$\newcommand{\oiint}{\iint\limits_{\partial V}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\;\;\;\subset\!\supset }$

Imagine a ball of dust of total mass $M$ and radius $R$. The total mass of the ball remains uniform as $R$ changes. We will use $r$ as the radial coordinate, so that in general $0<r<R$.

Write down the gravitational potential $V(r)$ inside and outside the ball (using Gauss' theorem for gravity), as defined here.

For the outside, I have the following: $$\oiint \mathbf{g} \cdot d\mathbf{A} = -4\pi GM \implies \int_S \mathbf{g} \cdot {\hat {\mathbf {n} }} dS = g(r) \int_S dS = g(r) 4\pi r^2 = -4\pi GM \implies g(r) = -\frac{GM}{r^2}. \; V(r)={\frac {U_g}{m}}={\frac {1}{m}}\int \limits _{\infty }^{r} F_g dr ={\frac {1}{m}}\int \limits _{\infty }^{r}{\frac {GmM}{r^{2}}}dr=-{\frac {GM}{r}}$$

I believe this is the right answer, though I'm not 100% on the work. However, I tried to do some research on such an equation for gravitational potential inside of the ball. I eventually found $$\frac{GMr^2}{2R^3} - \frac{3GM}{2R}$$ somewhere on Wikipedia, but I can't find the page for it nor do I understand the derivations. I read that it had something to do with Poisson's equations? However, I'm not too familiar with the Laplacian, so I was wondering if I could get an explanation on all these notions.

  • $\begingroup$ Sorry for the syntax error, I forgot that the {esint} package is not integrated into StackExchange. \oiint refers to a closed surface integral. $\endgroup$ Commented Jul 14, 2021 at 6:47
  • $\begingroup$ Hello! I have (kind of) made the \oiint work using the method from this post. See \oiint doesn't seem to work for more information. $\endgroup$
    – jng224
    Commented Jul 14, 2021 at 8:35

1 Answer 1


You can complete your task using relatively elementary Physics. Here's how...

(a) You have already used Gauss's law, with a Gaussian surface of radius $r\ (> R)$, to show that a spherically symmetric body behaves, as 'seen' from a point outside it, as if all its mass were concentrated at its centre.

(b) You can also use Gauss's law, with a Gaussian surface of radius $r\ (<R)$, to show that, inside the spherical mass distribution, $$g(r) =\ –\frac{GMr^3/R^3}{r^2}=\ –\frac{GMr}{R^3}$$ In other words the mass, $Mr^3/R^3$ in a uniform distribution, that is closer than $r$ to the centre behaves as if it were concentrated at the centre. The mass 'outside' $r$ has no gravitational effect at $r$.

(c) It is now very easy to calculate the work done when you change your position within the uniform spherical distribution. So you can calculate the work going from $r\ (<R)$ to infinity (or vice versa), because you can add the work going from $r\ (<R)$ to $R$ to that in going from $R$ to infinity, which you know from (a)! This gives you the expression that you "eventually found".


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