# Stability Condition for Circular Orbit

1. Apr 24, 2014

### cpburris

1. The problem statement, all variables and given/known data

Show that the stability condition for a circular orbit of radius a, i.e.

$f(a) + \frac{a}{3} (\frac{df}{dr})_{r=a} < 0$

is equivalent to the condition

$\frac{d^2V(r)}{dr^2} > 0$

for r=a where V(r) is the effective potential given by

$V(r) = U(r) + \frac{ml^2}{2r^2}$

3. The attempt at a solution

I understand fully why they are equivalent, and I would have no problem proving individually how each is a condition for stability, but analytically I really don't know how to show the two are equivalent. I'm not even sure what the question is asking. I tried just setting

$-\frac{d^2V(r)}{dr^2} = f(a) + \frac{a}{3} (\frac{df}{dr})_{r=a}$

and do something from there, but it didn't get me anywhere.

2. Apr 28, 2014

### BruceW

I'm guessing that you already know what $f(r)$ is as a function of the potential? So if you use this definition, you could write out $f(a) + \frac{a}{3} (\frac{df}{dr})_{r=a} < 0$ in terms of the potential instead, and start to see how it could be similar to the other equation.