Download Chapter 25 Celestial Mechanics

point is equal to L . (a) Find the effective potential energy and make sketch of effective
potential energy as a function of r . (b) Indicate on a sketch of the effective potential the
total energy for circular motion. (c) The radius of the particle’s orbit varies between r0
and 2r0 . Find r0 .
Solution: a) The potential energy, taking the zero of potential energy to be at r = 0 , is
r
b
U (r) = − ∫ (−br ′ 3 ) dr ′ = r 4
0
4
The effective potential energy is
U eff (r) =
L2
L2
b
+
U
(r)
=
+ r4 .
2
2
4
2mr
2mr
A plot is shown in Figure 25.13a, including the potential (yellow, right-most curve), the
term L2 / 2m (green, left-most curve) and the effective potential (blue, center curve). The
horizontal scale is in units of r0 (corresponding to radius of the lowest energy circular
orbit) and the vertical scale is in units of the minimum effective potential.
b) The minimum effective potential energy is the horizontal line (red) in Figure 25.13a.
(b)
(a)
Figure 25.13 (a) Effective potential energy with lowest energy state (red line), (b) higher
energy state (magenta line)
c) We are trying to determine the value of r0 such that U eff (r0 ) = U eff (2r0 ) . Thus
L2
b 4
L2
b
+
r
=
+ (2r0 )4 .
0
2
2
mr0 4
m(2r0 ) 4
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