Worksheet 8, Problem 2: Total Potential Energy of a Cluster

Consider a spherical distribution of particles, each with a mass mi and a total (collective) mass M, and a total (collective) radius R. Convince yourself that the total potential energy, U, is approximately \(U\approx -\frac{GM^2}{R}\)
You can derive or look up the actual numerical constant out front. But in general in astronomy, you don’t need this prefactor, which is of order unity. 

\(M\) is the total cluster's mass

\(\rho\) is the density of the cluster

\(R\) is the radius of the cluster

\(r\) is the distance from a particle to the center the cluster

\(G\) is Newton's gravitational constant

\(U\) is the total potential energy of the cluster

Let's begin with an equation for \(U\):

\[dU=-\frac{GMdM}{r}\]

This is the regular gravity potential equation, but because the mass is distributed at different distances from the center of the cluster, it will experience a different gravity potential.  Let's find an expression for \(dm\) in terms of \(r\).  

Let's slice the cluster into hollow spheres stacked inside one another, each with depth \(dr\).  Each slice's mass can be given by:

\[dM=4\pi r^2\rho dr\]

This makes sense, since each has a depth \(dr\) and its length and width are given by the surface area of the shell. This infinitesimal volume should be multiplied by the density \(\rho\) to get mass.  Next let's explore the total mass in terms of \(r\) since each particle only experiences a gravity once from the mass closer to the center than itself.

\[M=\frac{4}{3}\pi r^3 \rho\]

This also is logical because the total mass inside a shell is equal to the volume of that sphere times the density.

So, substituting in these new terms we get:

\[dU=-\frac{16G\pi^2 r^4 \rho^2 dr}{3}\]

Let's integrate from 0 to \(R\) since this is our radius' range.

\[-\frac{16G\pi^2 \rho^2}{3} \int^R_0 r^4 dr=-\frac{16G\pi^2 \rho^2}{15}r^5|^R_0=-\frac{16G\pi^2 \rho^2}{15}R^5\]

Next let's get \(\rho\) out of this expression.  We know that it is density so:

\[\rho =\frac{M}{\frac{4}{3}\pi R^3}\]

Plugging this into our equation for \(U\) we get:

\[U=-\frac{16G\pi^2 (\frac{M}{\frac{4}{3}\pi R^3})^2}{15}R^5=-\frac{144GM^2pi^2R^5}{240pi^2R^6}\]

This mess simplifies beautifully to our answer:

\[\boxed{U=-\frac{3GM^2}{5R}}\]

This problem done in collaboration with G. Grell and S. Morrison.