#### Transcript Two-Body, Central Force Problems

```Physics 430: Lecture 18
Two-Body Central Force
Problems
Dale E. Gary
NJIT Physics Department
8.1 Central Force Problem
We have already met the central force problem in both electrostatics and
gravitation. A key point that governs the behavior is that there is no
transverse force, hence angular momentum is conserved.
 This is only strictly true for two-body problems, and for spherically symmetric
masses or charges. As soon as a third body is added that is not on a line
between the first two, a transverse force appears. And even bodies as
nearly spherical as the Earth and the Moon nevertheless exert tidal forces
that cause an exchange of momentum.
 But well-separated planets and stars interact to an excellent approximation
through two-body central forces, as do a proton and electron (the hydrogen
atom) or diatomic molecules. Even when quantum effects are important,
those effects are described in language borrowed from the two-body central
force problem, so this is an important problem to study.
 For definiteness, we will restrict ourselves to the gravitation problem, but
keep in mind that these ideas apply directly to any central force problem.

November 3, 2009
The Gravitation 2-Body Problem

We have two gravitating bodies of mass m1 and m2, at positions r1 and r2.
The potential energy is
Gm1m2
U (r1 , r2 )  -

r1 - r2
.
A key point is that this depends only on the separation between the masses,
not on r1 and r2 separately. Thus, the system is translationally invariant. In
addition, as we saw in chapter 4, if a conservative force is central, then U is
independent of the direction of (r1 - r2) as well. It depends only on the
magnitude r1 - r2 :
U (r1 , r2 )  U ( r1 - r2 ).

To make things simpler, we introduce a new variable, r = r1 - r2 , which is
the position of body 1 relative to body 2. U depends only on the magnitude
of r, i.e.
U  U (r ).

In terms of Lagrangian mechanics, we have for the two-body problem:
L  T - U  12 m1r12  12 m2r22 - U (r ).
November 3, 2009
8.2 CM, Relative Coordinates, and
Reduced Mass
1
You should already be familiar with finding the center of mass R 
M
 For two masses, it is trivial:

R
mr .
m1r1  m2r2 m1r1  m2r2

.
m1  m2
M
i i
i
As we saw in Chapter 3, the CM is along the line joining the two masses, and
the distances of the masses from the CM is in the ratio m1/m2. If m2 » m1, for
example, then the CM is very close to body 2.
 Recall also from section 3.3 that the total momentum of two bodies is the
same as if the total mass were concentrated at the CM location and following
the motion of the CM: P  MR.
 Your first homework problem (Prob. 8.1) will ask you to show that r1 and r2 ,
in terms of CM R and relative position r, are
m
m
Notice the symmetry,
r1  R  2 r and r2  R - 1 r.
but with a change in sign
M
M
 Putting these into the kinetic energy
2
2

m 
m 


T  12 m1r12  12 m2r22  12 m1  R  2 r   12 m2  R - 1 r  .
M 
M 


November 3, 2009
Reduced Mass

Multiplying the squares, we find that the cross terms cancel and we are left
with the simple result:
T  12 MR 2  12 mr 2 .
where I have used the symbol m for the reduced mass:
mm
m 1 2 .
m1  m2

This is an important quantity that you will see again, so you should get used
to it. Another way to write it is:
Note, m is less than the
1 1
1
  .
smaller of m1 and m2.
m m m
1
2
It has the dimensions of mass, and ranges from m1 when m2  , to m2
when m1  , and is m/2 when m = m1 = m2. Thus, it tends toward the lower
of the two masses. The reduced mass of the Sun-Earth system is almost
exactly the mass of the Earth.
CM and relative coords
 Returning to the Lagrangian, we now find:
make good generalized
L  T - U  12 MR 2  12 mr 2 - U (r )
coords, and split the problem
 LCM
 Lrel .
into two parts.



