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?

  • $\begingroup$ 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

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.


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