Saturday, March 7, 2015

Worksheet 8, Problem 1: Virial to Kepler


For a planet of mass m orbiting a star of mass M*, at a distance a, start with the Virial Theorem and derive Kepler’s Third Law of motion. Assume that m << MRemember that since m is so small, the semimajor axis, which is formally a = ap +areduces to a = ap.

\(M\) is the Mass of the Star
\(m\) is the Mass of the Planet
\(a\) is the distance between the Planet and the Star
\(G=6.7\times 10^8 \frac{cm^3}{gs^2}\) is Newton's gravitational constant
\(P\) is the period of the Planet
\(K\) is Kinetic Energy
\(U\) is Gravity Potential Energy
\(v\) is the planet's velocity







First let's start with the virial theorem.
\[K=-\frac{1}{2}U\]
We also know that \(K=\frac{1}{2}mv^2\) and that \(U=-\frac{GmM}{a}\).
Therefore:
\[\frac{1}{2}mv^2=\frac{GmM}{2a}\]
We can cancel the 1/2 and m.
\[v^2=\frac{GM}{a}\]
A planet's period is the time it takes to travel all the way around the planet so, \(P=\frac{2\pi a}{v}\). We can rearrange this to say that \(v=\frac{2\pi a}{P}\).
Substituting this in for v we get:
\[\frac{4\pi^2 a^2}{P^2}=\frac{GM}{a}\]
Simplifying and solving for P, we get:
\[\frac{4\pi^2 a^3}{P^2}=GM\]
\[\boxed{P^2=\frac{4\pi^2 a^3}{GM}}\]
Which is Kepler's Third law of Planetary Motion.

This problem was solved in collaboration with G. Grell and S. Morrison.

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