
195Motion Under a Central Force
© 2010 Taylor & Francis Group, LLC
and
Ak rkekrr e
r
=−
−
13
2
µθ
θ
.
From these two equations, we have
AAAk
l
r
kk
22
2
2
22
=⋅=−
+
µ
. (6.88)
For a circular orbit, A and
must vanish at r = r
0
, and then Equation 6.88 reduces to
l
r
2
0
µ
=
. (6.89)
We now perturb the orbit, letting r → r
0
+ x and
in Equation 6.88:
kA
l
rx
x
22
2
0
2
=
+
−
+
µ()
. (6.90)
Assuming that x/r
0
≪ 1 and using Equation 6.89 twice, we approximate Equation 6.94 as
kA kxrkrx
22 2
0
2
0
2
()/ µ
. (6.91)
Because the particle is still moving in the same force eld, we have
A
; thus, from Equation
6.91, we have
22
2
3
kAAkrx xkxr
...