I'm having trouble getting the general solution of this differential equation.
The gravitational equation of motion is, for constants $M$ and $G$ and position vector $\vec{r}$,
$$\frac{d^2}{d t^2}\vec{r} = -\frac{MG}{r^2}\hat{r}
$$
By using 2D polar coordinates (one angle $\theta$ and one "distance from origin" $r$), one can calculate $\frac{d^2}{dt^2}\vec{r}$ by taking two time derivatives of $\vec{r}=r\hat{r}$. The hat notiation $\hat{r}$ is the unit vector pointing from $\vec{r}$ in the direction of increasing $r$. Similarly, $\hat{\theta}$ is the unit vector pointing from $\vec{r}$ in the direction of increasing $\theta$. By appropriate application of the chain rule of derivatives (remembering to take derivatives of the unit vectors themselves as well), one can derive that (dot means time-dirivative)
$$\frac{d^2}{d t^2}\vec{r} = -\frac{MG}{r^2}\hat{r} = \hat{\theta}(2\dot{r}\dot{\theta}+r\ddot{\theta})+\hat{r}(\ddot{r}-r\dot{\theta}^2)
$$
Which means
$$\begin{equation} -\frac{MG}{r^2}=\ddot{r}-r\dot{\theta}^2 \tag{1} \end{equation}
$$
and
$$\begin{equation} 0=2\dot{r}\dot{\theta}+r\ddot{\theta} \tag{2} \end{equation}
$$
The right hand side of the last equation (2) turns out to be the time derivative of the angular-momentum-per-unit-mass $h$ (to a factor of $r$):
$$\frac{d}{d t}h=\frac{d}{d t}(r^2\dot{\theta})=2r\dot{r}\dot{\theta}+r^2\ddot{\theta}=0
$$
Showing that h is constant in time.
My question is, using what has been laid out, how does one combine the two differential equations (1) and (2) to get something that looks like it can be solved? I have looked at references, but certain steps in the derivations seem to lack explanation.
From what I've seen, it seems important to eliminate $t$ from the equations (1) and (2) to yield an equation with only $r$ and $\theta$.
Thank you for any help, I have been tearing my hair out over this.
—Update—
It appears the substitution $u=\frac{1}{r}$ leads to a $\ddot{r}$ in terms of $\frac{d^2 u}{d^2 \theta}$.
Best Answer
Use the substitution $u=\frac 1r$ to get a differential equation involving $u$ and $\theta$
$$\frac{dr}{dt}==-\frac{1}{u^2}\frac{du}{d\theta}\frac{d\theta}{dt}=-h\frac{du}{d\theta}$$
Where $h=r^2\dot\theta$. Now differentiate again:
$$\frac{d^2r}{dt^2}=\frac{d}{dt}(-h\frac{du}{d\theta})=-h\frac{d}{d\theta}(\frac{du}{d\theta})\frac{d\theta}{dt}=-h^2u^2\frac{d^2u}{d\theta^2}$$