The Shell Theorem - A Gravitational Analog to Gauss’ Law

Below I will motivate the Shell Theorem, a powerful and useful result that greatly simplifies gravitational problems involving spherical mass distributions. To illustrate its importance, I will start with a simple example and then present a rigorous proof. This approach will help build intuition and show why the theorem is so widely applied in classical mechanics and astrophysics.

Dropping to center of earth

How long does it take for an object to fall to the Earth’s center, and what motion will it undergo afterward?

The force acting on the ball is

\[F=G\frac{M'm}{r^2},\]

where:

\[M'=M\frac{r^3}{R^3}.\]

Then, we have

\[\begin{aligned} m\ddot{r}&=G\frac{Mm}{R^3}r \\ \implies \ddot{r}&=G\frac{M}{R^3}r, \end{aligned}\]

where we immediately recognise this as a simple harmonic motion with frequency

\[\omega=\sqrt{\frac{GM}{R^3}}.\]

This corresponds to a period of

\[T=2 \pi \sqrt{\frac{R^3}{GM}}.\]

Dubious Assumption?

The astute reader will notice that this assumes the gravitational force is only contributed from the enclosed mass. Why is this true?

Gauss’ Law for Gravitation

Gauss’ Law for electromagnetism (Maxwell’s first equation) is well known, but Gauss’ Law for gravitation is equally fundamental.

It states that the divergence of the gravitational field \(g\) is proportional to the negative of the mass density \(\rho\):

\[\nabla \cdot \mathbf{g} = -4\pi G \rho\]

We can derive this from Newton’s law of universal gravitation, where the gravitational field at position \(\mathbf{r}\) due to a mass distribution \(\rho(\mathbf{s})\) is:

\[\mathbf{g}(\mathbf{r}) = -G \int \frac{\rho(\mathbf{s}) (\mathbf{r} - \mathbf{s})}{|\mathbf{r} - \mathbf{s}|^3} \, d^3\mathbf{s}.\]

Taking the divergence of \(\mathbf{g}(\mathbf{r})\), we have

\[\begin{aligned} \nabla \cdot \mathbf{g}(\mathbf{r}) &= -G \int \rho(\mathbf{s}) \left[ \nabla \cdot \frac{\mathbf{r} - \mathbf{s}}{|\mathbf{r} - \mathbf{s}|^3} \right] d^3\mathbf{s} \\ &= -4\pi G \int \rho(\mathbf{s}) \delta(\mathbf{r} - \mathbf{s}) \, d^3\mathbf{s} \\ &= -4\pi G \rho(\mathbf{r}), \end{aligned}\]

where use used the identity

\[\nabla \cdot \left(\frac{\mathbf{r} - \mathbf{s}}{|\mathbf{r} - \mathbf{s}|^3} \right) = 4 \pi \delta ( \mathbf{r} - \mathbf{s}).\]

Hence, we arrive at the differential form

\[\nabla \cdot \mathbf{g} = -4\pi G \rho.\]

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Integral Form (Using the Divergence Theorem)

Applying the divergence theorem to the differential form, we obtain the integral form of Gauss’ Law for gravitation:

\[\oint_{\partial V} \mathbf{g} \cdot d\mathbf{A} = \int_V (\nabla \cdot \mathbf{g}) \, d^3\mathbf{r} = -4\pi G \int_V \rho(\mathbf{r}) \, d^3\mathbf{r} = -4\pi G M_{\text{enc}}.\]

This result tells us that the total gravitational flux through a closed surface depends only on the mass enclosed by that surface—just as in electrostatics, where the electric flux depends only on the enclosed charge.

Exercises for the reader

• Show that \(\nabla \cdot (\frac{\mathbf{r}}{\lVe rt\mathbf{r}\rVert})^3=\left(\frac{1}{r^2}\right)(0)=0\) for \(r \neq 0\). Use \(\nabla \cdot \mathbf{A}=\frac{1}{r^2} \frac{\partial\left(r^2 A_r\right)}{\partial r}+\frac{1}{r \sin \theta} \frac{\partial}{\partial \theta}\left(A_\theta \sin \theta\right)+\frac{1}{r \sin \theta} \frac{\partial A_{\varphi}}{\partial \varphi}.\)

• Use divergence theorem to show what happens at \(r=0\). Recall that \(d \mathbf{A} = a^2 \hat{\mathbf{r}} \ d \Omega\), where \(d \Omega = \sin \theta \ d \theta \ d \phi\), with \(\theta \in [0, \pi)\) and \(\phi \in [0, 2 \pi).\)