Deriving Kepler's law

For worksheet 1, we used Kepler's law

$$P^2=\frac{4\pi^2a^3}{GM}.$$

P is the period, a the orbital separation (actually the distance to the mutual center of mass (barycenter)), G Newton's constant, M the Earth's mass.  How do you get this?

We can use the test case of a circular orbit to derive it.  While this is not the most general (in general, under gravity closed orbits must be ellipses; a circle is an ellipse with zero eccentricity), a circular orbit must satisfy this law, and so can be used to obtain it.

We balance centrifugal force with gravity (*):

$$\frac{mv^2}{a}=\frac{GMm}{a^2}.$$

m is the mass of the smaller object orbiting; here we approximate that $$m\ll M,$$ so the mutual center of mass is about at the Earth's center and we can just use the Earth's mass as the mass being orbited (the really correct thing to do is to use the reduced mass $$\mu=\frac{Mm}{M+m},$$ which becomes approximately $$M$$ here).

Solving the force balance equation (*) out for v, we get
$$v=\left(\frac{GM}{a}\right)^{1/2}.$$

Now, the orbital period is $$P=2\pi a/v,$$

so substituting in v we find

$$P=\frac{2\pi a^{3/2}}{GM},$$
and finally squaring both sides we get Kepler's law.

In class, Prof. Johnson mentioned that the factor of $$4\pi^2$$ is hard to remember.  It should be easy now: it is actually just coming from the fact that the period is proportional to the circumference, $$2\pi a,$$  and so $$P^2$$ will have a $$4\pi^2$$ in it.