# Kepler’s Third Law

Kepler’s Third Law relates the period of revolution of a planet, denoted by $T$, to the semimajor axis of the orbit of that planet, denoted by $a$.  The square of $T$ is directly proportional to the cube of $a$.
We know that the rate of change the area swept out by a radius is constant so
Our derivation consists of finding the appropriate expressions for $A$ and $DA$.  Recall that
and
We have the final form for $A$ but will have to work a while more on $DA$. In doing this, we will find an expression for $H$ that is more meaningful to the geometry of an ellipse.
Consider the orbit equation derived in a previous section:
In the ellipse below, we have picture the point in the orbit of the planet when $\nu=\frac{\pi}{2}$. This is when the segment $SP$ is perpendicular to the major axis of the ellipse.
Let $p$ be the length of the segment $SP$.  So $p=\frac{H^2}{\mu}$ from the equation for $r$ with $\nu=\frac{\pi}{2}$.
We may compute $p$ another way from the ellipse below:
By the Pythagorean Theorem
and solving this for $p$ gives us
so
Therefore,

and so
Therefore,
so that finally we have Kepler’s Third Law