1
$\begingroup$

I want to prove to show that the change of the rotation angle of a body in a two-body-problem is exactly $\Delta \phi = 2\pi$.

I know that the whole energy of the system is given by $$ E = \left(\frac{1}{m_1}+\frac{1}{m_2}\right)\left(p_r^2+\frac{L^2}{r^2}\right)-\frac{Gm_1m_2}{r},$$ where $p_r$ is the inertia associated with $r$ and $L$ is the angular momentum. Furthermore, the orbit is given by $$\phi+ \frac{\partial S_r}{\partial L} = const.,$$ where $$S_r = \int \sqrt{E\left(\frac{1}{m_1}+\frac{1}{m_2}\right)^{-1} + \frac{Gm_1m_2}{r}-\frac{L^2}{r^2}}dr.$$ Therefore $$\Delta \phi = - \frac{\partial \Delta S_r}{\partial L},$$ where $\Delta S_r$ is the change of $S_r$. So all I have to do is to show that this expression equals $2\pi$. How can I do this?

$\endgroup$
  • $\begingroup$ This should help en.wikipedia.org/wiki/Bertrand%27s_theorem $\endgroup$ – Alexander Apr 4 '16 at 23:01
  • $\begingroup$ But how can this help to show that the last term equals $2\pi$ $\endgroup$ – Razupaltuff Apr 4 '16 at 23:08
  • $\begingroup$ It wasn't closed orbit unless it's $2 \pi$ $\endgroup$ – Alexander Apr 4 '16 at 23:09
  • $\begingroup$ I know. But all I want to do is to show that the last expression equals $2\pi$, because that would mean the orbits are closed. $\endgroup$ – Razupaltuff Apr 4 '16 at 23:11
2
$\begingroup$

Take the orbit equation (using $u=\frac1r$):

$$u(\theta)=A\left(1+e\cos\theta\right) $$

Where $A$ is some constant that you can work out from the differential equation if desired, and $e$ is the eccentricity of the orbit. Now, what happens when $\theta$ goes through an angle of $2\pi$? Notice that we would not have returned to the original radius if we had a some potential that changed the frequency inside the cosine term as often happens with some potentials. Note that I am assuming the orbit is an ellipse. In regards to your integral...use Mathematica or look up the details in any classical mechanics textbook such as Goldstein or Taylor.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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