Blog #12 - Deriving Kepler's 3rd Law

Kepler's 3rd Law 

For a planet of mass m orbiting a star of mass \(M_{\star} \) at a distance of a in a circular orbit, start with the virial theorem and derive Kepler's Third Law of Motion. Assume that \(m << M_{\star} \) and remember that, since m is so small, the semimajor axis, which is formally \(a = a_p + a_{\star} \) reduces to \(a = a_p \). 

We begin with the virial theorem that states \[<t> = -\frac{1}{2}<U> \space \rightarrow \frac{mv^2}{2} = \frac{GM_{star}m}{2a} \]From this theorem we can divide out \(m\) and constants to get an expression for \(v^2\) as \[v^2 = \frac{GM_{star}}{a}\]This gives us a velocity in units \([\frac{m}{s}]\). We need to find the relationship between velocity and period with units [s]. We know that a period is the time that it takes a body to completely through its orbit, which is, in this case, circular. The distance traveled for a circular orbit is simply \[d = 2 \pi a \] and we know that time (period) is \[t = \frac{d}{v} \] So we divide our two terms to get the relationship between period and semimajor axis. \[P = \frac{d}{v} = \frac{2 \pi a}{\sqrt{\frac{GM_{\star}}{a}}} \]Squaring this equation we get the relationship\[P^2 = \frac{4 \pi^2 a^3}{GM_{\star}} \]Or more commonly, Kepler's 3rd Law of Motion \[P^2 \propto a^3 \]