Laplace-Runge-Lenz Vector

${\overrightarrow{F}}_e=\hat{r}f(r)$. Let’s compute $\dot{\overrightarrow{p}}\times \overrightarrow{L}=\cdots =-mf(r)r^2(\frac{\dot{\overrightarrow{r}}}{r}-\frac{\overrightarrow{r}\dot{r}}{r^2})$. $\frac{d}{dt}(\overrightarrow{p}\times \overrightarrow{L})=\dot{\overrightarrow{p}}\times \overrightarrow{L}+\overrightarrow{p}\times \dot{\overrightarrow{L}}=\dot{\overrightarrow{p}}\times \overrightarrow{L}=-mf(r)r^2(\frac{\dot{\overrightarrow{r}}}{r}-\frac{\overrightarrow{r}\dot{r}}{r^2})$.

Assume $f(r)=-\frac{k}{r^2}$ with $k$ positive. Then, $\frac{d}{dt}(\overrightarrow{p}\times \overrightarrow{L})=mk\frac{d}{dt}\left(\frac{\overrightarrow{r}}{r}\right)$. So, $\frac{d}{dt}(\overrightarrow{p}\times \overrightarrow{L}-mk\frac{\overrightarrow{r}}{r})=0$, hence a conserved quantity.

The vector is then horizontal. Consider, $\overrightarrow{A}\cdot \overrightarrow{r}=Ar\cos \phi$.

Let $\overrightarrow{A}=\overrightarrow{p}\times \overrightarrow{L}-mk\frac{\overrightarrow{r}}{r}=\overrightarrow{\mathbb{C}}$. Then, $\overrightarrow{A}\cdot \overrightarrow{r}=\overrightarrow{r}\cdot (\overrightarrow{p}\times \overrightarrow{L})-mk\frac{\overrightarrow{r}\cdot \overrightarrow{r}}{r}=\overrightarrow{L}\cdot (\overrightarrow{r}\times \overrightarrow{p})-mkr={\ell }^2-mkr$. Then, $r=\frac{{\ell }^2/(mk)}{1+\frac{A}{mk}\cos \phi }$. To determine $A$, we compute $\overrightarrow{A}\cdot \overrightarrow{A}=(\overrightarrow{p}\times \overrightarrow{L}-mk\overrightarrow{r}/r{)}^2=p^2{\ell }^2-2mk\frac{{\ell }^2}{r}+m^2{k}^2$ So, $\frac{A^2}{2m{\ell }^2}=\frac{p^2}{2m}-\frac{k}{r}+\frac{m{k}^2}{2{\ell }^2}=E+\frac{m{k}^2}{2{\ell }^2}$ Hence, $A=\sqrt{2m{\ell }^{2}E+m^2{k}^2}$. So, $r=\frac{{\ell }^2/(mk)}{1+\sqrt{1+\frac{2E{\ell }^2}{m{k}^2}}\cos \phi }$