# Homework Help: Angular velocity of circular orbit, small oscillations

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1. Jul 25, 2016

### Dazed&Confused

1. The problem statement, all variables and given/known data
The potential energy of a particle of mass $m$ is $V(r) = k/r + c/3r^3$ where $k<0$ and $c$ is a small constant. Find the angular velocity $\omega$ in a circular orbit of radius $a$ and the angular frequency $\omega'$ of small radial oscillations about this circular orbit. Hence show that a nearly circular orbit is approximately an ellipse whose axes precess at an angular frequency $\Omega \approx (c/|k|a^2)\omega^2$
2. Relevant equations
$$m\ddot{r} -mr\dot{\theta}^2 = -\frac{dV}{dr}$$

3. The attempt at a solution
So the first part is just setting $\ddot{r}$ to zero so that
$$-\omega^2 = \left. -\frac{1}{ma}\frac{dV}{dr} \right|_a = \frac{k}{ma^3} + \frac{c}{ma^5}$$
which is the answer given. For the second part I Taylor expanded $$\omega^2r -\frac{dV}{dr}$$
about $a$ took the second term to be minus the angular frequency squared. Thus I get
$$\omega'^2 = -\omega^2 + \left. \frac{d^2V}{d^2r}\right|_a$$

however with this I get a result different from the answer. I'm not sure where the mistake lies.

2. Jul 25, 2016

### Ken G

The acceleration is minus omega prime squared, not omega prime squared. Is that your problem? Note that if V is 1/r, omega prime must = omega, that's a true ellipse.

3. Jul 25, 2016

### Dazed&Confused

Sorry I think I wasn't clear. I know the acceleration is $-\omega'^2$ so in the answer for $\omega'^2$ I swapped the sign. Is this what you mean?

4. Jul 25, 2016

### Ken G

If the sign error is not the mistake, then everything else looks right. Note that the answer given is the difference between omega and omega prime, as that will be the precession frequency.

5. Jul 25, 2016

### Dazed&Confused

Ok. The answer given in the book is
$$\omega'^2 = -\frac{k}{ma^3} +\frac{c}{ma^5}$$

6. Jul 25, 2016

### Ken G

Ok, then what you should get is that the difference between omega and omega prime is the second derivative of V at r=a divided by 2*omega, and I presume that comes out what is given in the question.

7. Jul 25, 2016

### Dazed&Confused

Well I'm not sure how they get $\omega'$. Do you have any ideas?

8. Jul 25, 2016

### Ken G

9. Jul 25, 2016

### Dazed&Confused

Well I'm expanding in terms of $r$ so that it's always first order in $c$.

10. Jul 25, 2016

### Dazed&Confused

I get $$\omega'^2 = \frac{3k}{ma^3} + \frac{5c}{ma^5}$$

11. Jul 25, 2016

### Ken G

Perhaps the problem is in the effective potential you are using. You are effectively in a rotating reference frame with a centrifugal potential that is proportional to r. But that means when you perturb r, the particle is given extra angular momentum so it continues to corotate with the rotating frame. What should actually happen is the particle should maintain its angular momentum. So the effective potential actually looks like L2/r, not proportional to r like a centrifugal potential. Look up the equation for r motion that conserves L, and use that as the effective potential instead, maybe there's a lowest order correction induced in effect by the L-conserving coriolis force.

12. Jul 25, 2016

### Dazed&Confused

Yes thank you that did the trick. I will attempt the final part now.

13. Jul 25, 2016

### Dazed&Confused

I figured I may as well post what I did:
$$\omega'^2 - \omega^2 = (\omega'-\omega)(\omega+\omega') \approx (\omega'-\omega)(2\omega)$$
so that
$$\omega'-\omega \approx \frac{\omega'^2 - \omega^2}{2\omega} \approx \frac{c\omega^2}{|k|a^2\omega} = \frac{c}{|k|a^2}\omega$$
as terms with $c^2$ are negligible.