A planet orbits a star in an elliptical orbit: Kepler's second law

Question:
A planet orbits a star in an elliptical orbit. The distance at aphelion is $2a$ and the distance at perihelion is $a$. Find the ratio of the planets speed at perihelion to that at aphelion.

Answer:
According to Kepler's second law, the area swept out per unit time by a radius from the star to a planet is constant. The area can be expressed by
\[
dA = \frac{1}{2}r rd\theta
\]
Take the derivative in terms of time.
\[
\frac{dA}{dt}=\frac{1}{2}r^2 \frac{d\theta}{dt}
\]
$\frac{d\theta}{dt}$ is the angular velocity, $\omega$. We know angular momentum $L=mr^2\omega$. Thus,
\begin{eqnarray}
\frac{dA}{dt} &=& \frac{1}{2}r^2\omega  \\
  &=& \frac{L}{2m}
\end{eqnarray}
This shows that the equal area per unit time indicates constant angular momentum, $L$. We get back to the original expression of the angular momentum, $L=mr^2\omega$. We also know that $v=r\omega$ and $v$ is the tangential speed, so $L=mvr$. Compare the angular momenta at perihelion and aphelion.
\[
L = mv_{\mathrm{p}}a = mv_{\mathrm{a}}2a
\]
Therefore the ratio of the speeds is
\[
v_{\mathrm{p}}:v_{\mathrm{a}}=2:1
\]
We can state that Kepler's second law is essentially equal to the conservation of angular momentum. The more radius, the slower the planet has. The less radius, the faster the planet gets.

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