## Fun with dimensional analysis 4 – Kepler’s Third Law

Given a body orbiting about a much larger body (so that $\displaystyle M >> m$) and assuming nearly circular orbits, we wish to know how the period $\displaystyle \tau$ varies as the distance $\displaystyle r$ of the small body (planet) from the larger body (star). The relevant physics comes from Newton’s Law of Gravitation:

$\displaystyle m \frac{d^2 \vec{r}}{dt^2} = - \frac{GMm}{r^2} \hat{r} \Rightarrow \vec{a} = -GM \frac{ \hat{r}}{ r^2} \\ \Rightarrow \lbrack a \rbrack = \lbrack GM \rbrack \lbrack r^{-2} \rbrack \Rightarrow \lbrack GM \rbrack = L^3 T^{-2}$

So we have

$\displaystyle \tau \sim (GM)^\alpha r^\beta \Rightarrow \lbrack \tau \rbrack = \lbrack GM \rbrack^\alpha \lbrack r \rbrack^\beta \Rightarrow T = L^{3 \alpha}T^{-2 \alpha} L^\beta \\ \Rightarrow 3\alpha + \beta = 0 \mbox{ and } -2\alpha =1 \Rightarrow \alpha = -\frac{1}{2} \mbox{ and } \beta = \frac{3}{2}$

Now we drop the $\displaystyle GM$ term from the equation since it’s just a constant and we arrive at

$\displaystyle \tau \propto r^{ \frac{3}{2} } \mbox{ or } \tau^2 \propto r^3$

Which is the result we now call Kepler’s Third Law. Interestingly, if the units are chosen in terms of astronomical units of distance and solar units of mass, the proportionality is exact and we have $\displaystyle \tau^2 = r^3$.