Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In solving an exercise I had to find the equation of the quasi-circular orbits of an object with the potential $V(r)=-\alpha r^{-1-\eta}$ and I expressed it as: $$r(\phi)=\frac{r_c}{1+\epsilon \cos(\phi\sqrt{1-\eta})}$$ Where $r_c$ is the radius of the circular orbit and $\epsilon$ depends on the initial conditions. Now (among other things) I am asked about the period of the motion. I thought that in order to find the period I should integrate $\phi(t)$ using the conservation of angular momentum $L$ in the form $\dot\phi(t)=\frac{L}{mr^2(\phi)}$. This integration isn't easy at all and, in my opinion, can only be approximated.

However, the author of the exercise wrote that the period can be found easily by $mr_c^22\pi/T=L$ but he doesn't explain why. My question is where does this formula come from and whether it is exact or just an approximation.

share|improve this question
    
That's because $mr^2\omega$ is the angular momentum. –  Ron Maimon Jul 10 '12 at 20:19
    
@RonMaimon I know, but $\omega(t)$ is a function of time, so it is not necessarily $2\pi/T$ –  Ralph Jul 10 '12 at 23:32
    
The fellow is ignoring the non-circularity to first order. –  Ron Maimon Jul 11 '12 at 2:38

2 Answers 2

up vote 4 down vote accepted

I think "quasi-circular" is a misleading name for this problem. Perhaps "quasi-elliptical" would be better? I say this because this problem does, in fact, contain a closed circular orbit (the radius of which you have called $r_c$). For that orbit you can find the period using Kepler's second law, which gives the result you show.

An interesting way to approach this problem is to consider only small radial oscillations about the circle. Find the oscillation frequency by expanding the effective potential about the minimum. See how this is related to the frequency of the circular orbit. What happens when $\eta\rightarrow 0$? You should find that in that case (inverse square force law), the small-oscillation frequency is identical to the circular orbit frequency. This is another way of seeing that the orbits must be ellipsoidal. But, for non-zero $\eta$, this is no longer the case. For small $\eta$, you instead get near-ellipses that just fail to close. These precessing ellipses are described by the radial formula you gave.

share|improve this answer

Your problem is one of a potential that depends only on radius. Newton proved that with these sorts of problems, angular momentum is conserved. Your instructor used this well known fact.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.