Kepler's laws

The orbit of a planet is an ellipse with the sun at one of the two foci.
A line segment joining a planet and the sun sweeps out equal areas

during equal time intervals.

The planet’s orbital period $T$ is related to the semi-major axis of its

orbit $a$ by $T^2 \propto a^3$ .

Kepler’s three laws follow from Newton’s law of gravitation: that the force between two bodies of mass $m_1$ and $m_2$ is

\begin{align*} \mathbf F_G = -\frac{Gm_1m_2}{r^2} \mathbf{\hat r}. \end{align*}

We consider the motion of the two bodies in the center of mass frame. Let $r$ be their separation and $M$ be their total mass. We see that the distances from the COM are $rm_2/M$ and $rm_1/M$ . The Lagrangian is then

\begin{align*} L = \frac12 m_1 \left[ \left(\frac{m_2\dot r}{M}\right)^2 + \left(\frac{m_2r\dot\theta}{M}\right)^2 \right] \frac12 m_2 \left[ \left(\frac{m_1\dot r}{M}\right)^2 + \left(\frac{m_1r\dot\theta}{M}\right)^2 \right] + \frac{Gm_1m_2}{r}. \end{align*}

Following some algebra, we define the reduced mass $\mu = m1_m2/M$ to write

\begin{align*} L = \frac12 \mu(\dot r^2 + r^2\dot\theta^2) + \frac{Gm_1m_2}{r}. \end{align*}

We see there is no $\theta$ dependence so $\partial L/\partial\dot\theta$ is conserved. In this case that is the angular momentum, $l = \mu r^2 \dot\theta$ . This is enough to prove Kepler’s 2nd law.

The area swept out in some slice of the orbit is approximately $\d A \approx 1/2\, (r+\d r)r \d\theta \approx 1/2\,r^2 \d\theta$ .

Dividing by $\dt$ we get $\dot A = 1/2\,r^2 \dot\theta$ . We substitute in the angular momentum $l$ to show $\dot A = l / 2\mu$ is constant.