So the eccentricity can be written in this form $$\mathbf{e}=\frac{\mathbf{A}}{mk} =\frac{1}{mk}(\mathbf{p}\times\mathbf{L})-\hat{\mathbf{r}}$$ but I cannot find a proof or figure it out on my own.
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$\begingroup$ Hello! It is preferable to type out screenshots or images of text; for formulae, one can use MathJax. Thanks! $\endgroup$– jng224Commented Nov 24, 2021 at 15:56
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$\begingroup$ If you understood the trivial derivation of Kepler orbits, you just see, by inspection, that |A| is proportional to the eccentricity... $\endgroup$– Cosmas ZachosCommented Nov 24, 2021 at 16:08
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1 Answer
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The force is $-mkr^{-2}\hat{r}$; the special angular momentum is $h=r^2\dot{\theta}$. Since $\vec{L}=mh\hat{k}$ is conserved,$$\frac{d}{dt}\left(\vec{p}\times\vec{L}\right)=\vec{F}\times\vec{L}=m^2khr^{-2}\hat{\theta}=m^2k\dot{\theta}\hat{\theta}=m^2k\frac{d}{dt}\hat{r}.$$This implies $\vec{A}:=\vec{p}\times\vec{L}-mk\hat{r}$ is conserved. Since$$Ar\cos\theta=\vec{A}\cdot\vec{r}=\vec{r}\cdot\vec{p}\times\vec{L}-mkr=L^2-mkr\implies r=\frac{L^2}{mk}\frac{1}{1+\frac{A}{mk}\cos\theta},$$the eccentricity is $\frac{A}{mk}$. Thus $\frac{\vec{A}}{mk}$ is a vector of length $|e|$.
