# Internal and external gravitational potentials $V(r)$ of a spherical object of matter of constant total mass $M$, and variable radius $R$

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

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.

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.

• Sorry for the syntax error, I forgot that the {esint} package is not integrated into StackExchange. \oiint refers to a closed surface integral. Jul 14, 2021 at 6:47
• 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. Jul 14, 2021 at 8:35

(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 $$r\ (, to show that, inside the spherical mass distribution, $$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\ ( to infinity (or vice versa), because you can add the work going from $$r\ ( to $$R$$ to that in going from $$R$$ to infinity, which you know from (a)! This gives you the expression that you "eventually found".