Worksheet 8, Problem 2: Going Virial

Consider a spherical distribution of particles, each with a mass \(m_i\) and a total (collective) mass \(\sum_{i}^{N} m_i = M\), and a total (collective) radius R. Convince yourself that the total potential energy, U, is approximately \[U \approx - \frac{GM^2}{R} \]
To begin this problem, let's start by drawing our system:

Now, let's recall Kepler's Law: \[ \frac{p^2}{R^3} = {4 \pi^2}{GM} \] Rearranging terms, we achieve:  \[ \frac{GM^2}{R} = {4 \pi^2 R^2 M}{p^2} \] Next, we'll want to consider what assumptions we can make about this system. First, we know that the volume of a sphere is represented by: \(V = \frac{4}{3} \pi r^3\). From this, we can draw that for a sphere, the majority of the volume is achieved at the greatest r values (near the surface). For this derivation, we'll assume that the majority of the particles are also at an average distance of R. Based on this, we can approximate the average period of the particles to be: \[average \, period = \bar{p} = \frac{2 \pi R}{\bar{v}} \] Next, we can use this to solve for average velocity, \( \bar{v} \). \[ \bar{v} = \frac{2 \pi R}{\bar{p}} \] To relate this average velocity of a particle to energy, we can now consider the kinetic energy, K, for a particle: \[ K = \frac{1}{2} m \bar{v}^2 \] Plugging in our average velocity of all particles, we get the total kinetic energy: \[ K = \frac{1}{2} M (\frac{2 \pi R}{\bar{p}})^2  = \frac{2M \pi^2 R^2}{\bar{p}^2} \] Next, we can apply the virial theorem. We recall this as: \[K = \frac{1}{2}U\] \[or\] \[U = -2K\] Applying this to our K equation from above, we achieve: \[U =   \frac{4M \pi^2 R^2}{\bar{p}^2} \] Recalling our rearranged form of Kepler's Law, we notice that this expression is equal to: \[ - \frac{GM^2}{R} \] And thus, our assertion is confirmed! The total potential energy, U, is approximately \[U \approx - \frac{GM^2}{R} \]

Acknowledgements: Johnathan Budd & Willie Pirc Team EE