####
Statement of Kepler's Third Law

From observations collected over many centuries, and especially data compiled by
the Danish astronomer Tycho Brahe, Kepler deduced a relationship between the
orbital period and the radius of the orbit. Precisely:

* the square of the period of an orbit is proportional to the cube of the
semimajor axis length $a$.*

Although Kepler never expressed the equation in this way, we can write down the
constant of proportionality explicitly. In this form, Kepler's Third Law
becomes the equation:
\begin{equation} T^2 = \frac{4\pi^2 a^3}{GM} \end{equation}
where $G$ is the Gravitational Constant
that we
shall encounter in Newton's Law, and $M$ is the mass about which the planet is
rotating (usually the sun for our purposes). This relationship is extremely
general and can be used to calculate rotational periods of binary star systems
or the orbital periods of space shuttles around the earth.

####
A problem involving Kepler's Third Law

The orbit of Venus around the sun is roughly circular, with a period of 0.615
years. Suppose a large asteroid crashed into Venus, instantaneously
decelerating its motion, such that it was thrown into an elliptical orbit with
aphelion length equal to the radius of the old orbit, and with a smaller
perihelion length equal to $98 \times 10^6$ kilometers. What is the period
of this new orbit?

First we must calculate the radius of the original orbit:
\begin{eqnarray*}
r &=& \left(\frac{GM_sT^2}{4\pi^2}\right)^{1/3} \\
&=& \left(\frac{6.67\times 10^{-11}\times 1.989 \times 10^{30} \times
(1.94 \times 10^7)^2}{4\pi^2}\right)^{1/3} \\
&=& 108 \times 10^9 \rm{meters}
\end{eqnarray*}
where $1.94 \times 10^7$ is the period expressed in seconds. The period of the
new orbit is once again given by Kepler's Third Law but now with the
semimajor axis length $a$ replacing $r$. This length is given by half the sum
of the aphelion and perihelion lengths:
\begin{equation}
a = \frac{(98 + 108) \times 10^9}{2} = 103 \times 10^{9} \rm{meters}
\end{equation}
The new period is then given by:
\begin{eqnarray*}
T_{new} &=& \sqrt{\frac{4\pi^2a^3}{GM_s}} \\
&=& \sqrt{\frac{4\pi^2 \times (103 \times 10^9)^3}
{6.67 \times 10^{-11} \times 1.989 \times 10^{30}}} \\
&=& 1.80 \times 10^7 \rm{secs}
\end{eqnarray*}
Although the asteroid slowed the planet down, we see that it now circles the sun
in a shorter time. This is because the collision caused the planet to move
faster at the perihelion, shortening the overall orbital distance.