I have this equation:

$$\ddot{r}=-\frac{G M}{r^{2}}$$

I have to show that the solution of $r(t)$ can be parametrized by $\theta$ like this:

$$ r=A(1-\cos \theta), \quad t=B(\theta-\sin \theta), \quad A^{3}=G M B^{2} $$

I could probably show by direct substitution that this parametrization is correct using double chain rule. Assuming I don't know what the solution is going to be, how could I go about obtaining this parametrization?

I can multiply both sides by $\dot{R}$:

$$\dot{R} \ddot{R}=-\frac{G M\dot{R}}{R^{2}}$$

which leads me to:

$$\frac{d}{d t}\left[\frac{1}{2} \dot{R}^{2}-\frac{G M}{R}\right]=0$$

which gives me:

$$\frac{d r}{d t}=\frac{2 G M}{r}+C.$$

I don't know how to continue to get to the parametrization.

  • $\begingroup$ Does this helps math.stackexchange.com/q/2140677/399263 ? $\endgroup$
    – zwim
    Mar 30, 2021 at 16:49
  • $\begingroup$ Hmm, I don't see how that helps with the above issue but there is a very nice answer which I did not know about, so thank you! $\endgroup$
    – zabop
    Mar 30, 2021 at 16:52
  • $\begingroup$ Oh wait, it might help actually. $\endgroup$
    – zabop
    Mar 30, 2021 at 16:53
  • $\begingroup$ $M$ and $r$ in the answer have very similar form of what I need. My $\theta$ is their $E$. $\endgroup$
    – zabop
    Mar 30, 2021 at 16:54


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