I am reading Kolenkow and Kleppner's Classical Mechanics and they have tried to calculate the gravitational force between a uniform thin spherical shell of mass $M$ and a particle of mass $m$ located at a distance $r$ from the center.

The shell has been divided into narrow rings.$R$ has been assumed to be the radius of the shell with thickness $t$ ($t<<R$). The ring at angle $\theta$ which subtends angle $d\theta$ has circumference $2\pi R\sin\theta$. The volume is $dV=2\pi R^2t\sin \theta d\theta$ and its mass is $pdV=2\pi R^2t\rho\sin\theta d\theta$. If $\alpha$ be the angle between the force vector and the line of centers, $$dF=\frac{Gm\rho dV}{r'^2}\cos\alpha$$ where $r'$ is the distance of each part of the ring from $m$.

Next, an integration has been carried out using $\cos\alpha=\frac{r-R\cos\theta}{r'}$ and $r'=\sqrt{r'^2+R^2-2\pi R\cos\theta}$.

Question: I would like to avoid these calculations and I was wondering if there exists a better solution.

  • $\begingroup$ I'm flabbergasted every time I see ($\theta$,$\alpha$). Doing things without spherical symmetry is very hard. We do have questions about field/potential around an arbitrary ring, so in theory you could integrate that but it would be hilariously round-about since you'd be using the result from a way harder problem to solve a very simple problem. when you wrote $dF$, you would need a double-integral to make this work, $d\theta$ and $d\alpha$. Your expression for $dV$ in that context is almost certainly wrong. $\endgroup$ Commented Mar 9, 2013 at 19:17

2 Answers 2


This is basically a proof of the Shell theorem

Simple. First, you must know that gravitational forces superpose.

Ok. First, calculate the force at a distance $r$ from the origin for a sphere of mass $M$ and radius $R$. You ought to get the result that it is $\frac{GMm}{r}$ if $r\geq R$, 0 otherwise. This is a standard textbook formula.

Ok, now, pretend the shell is made up of a larger sphere with positive mass and a smaller sphere with "negative" mass that cancels out all the mass except a shell of thickness $t$ on the outside. If you feel uncomfortable doing this, convert it into an electrostatics problem, calculate the formula with a similar method, convert it back.

Knowing $M,R,t$, you can calculate the masses and radii of the component spheres. Both have equal magnitudes of density.

Now, calculate the force at $r$, and superpose the two, taking care to subtract the force of the inner "negative" sphere.

That's it. You have (ab)used the principle of superposition to calculate this force, which is the final answer (it will have three cases -- the gravitational field inside will be 0).

The harder you try to avoid integration, the more fun newtonian mechanocs becomes :)

I'm on mobile right now (hard to add math) if you want I'll put more math in the answer tomorrow.

  • 1
    $\begingroup$ The question is discussing a proof of the shell theorem. $\endgroup$
    – user4552
    Commented Jun 7, 2013 at 20:57
  • $\begingroup$ @BenCrowell: Right. And my answer gives exactly that, read it more carefully (Though I had made a mistake in mentioning the ST was "needed", fixed. Thanks for the headsup). I didn't want to give the full mathematics due to the HW policy, but I gave a method which lets you derive the ST with relative ease. $\endgroup$ Commented Jun 8, 2013 at 5:12
  • 1
    $\begingroup$ Your force is missing a negative, and should depend on 1/r^2 rather than 1/r. Also the force inside the solid sphere is proportional to r, not 0. $\endgroup$ Commented Apr 2, 2014 at 10:41

Yes, there does exist an easier method though the reasoning is quite subtle. Because you are reading Kleppner, I hope we can take for granted the formula that the field outside a spherical mass is exactly the same as due to an equal pt. mass kept at the center of the shell (you can also use the Gauss's law for gravitation to prove this).

Now, we make a very intelligent use of Newton's third law. We are to calculate the force exerted on a sphere of mass M by a particle of mass m located at a distance r from the center. Instead of doing that, we find the force exerted by the sphere on mass m. But as long as its field is being concerned, the sphere just behaves as an equal point mass kept at the center (as given above). So, the force on mass m is simply GMm/r^2. This, by Newton's Third Law, is also the force of mass m on M.

Hope you have got your answer.

  • $\begingroup$ You knew the sphere behaves as a point mass and so used that. That is not explaining OP's question and rather using the formula to prove that formula kind of thing $\endgroup$
    – Ham Lemon
    Commented Nov 4, 2023 at 12:20

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