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I was looking into the formal proof of Kepler's laws of planetary motion and I happened to come upon this equation for the eccentricity of the orbit:

$$e=\sqrt{\frac{2EL^{2}}{G^{2}M^{2}m^{3}}+1}$$

For the proof of Kepler's first law I was studying, it seems necessary to prove that e is always $\ge0$ in order to solve this integral: $$\theta=-\int_{ }^{ }\frac{1}{\sqrt{\frac{e^{2}}{r_{0}^{2}}-\left(u-\frac{1}{r_{0}}\right)^{2}}}du $$ Where $u=1/r$ and $r_0 = \frac{L^2}{GMm^2}$.

Since $e$ is always $\ge0$, $\frac{2EL^{2}}{G^{2}M^{2}m^{3}}+1$ must also be $\ge0$. I have tried to prove this by substituting $$E=\frac{1}{2}mv^{2\ }-\frac{GMm}{r}$$ as well as $$E=\frac{1}{2}m\dot{r}^2 + \frac{1}{2}mr^2\dot{\theta}^2-\frac{GMm}{r}$$ but so far I've been unable to prove it. Could anybody help me or at least point me in the general direction?

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    $\begingroup$ What would an ellipse with negative eccentricity look like? $\endgroup$
    – J...
    May 31 '21 at 10:09
  • $\begingroup$ When I plugged it into the equation in Desmos, it looked pretty much the exact same with $-1<e<0$ being an ellipse, $e=-1$ being a parabola and $e<-1$ showing a hyperbola. $\endgroup$ May 31 '21 at 10:56
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By definition, $e$ is a ratio of two lengths. Since these are $\ge0$, so is $e$. Meanwhile, $$1+\frac{2EL^2}{G^2M^2m^2}=1+\frac{2Er_0}{GM}$$ is $\ge0 \iff E\ge-\frac{GM}{2r_0}$, which is true for closed orbits by the virial theorem.

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    $\begingroup$ Is true also for open orbits as $E>0$ $\endgroup$ May 30 '21 at 21:47
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You start with the energy

$$E=\frac m2 \vec v\cdot \vec v-\frac{G\,m\,M}{|\vec R|} $$

with:

$$\vec R=r(\varphi)\,\begin{bmatrix} \cos(\varphi) \\ \sin(\varphi) \\ \end{bmatrix} ~,\quad r(\varphi)=\frac{p}{1+e\cos(\varphi)}~,p=\frac{h^2}{G\,M}$$

$$\vec v=\frac{\partial\vec R}{\partial\varphi}\,\dot\varphi~, \quad \dot{\varphi}=\frac{h}{R^2}~, \quad h=\frac{L}{m}$$

Thus the energy is:

$$E=\frac 12\,{\frac { \left( {e}^{2}-1 \right) {M}^{2}{m}^{3}{G}^{2}}{{L}^{2}} } $$

solving this equation for $e^2$ you obtain

$$\boxed{e^2=\frac{2\,E\,L^2}{m^3\,M^2\,G^2}+1}$$


How you obtain $~r(\varphi)$ \begin{align*} &\text{start again with the energy }\\ &E=\frac{m}{2} \left( {{\dot{r}}}^{2}+{r}^{2}{\dot\varphi }^{2} \right) +U \left( r \right)\\ &\text{with}\\ &\dot{\varphi}=\frac{h}{r^2}\qquad,\frac{dr}{dt}=\frac{dr}{d\varphi}\dot{\varphi}\\ &E=\,m{h}^{2} \left( {\frac {{{\left( \frac{dr}{d\varphi}\right)}}^{2}}{{r}^{4}}}+\frac{1}{r^2} \right) +U \left( r \right)\\ &\text{solving this equation for $~\frac{d\varphi}{dr}~$ and with } ~ h^2=\frac{L^2}{m^2} \\ &\frac{d\varphi}{dr}=\frac{1}{r^2}{\frac {1}{\sqrt {2\,{\frac {m \left( E-U \left( r \right) \right) }{{L}^{2}}}-\frac{1}{r^2}}}}\\ &\text{the potential energy $~U(r)=-\frac{G\,m\,M}{r}~$ and for eliptical path $~E=0~$ you obtain}\\ &\frac{d\varphi}{dr}=\frac{1}{r^2}{\frac {1}{\sqrt {2\,{\frac {G\,M{m}^{2}}{r{L}^{2}}}-\frac{1}{r^2}}}}\\ &\Rightarrow\\ &\varphi=2\,\arctan \left( {\frac {\sqrt {2\,G\,M{m}^{2}r-{L}^{2}}}{L}} \right)\\ &\text{or}\\ &r={\frac {{L}^{2}}{G\,M{m}^{2} \left( \cos \left( \varphi \right) +1 \right) }}\\ &\boxed{r(\varphi)=\frac{p}{1+\cos(\varphi)}} \end{align*}

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    $\begingroup$ Very nice explanation. I have voted up all :-) $\endgroup$
    – Sebastiano
    May 30 '21 at 21:44
  • $\begingroup$ Doesn't this proof require you to already know r as a function of angle? I wanted to prove that $e\ge0$ so that I could then solve the integral and find r in terms of angle. $\endgroup$ May 31 '21 at 1:30
  • $\begingroup$ @ShudheshVelusamy if e > 1 then the energy E > 0 you obtain hyperbolic path, e < 1 elliptic path e=1 parabolic path $\endgroup$
    – Eli
    May 31 '21 at 6:00
  • $\begingroup$ @ShudheshVelusamy I put more information how to obtain elliptical path $\endgroup$
    – Eli
    May 31 '21 at 15:50
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    $\begingroup$ @Eli Thank you! $\endgroup$ Jun 1 '21 at 8:46
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A more trivial proof is by using the eccentricity vector (which is a vector with the same direction as the semi-major axis and whose modulus equals the eccentricity of the conic)

$$\mathbf{e}\equiv\frac{\mathbf{A}}{mk}=\frac{1}{mk}(\mathbf{p}\times \mathbf{L})-\hat{\mathbf{r}}$$ $$|\mathbf{e}|\equiv\frac{|\mathbf{A}|}{mk}\geq0$$


~References

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