Movement of planets (Kepler's Law)

Question

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! :)

Answer 1

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.

Answer 2

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 $\cos{\theta}$.