November 3, 2009
8.3 The Equations of Motion



With the Lagrangian L  T - U  12 MR 2  12 mr 2 - U (r ) , we can now write
down the equations of motion. The CM equation is trivial:
MR  0
or
R = const.
This should be no surprise, because recall that our 2-body problem is an
isolated system, hence no outside forces are acting (Newton’s 1st Law). In
the language of our previous lecture, the CM coordinate is ignorable.
Another way of looking at it is that the Lagrangian does not depend on R,
which expresses a conservation law (conservation of momentum).
 The Lagrange equation for the other coordinate, the relative position r, gives
the separate equation of motion:

m r  -U (r ),
which we should pause to think about. This is the equation of motion for a
single free particle of mass m (reduced mass) subject to potential energy
U(r). In the case of two stars, for example, neither one orbits the other, but
rather they orbit their common center of mass. Nevertheless, we can treat
the problem AS IF a single star of reduced mass m were orbiting at a
distance r = |r1 - r2| from the other, fixed (unmoving) star. And note that
EITHER star can be the fixed one, not just the more massive one.
November 3, 2009
The CM Reference Frame
Since we have seen that the velocity of the CM is constant, we can change to
a frame moving with this constant velocity so that, in this alternate inertial
frame, R = 0.
2
1
 In the CM frame, the Lagrangian is just L  2 m r - U ( r ) and the problem is
reduced to a one-body problem. It is important to understand this pseudosingle body system.
 The “single body” is of
reduced mass m, and

m1 r O
1
r2
CM
r1
m2
r = |r1- r2|
m

r2
O
Arbitrary
Origin atorigin
CM
Path relative to CM

Equivalent one-dimensional problem
the center of its orbit is
the other body NOT the
CM.
This is true even
though the choice of
origin that led to the
above Lagrangian is the
CM.
Note, as m1   ,
m2  m and this
becomes more
accurate.
November 3, 2009
Conservation of Momentum

In the Lagrangian L  12 m r 2 - U (r ), the r term has to be evaluated in polar
coordinates. Recall that
d
ˆ
r

dt
Thus, the Lagrangian becomes

rrˆ  rrˆ  rff .

L  12 m r 2  r 2f 2 - U (r ),
which is independent of f, from which you can immediately see that angular
momentum is conserved:
L
 m r 2f  const  l. (angular momentum)
f
This is for our pseudo-one-body problem, so let’s see if this holds up when
we look at the two-body problem with origin at the CM.
 The total angular momentum for the two bodies is

L  r1  p1  r2  p2  m1r1  r1  m2r2  r2 .

But recall

You will find upon substitution that L  r  mr, which is that for one body.
r1  R 
m2
m
r 2r
M
M
and r2  R -
m1
m
r  - 1 r.
M
M
November 3, 2009
The Two Equations of Motion


f


 m r f  const  l. (angular momentum)
The radial equation, in turn, is L  m rf 2 - dU  m r .
r
dr
Notice that this equation depends on f, so the two equations are coupled, but
in the nicest possible way since we can simply replace
l
2
 l  dU
l
dU

.
2 
3
dr
 m r  dr m r
and find m r  m r 


We just saw that the f equation for the Lagrangian L  12 m r 2  r 2f 2 - U (r ),
L
is
2
f
2
mr 2
Notice that this depends only on r, and leads us to think about an even
simpler, one-dimensional problem. Notice that the above equation is a force
equation, so each term is a force. The first term on the right can be identified
as the “centrifugal force” Fcf. There is nothing to stop us, mathematically,
from writing this as the (1-d) gradient of a potential energy:
l2
d
Fcf  -U cf  3  - U cf
mr
dr
l2
 U cf 
.
2m r 2
November 3, 2009
8.4 The Equiv. 1-D Problem

Writing this term as the gradient of a potential, the equation of motion becomes
mr  -
dU dU cf
d
d
 - U (r )  U cf (r )   - U eff (r ),
dr
dr
dr
dr
where Ueff is the effective potential energy, i.e. the sum of the actual potential
energy U(r) and the centrifugal potential energy Ucf(r):
U eff

As written, this is actually a correct equation for any two-body central force
problem. For the specific case of a gravitational potential energy, we have
U eff

l2
 U (r ) 
.
