Worksheet 2.1, Problem 4: Kepler's Law on a Desert Island

The problem: Using dimensional analysis alone, what is the form of Kepler’s law, which relates the period, total mass and separation of a two-body gravitational orbit? Use only the units and a tad of physics, e.g. what constants are likely involved?

What we have:

$$P \: \alpha \: M\cdot a$$ where P is period, M is mass, and a is the separation of a two-body gravitational orbit.

What we know:

Because gravity is involved, we know G will be used, which is $G = 6.67 x 10^{-12}\frac{cm^3}{gs^2}$.

We also know the units:

Period: T (time), M: M (mass), a: L (distance) and finally G: $\frac{L^3}{MT^2}$ (a combination of the three).

So we can set up a dimensional analysis problem with only the exponents to work out:

$$P^w=M^x\cdot a^y\cdot G^z       \rightarrow      T^w=M^x\cdot L^y\cdot (\frac{L^3}{MT^2})^z$$

For the units of time (T) to balance out on both sides of the equation, w must equal 2 and z must equal -1. With this inversion, for the units for distance to work out (L), y must equal 3. And finally, for mass to cancel (M), x must equal -1.

In conclusion, w=2, x=-1, y=3, and z=-1, which when put back into the first equation gives:
$$P^2=M^{-1}\cdot a^3\cdot G^{-1}$$

which cleans up to:
$$P^2=\frac{a^3}{M\cdot G}$$

All done!

P.S. Because I do not live on a desert island, I looked up the formula for Kepler's law:
$$P^2=\frac{4\pi^2 a^3}{MG}$$
and I was only off by a constant!

Many thanks to Sean and Eden for working on this worksheet with me.