3 Problems 1
An example question from the first problems class.
3.1 Gravitational potential
A uniform spherical mass, \(M\), has radius \(R\).
Calculate the Newtonian gravitational potential function \(\Phi(r)\), from the Poisson equation in the form
\[ \frac{1}{r^2} \frac{d}{dr}\left( r^2 \frac{d \Phi}{dr} \right) = 4 \pi G \rho \]
appropriate for a spherically-symmetric object, where \(\rho\) is the (constant) density of the mass.
The Poisson equation
\[ \frac{1}{r^2} \frac{d}{dr}\left( r^2 \frac{d \Phi}{dr} \right) = 4 \pi G \rho \]
for a Newtonian potential can be integrated once as
\[ r^2 \frac{d \Phi}{dr} = \int 4 \pi G \rho r^2 \, dr + C_1 \]
where \(C_1\) is a first integration constant (this is an indefinite integral). For a constant-density object of radius \(R\) there are two cases to consider — with \(r \le R\) and with \(r \ge R\).
For \(r \le R\),
\[ r^2 \frac{d \Phi}{dr} = C_1 + \frac{4}{3} \pi G \rho r^3 \]
where \(C_1\) is the first constant of integration. Then
\[ \Phi(r) = \int \frac{C_1}{r^2} \, dr + C_2 + \frac{4}{3} \pi G \rho \int r \, dr = -\frac{C_1}{r} + C_2 + \frac{2}{3} \pi G \rho r^2 \, .\]
The \(C_1\) term represents the effect of a point mass at the centre of the body, but we don’t have one here, so \(C_1 = 0\). \(C_2\) will be found by matching solutions with the exterior solution: being a constant added to the potential, it doesn’t represent a gravitational force.
At \(r \ge R\),
\[ r^2 \frac{d \Phi}{dr} = D_1 + \frac{4}{3} \pi G \rho R^3 \]
where we apparently need \(D_1\) as a constant of integration. Rearrange and integrate again,
\[ \Phi(r) = -\frac{D_1}{r} - \frac{4}{3} \frac{\pi G \rho R^3}{r} + D_2 \, .\]
Here we can set \(D_2 = 0\), since by convention we put \(\Phi \rightarrow 0\) as \(r \rightarrow \infty\). \(D_1\) can also be taken as zero since it also represents a gravitational potential from a point mass at \(r = 0\), and none is present.
It remains to match the solutions at \(r = R\), so that there isn’t an infinite force felt there. This means that
\[ C_2 + \frac{2}{3} \pi G \rho R^2 = - \frac{4}{3} \pi G \rho R^2 \]
so
\[ C_2 = - 2 \pi G \rho R^2 \]
and the overall solution is
\[ \Phi(r) = \begin{cases} - 2 \pi G \rho R^2 \left( 1 - \frac{1}{3} \left( \frac{r^2}{R^2} \right) \right) & r \le R \\ - \frac{4}{3} \frac{\pi G \rho R^3}{r} & r \ge R \end{cases} \]
which can be written in terms of mass as
\[ \Phi(r) = - \frac{G M}{R} \begin{cases} \frac{3}{2} - \frac{1}{2}\left( \frac{r^2}{R^2} \right) & r \le R \\ \frac{R}{r} & r \ge R \end{cases} \]
The maximum (negative) potential is at \(r = 0\), and has value
\[ \Phi_0 = -\frac{3}{2} \frac{G M}{R} \, . \]