2
2m r
Gm1m2
l2

.
2
r
2m r
Notice that one term of this is negative, while the
other is positive. A schematic plot is shown a right,
which you may recognize from Chapter 4 (and the
exam!).
U
~ l2/r2
r
Ueff
~ -1/r
November 3, 2009
The Equivalent 1-D Problem-2

curve applies here as well, although there are “nuances.”
rmin
U
rmax
r
m
~ l2/r2
rmin
E>0
As before, if the total energy is less than zero
then the object (planet, star, comet…) has two
E<0
rmin
turning points, rmin and rmax. However, the actual
rmax
motion is not purely radial, but rather the object
~ -1/r
is moving in an ellipse. The “force” that turns the
object at the inner point is the irreducible angular momentum of the object.
 The object can get no closer to the central star than rmin, because to do so
would require losing some angular momentum.
 How would you interpret the equilibrium (the minimum point in the Ueff curve)?

November 3, 2009
r
Conservation of Energy

Going back to the radial equation of motion:
mr  
d
U eff (r ),
dr
U
Multiplying both sides by dr/dt, this can be written

~ l2/r2
rmin
E>0

d 1 2
dr d
d
m rr 
m
r

U
(
r
)

U eff (r ),
eff
dt 2
dt dr
dt
or, 12 m r 2  U eff (r )  const.
r
rmin
This is just an expression of conservation of energy,
but notice that this says the radial kinetic energy
plus the effective potential energy is a constant (in
fact it is just E).
 We can see that the effective potential energy is made
of the true potential energy and the angular kinetic energy!


1
2
E<0
rmax
~ -1/r
m r 2  U eff (r )  12 m r 2  12 m r 2f 2  U (r )  E.
One last point—we developed this for an inverse square law, in which case finite
orbits, which we will see are ellipses, close on themselves. However, for some
other force law the orbit does not close.
November 3, 2009
Equation of the Orbit
The radial equation of motion provides the position r and a function of t, but if
we want the equation of the orbit we need r as a function of f.
 We start with the equation of motion in terms of forces, and transform it using a
couple of (quite non-obvious) tricks:
l2

m r  F (r ) 


mr
3
.
The first trick is to change variables from r to u = 1/r. The second trick is to
convert the differential operator d/dt in terms of d/df:
d df d
d
l d lu 2 d

f
 2

.
dt dt df
df m r df m d f
Hang on to your hats—let’s find out what r becomes:
d
lu 2 d 1
l du
r  (r ) 
dt
m df u
m df
so
d
lu 2 d  l du 
l 2u 2 d 2u
r  (r ) 
.
- 2
2
dt
m df  m df 
m df
l 2u 2  2u
l 2u 3
-m 2
 F (r ) 
2
m f
m
or
u(f )  -u (f ) -
m
F (r ).
2
2
l u (f )
November 3, 2009
for a Free Particle

Before we get into the solution of this equation for the 2-body problem, which
we’ll do next time, let’s look at its solution for a free particle. Clearly, for a free
particle the force F(r) = 0, so we have:
u(f )  -u (f ),
which is the oscillation equation whose general solution is
u (f )  A cos(f - d ),
where A and d are arbitrary constants (and notice that w = 1).
 Remember that u = 1/r, so we’ll rename A = 1/ro. The above solution then
becomes:
r (f ) 

ro
1

.
u (f ) cos(f - d )
Believe it or not, this is the polar coordinate equation
for a straight line, as you can see in the plot at right.
Here Q is the point of closest approach, where f  d and
therefore r = ro. At any other point, P, ro = r cos(f - d.
P
r
f
ro Q
d
November 3, 2009
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