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Two Bodies
Two-Body System

A two-body system can be
defined with internal and
external forces.
• Center of mass R
• Equal external force
  int  ext

m1r1  F1  F1
  int  ext

m2 r2  F2  F2

Add to get the CM motion
  ext  ext
MR  F1  F2

Subtract for relative motion
 int  int
r  r  F1  F2
1
2
m1
m2
F2int
m2
r = r1 – r2
F2ext
r2
m1
R
F1int
r1
F1ext
Reduced Mass


The internal forces are equal
and opposite.
 int  int

r  r  F  F  ( 1  1 ) F int
1
2
m1
m2
m1 m2
Express the equation in
terms of a reduced mass m.
 int

r  r  ( m1  m2 ) F int  F
1
2
m1m2
m
•
•
m less than either m1, m2
m approximately equals the
smaller mass when the
other is large.
m
m1m2
 m2 for m1  m2
m1  m2
Central Force Equations

Use spherical coordinates.
• Makes r obvious from central
force.
• Generalized forces Qq = Qf = 0.
• Central force need not be from
a potential.

Kinetic energy expression
T  12 m (r 2  r 2q 2  r 2 sin 2 qf 2 )
d T T

 Qr
dt r r
d T T

0

dt q q
d T T

0

dt f f
Coordinate Reduction


T doesn’t depend on f
directly.
Constant angular momentum
about the polar axis.
• Constrain motion to a plane
• Include the polar axis in the
plane

Two coordinates r, q.
d T T

0

dt f f
d T
0

dt f
T
2
2
 constant

m
r
sin
q
f
f
T  12 m (r 2  r 2q 2 )
Angle Equation

T also doesn’t depend on q
directly.
• Represents constant
angular momentum
• Angular momentum J to
avoid confusion with the
Lagrangian

Change the time derivative
to an angle derivative.
d T T

0

dt q q
d T
0
dt q
T
2 

m
r
qJ

q
constant
d dq d
J d

 2
dt dt dq mr dq
Central Motion

Central motion takes place in a plane.
• Force, velocity, and radius are coplanar

Orbital angular momentum is constant.

If the central force is time-independent, the orbit is
symmetrical about an apse.
• Apse is where velocity is perpendicular to radius
Orbit Equation
d T T

 Qr
dt r r
J d T T

 Qr
2

mr dq r r
J d [ 12 m (r 2  r 2q 2 )] [ 12 m (r 2  r 2q 2 )]

 Qr
2

mr dq
r
r
J dr
J 2 J d J dr
J2
 mr ( 2 )  2
( 2
)  3  Qr
2
r dq
mr
r dq mr dq
mr
1 d 1 dr
1 mQr
(
)

 2
2
2
3
r dq r dq
r
J
Let u = 1/r
d 2u
mQr

u


dq 2
J 2u 2
Central Potential

The central force can derive
from a potential.
d T T
V

 Qr  
dt r r
r

Rewrite as differential
equation with angular
momentum.
J 2 V
mr  3 
0
mr
r

Central forces have an
equivalent Lagrangian.
2
J
L  12 mr 
V
2
2mr
2
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