Gravitational potential energy inside of a solid sphere [duplicate]

Question

I am self-studying classical mechanics. I came across a problem which required me to calculate the gravitational potential inside of a sphere. I found in one of my textbooks that the potential energy of a point P inside of a sphere is $$V=-\frac{GMm}{R}$$ where $M$ and $R$ are the mass and radius of the sphere. However, I also found this answer: Confusion over the gravitational potential energy inside a sphere in which the top answer gives a more complicated formula for the potential, which wouldn't agree when finding the PE. The one given in the link makes more sense to me, but I keep seeing different formulas for the same thing, and sometimes it's not specified whether it is the potential or the potential energy. I would like some clarification on what is going on (is the concentric sphere outside the point contributing to the potential or is it not; if so, why can't we just forget about the outer part of the sphere and just calculate it like P is on the surface of a smaller sphere?)

For context, I am working on solving the brachistochrone problem for a particle traveling between points through the Earth and need to find the velocity.

Answer 1

Potential energy is not a local concept: it's found by integrating the force from infinity to $r$, thus: any mass anywhere affects it everywhere.

The formula you gave is for a point source, not a sphere.

Since you're only concerned about the inside/surface of the sphere, the potential out in space is irrelevant. You can put the $0$ potential energy at $R$ so:

$$ V(R) = 0 $$

Then, take the force (per unit mass) at $r \le R$:

$$ g(r) = -G\frac{M(r)}{r^2} $$

where

$$M(r) = \frac 4 3 \pi r^3 \rho $$

is the mass inside the sphere of radius $r$. Spherically symmetric mass at larger radii do not contribute force.

Then compute a potential:

$$ V(r) = \int_{R}^{r \le R}g(r')dr' $$

which should be negative.

Answer 2

Let $\rho(r)$ be constant for $r\leq R$ and vanish for $r>R$. If $\rho(r)$ is radial, then $\Phi(r)$ will be as well. We can therefore consider the Possion equation in spherical coordinates: \begin{align*} \Delta\Phi(r) =\frac{1}{r}\frac{\mathrm d^2}{\mathrm dr^2}\big(r\Phi(r)\big) =4\pi G\rho &\Rightarrow \frac{\mathrm d^2}{\mathrm dr^2}\big(r\Phi(r)\big)=4\pi G\rho r \\ &\Rightarrow \frac{\mathrm d}{\mathrm dr}\big(r\Phi(r)\big)=2\pi G\rho r^2+a \\ &\Rightarrow r\Phi(r)=\frac{2}{3}\pi G\rho r^3+ar+b \\ &\Rightarrow \Phi(r)=\frac{2}{3}\pi G\rho r^2+\frac{b}{r}+a. \end{align*} (I didn't write $\rho(r)$ to avoid confusion when integrating, because it's constant for both $r\leq R$ and $r>R$.) We therefore get the family of solutions: $$\Phi(r)=\left\{\begin{array}{cc} &\frac{2}{3}\pi G\rho r^2+\frac{b}{r}+a & ;r\leq R \\ &\frac{d}{r}+c & ;r>R \end{array}\right.$$

We now have to find the parameters $a$, $b$, $c$ and $d$ to get the "physical" solution. Since a diverging gravitational potential at $r=0$ is an "unphysical" solution for $R>0$, we have $b=0$. As we also often take $\lim_{r\rightarrow\infty}\Phi(r)=0$ per convention in physics, we have $c=0$. (But you can also use $\Phi(R)=0$ like in the answer of JEB.)

Since $\Phi'(r)$ should be continuous at $r=R$, we have: $$\frac{4}{3}\pi G\rho R^3 \stackrel{!}{=}-\frac{d}{R^2} \Rightarrow d=-G\cdot\underbrace{\frac{4}{3}\pi\rho R^3}_{=M}=-GM.$$

Since $\Phi(r)$ should be continuous at $r=R$, we have: $$\underbrace{\frac{2}{3}\pi G\rho R^2}_{=\frac{GM}{2R}}+a \stackrel{!}{=}-\frac{GM}{R}=-\frac{4}{3}\pi G\rho R^2 \Rightarrow a=-2\pi G\rho R^2.$$

We therefore get the solution: $$\Phi(r)=\left\{\begin{array}{cc} &\frac{2}{3}\pi G\rho(r^2-3R^3) =GM\frac{r^2-3R^2}{2R^3} & ;r\leq R \\ &-\frac{GM}{r} & ;r>R \end{array}\right.,$$ whose inner solution is exactly the same as the link you've given (See here).