Movement of planets (Kepler's Law) Let $m_p$ be the mass of a planet and $m_s$ the mass of the sun. Then the following differential equation holds :
$$m_p x''(t) = -G m_p \cdot m \cdot \frac{x(t)}{|x(t)|^3}$$ with $ m = m_s + m_p$. The function $t \rightarrow x(t)$ describes the orbit of the planet in relation to the sun. I know that energy and angular momentum are conserved quantities and that the following holds true:
$$E = \frac{-Gm_p m}{|x|} + \frac{m_p}{2} |x'|^2 $$ and $$L = m_p x \times x'$$
Let $L$ be non zero. In this case, the curve of the orbit is in a plane which is spanned by the orthonormal base ${e_1, e_2}$. Let's define for an angle $\psi$ : $$e_r = cos \psi e_1 + sin \psi e_2$$
$$e_{\psi} = -sin \psi e_1 + cos \psi e_2$$
Let's use polar coordinates to consider the movement and let $x=re_r$.
How can the following system be deduced by the conservation laws? 
$$ r^2 \psi' = \frac{|L|}{m_p}$$ 
$$ (r')^2 + r^2(\psi')^2 - 2 \frac{Gm}{r} = 2 \frac{E}{m_p} $$
Thanks for any hint and help! :)
 A: It's just the restatement of conservation laws in polar coordinate. For example, the angular momentum $\vec{L} = m_p x \times x'$. Since $x = re_r$, 
$$
x' = (re_r)' = r'e_r+ r \frac{d}{dt}(\cos \psi e_1 + \sin \psi e_2) = r'e_r + r\psi'e_{\psi}.
$$
So you'll have
$$
\vec{L} = m_p(re_r)\times(r'e_r+r\psi' e_{\psi}) = m_pr^2\psi' (e_r\times e_{\psi}) \implies|L| = m_pr^2\psi'.
$$
And similar for conservation of energy $E$, that you just need to find the expression of $x''$ in polar coordinates.
A: Add up your forces, gravity and the centripetal force. 
$$m_p\ddot r=\frac{-GMm_p}{r^2}+m_pr\dot \theta^2$$
$$m_p\ddot r=\frac{-GMm_p}{r^2}+\frac{L\dot \theta}{r}$$
$$m_p\ddot r=\frac{-GMm_p^2\dot \theta}{m_p\dot \theta r^2}+\frac{L\dot \theta}{r}$$
$$m_p\ddot r=\frac{-GMm_p^2\dot \theta}{L}+\frac{L\dot \theta}{r}$$
substituting $L$ for $m_pr^2\dot \theta$
Divide both sides by $\dot \theta$. Note that $\frac{\ddot r}{\dot \theta}=\frac{d}{d\theta}\dot r$
$$m_p\frac{d}{d\theta}\dot r=\frac{-GMm_p^2}{L}+\frac{L}{r}$$
This suggests we change variables letting $\dot r = \frac{dr}{d \theta} \dot \theta$ by the chain rule. 
So $$m_p\frac{d}{d\theta}\frac{dr}{d \theta}\dot \theta=\frac{-GMm_p^2}{L}+\frac{L}{r}$$
And $$\frac{d}{d\theta}\frac{1}{r^2}\frac{dr}{d \theta}m_pr^2\dot \theta=\frac{-GMm_p^2}{L}+\frac{L}{r}$$
And $$\frac{d}{d\theta}\frac{1}{r^2}\frac{dr}{d \theta}L=\frac{-GMm_p^2}{L}+\frac{L}{r}$$
And $$\frac{d}{d\theta}\frac{1}{r^2}\frac{dr}{d \theta}L=\frac{-GMm_p^2}{L}+\frac{L}{r}$$
Divide both sides by $L$ and replace $\frac{1}{r^2}\frac{dr}{d \theta}$ with $\frac{-d\frac{1}{r}}{d \theta}$. 
Finally you get :
$$-\frac{d^2}{d \theta^2}\frac{1}{r}=\frac{-GMm_p^2}{L^2}+\frac{1}{r}$$
or:
$$\frac{d^2}{d \theta^2}(\frac{1}{r})+(\frac{1}{r})=\frac{GMm_p^2}{L^2}$$
This the differential equation for an ellipse having 3 solutions, the sum of a constant term, the real constant times $\sin{\theta}$ and another real constant times the $\cos{\theta}$.
