Proof that the eccentricity of a planet in orbit is always greater than or equal to zero 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?
 A: 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.
A: 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*}
A: 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

*

*Laplace-Runge-Lenz vector

*Eccentricity vector
