I came across this question during my physics class. Suppose we have a solid, spherical planet with mass $M$ radius $R$ s.t. the density of this planet is uniform everywhere, then what is the gravitational potential at its center. (Assume that there is no external force that affects this system)

Approach one:

Suppose $r<R$ is the distance of the point from the center of the planet, then we know that $$ V_r=-\dfrac{G\rho\frac{4}{3}\pi r^3}{r}=-\dfrac{4}{3}G\rho\pi r^2=-\dfrac{GMr^2}{R^3} $$ which means $V_r\propto r^2$, so that $V_0=0$ (the gravitational potential at the center is zero)

Approach two:

By definition, gravitational potential at a point is the work done per unit mass to move an object from infinitely far to that point, so $$ V_r=\int_\infty^r\vec{E}\cdot d\vec{l}=\int_\infty^R\dfrac{GM}{l^2}dl+\int_R^r -\dfrac{GM l}{R^3}dl=-\dfrac{GM}{R}-\left(\dfrac{GMr^2}{2R^3}-\dfrac{GM}{2R}\right)=-\dfrac{GM}{2R}-\dfrac{GMr^2}{2R^3} $$ suggesting that $V_0\neq 0$, which contradicts result obtained by the previous method.

Interestingly, if plug in $r=R$, then Approach one agrees with Approach two, so I'm kind of confused about which method is correct.

Can someone please explain what's wrong? Thanks in advance.


1 Answer 1


There is actually a mistake in both your methods, although you were closer with your second approach. In your first method, your formula simply isn't valid. The corollary of the shell theorem, that gravitational field inside a solid sphere is only dependent upon the part of the sphere closer to the centre than the point of consideration, which you seem to have tried to use, is for calculating $\vec g$ and not potential. So, you are basically not counting the work done by the outer layers of the ball in bringing point mass from a point just outside the sphere to the point at $r$ distance from centre.

In your second method, you have taken a wrong definition of potential. Potential at a point is the work done by $\mathbf {external}$ $\mathbf {agent}$ in bringing a unit mass particle from $\infty$ to that point. So take $V_r=-\int {\vec E. d\vec l}$. Keep in mind the direction of the field and the direction of elemental displacement. Your final answer should come out to be: $$V_r=-\frac {3GM}{2R}$$

  • $\begingroup$ I understand your method, but it's not the case. Your method only works for sphere but my question is about solid ball. Sphere does not include the interior but solid ball does. $\endgroup$
    – Kevin.S
    Commented May 18, 2021 at 5:54
  • $\begingroup$ I am talking about a solid ball too. I edited my answer for more clarity. $\endgroup$ Commented May 18, 2021 at 5:54
  • $\begingroup$ For a sphere, by which you mean a shell, $V_r=-\frac {GM}{R}$ because there is no field inside. $\endgroup$ Commented May 18, 2021 at 5:55
  • $\begingroup$ OK, probably I made some mistakes, but let me try to do it again before accepting your answer. Thank you for your attention and your explanation. :) $\endgroup$
    – Kevin.S
    Commented May 18, 2021 at 5:59
  • $\begingroup$ Sure, do read the edit once. $\endgroup$ Commented May 18, 2021 at 6:00

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