the reduced mass and traveling a path with
approximate radius 240,000 miles. The earth’s center travels a path with approximate radius 3,000
miles, and the moon a path wit approximate radius 237,000 miles. WE IGNORE TIDAL EFFECTS.
The angular momentum L is conserved so r and v are always perpendicular to its fixed direction
meaning that the motion is restricted to a plane. We will adopt plane polar coordinates. Note that r is to
be used rather than  as we are analyzing an effective single particle problem.
L (r ,  , r ,  )  ½m[r 2  r 2 2 ]  V (r )
 L   mr 2  L  0 . We conclude that the orbital angular momentum is a
  
constant of the motion. Store this information. The angular momentum and the mechanical energy are
constants of the motion.
The  equation yields
d
dt
The r equation yields
d
dt
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L
V
V
L
 r   mr  mr  r  mr  mr 2   r .
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Physics Handout Series: Basic Problem Solving
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The Two Body Problems with Central Forces and Kepler’s Orbits
Orbital Mechanics provides an excellent example of the power of changing the independent variable
and the dependent variable to transition from the DE form of the equation of motion to the DE for the
orbital path. That is: from an equation for r(t) and (t) to one for r().
Recall that under the influence of a central force, a mass orbits in a plane passing through the force
center and that angular momentum L is conserved. Adopting the polar coordinates r and , the
equations of motion take the form.
GM m
mar  Fr
 mr  mr2  
and m r 2   L , a constant
[DE.1]
2
r
Substituting from the conservation of angular momentum equation   L
, the radial equation
m r2
becomes:
2
GM
[DE.2]
r L 2 3 2
m r
r
Change to a new dependent variable u = r -1, and change the independent variable from t to  using the
chain rule and   d  L 2  L u 2 .
dt
m
mr
 
 
du 1
1 du d
du
 2
 L
m
dt
u d dt
d
2
d 
du 
d u d
r   L
 L
 L

m
m
m
dt 
d 
d 2 dt
After substitution into the radial equation of motion:
d 2u
G M m2
1

u

 RCirc
d 2
L2
r
 
 
 
2
d 2u
u
d 2
2
[DE.3]
where RCirc is the radius of a circular orbit for a mass m with angular momentum L.
1
 A cos(  0 )  1
This equation is easily solved to yield u( )  RCirc
r ( ) 
r ( )
. Inverting,
RCirc
RCirc

1  A RCirc cos(  0 ) 1   cos(  0 )
[DE.4]
The orbit equation is in a standard from for the conic sections and  is called the eccentricity. The value
2 RCirc is also called the latus rectum.
Center of Mass and Relative Angular Momentum: The total angular momentum of the collection of
particles is
L   ri  mi vi   ( R  ri )  mi (V  vi )
i
i
 R   mi V  R   mi vi   mi ri  V   ri  mi vi  R  MV   ri  mi vi
i
i
i
i
i
The factors in red vanish as the average position and velocity relative to that of the center of mass are
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Physics Handout Series: Basic Problem Solving
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The Two Body Problems with Central Forces and Kepler’s Orbits
zero.
L  R  MV   ri  mi vi LCM  Lrel .
i
The angular momentum of a system of particles decomposes into ‘the angular momentum associated
with the center of mass motion’ plus the angular momentum associated with the motion relative to the
center of mass.
One should note that these decompositions work for center of mass and relative to the CM values. These
decompositions are not valid if it is not for the CM values and values relative to the CM.
Special Case: The Two Body Problem - A system of two particles.
Newton’s Laws can be applied to solve for the motion of a single particle in
many cases, but the study of two interacting particles proves to be
fundamentally more difficult. The motion can be represented as the center of
mass motion plus the motion relative to the center of mass.
MA  Fexternal net
m1

r1
r2 
r1
m2
Initially, we will restrict our attention to cases in which there are no external
forces, but there is a velocity independent interaction between the masses.
r2
Further, we define the relative position of the two particles to be   r2  r1 . (A common choice is to
choose particle one to be the more massive of the pair.) Each particle is subject only to a force due to the
other particle so1:
m1r1  F (  )on1by 2 and m2 r2  F (  )on 2by1
Note that the internal forces depend on the relative position of the particles only. We adopt the shorthand
that F (  )on 2by1  F (  ) . In summary, m2 r2  F (  )on 2by1  m1r1 . Next we locate the center of mass and
express the particle positions in terms of R and  .
(m1  m2 ) R  m1r1  m2r2 ; r1  R  ( m2 [ m  m2 ])  ; r2  R  ( m1 [ m  m2 ]) 
1
1
Exercise: Verify that r1  R  ( m2 [ m1  m2 ])  and r2  R  ( m1 [ m1  m2 ])  .
mm
In the absence of external forces, m1r1  m1 R  ( m11 m22 )   0      F (  ) where  is called the reduced
mass of the pair of particles. 
Exercise: Consider m2 r2 to show that    F (  ) .
1
In Newton’s overdot notation, each ‘over dot’ represents a derivative with respect to time.
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The Two Body Problems with Central Forces and Kepler’s Orbits
In the absence of external forces, the problem reduces to the motion of a single particle of mass , the
reduced mass subject to the interaction force:
F (r2  r1 )  F (  ) .
   F ( )
Two Particle Center of Mass Theorems: K  KCM  ½  2 and L  R  MV     .
The Kepler Problem: The sun is particle one, and a planet (comet, meteor, …_) is particle two. The
displacement  runs from the center of the sun to the center of the planet. The influences of all other
bodies are ignored (no external forces). The situation is somewhat more complex when the motion of
earth’s moon is considered. The force on the moon due to the sun has a greater magnitude than that on
the moon due to the earth. The moon basically orbits the sun while the earth perturbs its motion to speed
up and slow down and to weave inside of and outside of the earth’s orbit. None-the-less, the problem
can be approximately represented as the motion of the earth-moon center of mass plus the motion of the
moon relative to the earth. As a start, the earth and moon have nearly the same orbit about the sun which
is the center of mass motion. The next large consideration is the relative motion of the moon and earth
which has the earth-moon gravitational interaction as its cause. Finally, one must account for the
difference in the sun’s gravitational field at the earth’s and moon’s locations as compared to that at the
center of mass. That is the tidal forces must be considered.
The center of mass/reduced mass relative motion decomposition converts a two particle relative motion
problem to two independent single body problems if the external force field is uniform over the volume
enclosing the two particles. If the external field is not uniform, tidal force corrections must be made.
Exercise: For two particle systems, show that K  KCM  ½  2 and L  R  MV     .
Tools of the Trade:
Review - -
Problems
1.)
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