When calculating the potential energy of an assembled sphere, for example gravitational binding energy for gravity or electrostatic energy of a sphere for the electric force, we usually start by calculating the potential energy required to bring the mass/charge that an infinitely thin shell has at a specific radius $r$ from infinity to that radius $r$. This procedure is then repeated for every shell by integrating from $0$ to the radius $R$ of the sphere.

Why would we not consider the change in potential energy for every spherical shell as the charges it consists of are getting closer to each other?

Edit: It seems my question is unclear, so I'll try to clarify what my confusion is about. Consider a shell at radius $r$ with mass $m_{shell} = 4 \pi r^2 \rho dr$. Taking this shell from infinity to $r$ is ascribed a change in potential of $$dU = - \frac{m_{shell} m_{inside}}{r}$$ But moving the shell from infinity to $r$ means we change the density of the shell from infinitesimally small to $\rho$, doesn't it? Shouldn't there be energy associated with that?

  • $\begingroup$ More on Newton's shell theorem. $\endgroup$
    – Qmechanic
    Commented Jun 29, 2021 at 20:34
  • $\begingroup$ Using gauss's law, find the electric field due to a spherical shell distribution a distance r from the center of the sphere, and try $r \to R$ $\endgroup$
    – g s
    Commented Jun 29, 2021 at 21:04
  • $\begingroup$ @Qmechanic I haven't dived really deep into derivations of Newton's shell theorem, but all relatively simple ones that I've seen didn't mention anything of what my question is about. Is my question unclear? $\endgroup$
    – reveance
    Commented Jun 29, 2021 at 21:05
  • $\begingroup$ Edited, hopefully makes it a bit clearer $\endgroup$
    – reveance
    Commented Jun 29, 2021 at 21:29

1 Answer 1


It might be easier to imagine pulling apart a body of constant density $\rho$ and taking each shell from radius $r$ to infinity (the resultant answer will be the negative of the binding energy).

Your formula $m_{shell} = 4 \pi r^2 \rho dr$ is the mass of the shell removed, of density, $\rho$ and thickness $dr$, this mass is taken to infinity and there is no need to consider any density changes.

  • $\begingroup$ Thanks for your answer. Why wouldn't you need to consider these density changes here though? Taking each shell from $r$ to infinity would mean you do extra work by decreasing the density of the shell, I don't think this is included in $dU = - \frac{m_{shell} m_{inside}}{r}$ because that formula holds for moving a point mass away from the interior, which you are in this case doing in addition to moving point masses away tangentially to other masses that make up the shell. $\endgroup$
    – reveance
    Commented Jun 29, 2021 at 22:23
  • $\begingroup$ Maybe its more clear to think about picking the shell apart piece by piece too, in addition to picking the sphere apart shell by shell. Then we have the gravitational force acting on the pieces we're moving to infinity from the remainder of the shell (in addition to the interior), which would be the same as the potential required to pick the shell apart (changing it's density to infinitesimally small) if I'm not mistaken? $\endgroup$
    – reveance
    Commented Jun 29, 2021 at 22:35
  • 1
    $\begingroup$ It's because each shell that is removed has infinitesimally small mass and so the reduction in density changing it's own GPE can be ignored (it'll be a $\delta m^2$ term instead of a $m\times \delta m$ term). $\endgroup$ Commented Jun 30, 2021 at 6:39
  • $\begingroup$ Thank you! Conceptually it's still kind of bugging me, but I know where to look for now, great! $\endgroup$
    – reveance
    Commented Jun 30, 2021 at 18:41

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