# 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

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