Imagine a spherical shell of uniform thickness and density. Let X equal the strength of its surface gravity. Now double the shell's size while keeping its thickness and density the same. What's happened to its surface gravity?

We know that the gravitational force exerted by an object is proportional to its mass. And if we're keeping the shell's thickness and density the same, then its mass is proportional to its surface area. And its surface area is proportional to the square of its radius. So its mass is proportional to the square of its radius. So doubling its radius would increase its mass, and when not taking the inverse square law into account, its surface gravity, by a factor of 4.

But we also know that the gravitational force exerted by an object is inversely proportional to the square of the distance from its center of mass (the aforementioned inverse square law). So when not taking the change in mass into account, doubling the shell's radius would decrease its surface gravity by a factor of 4.

But these two effects exactly cancel out, don't they? Or am I missing something? Does this mean that the surface gravity of a spherical shell of constant thickness and density would be the same regardless of its size? So a 1-foot-thick shell with a radius of 10 feet would exert the same gravitational force on an object on its surface as a 1-foot-thick shell of the same material that's 10 times the size of Earth?

  • $\begingroup$ Have you tried doing a separate calculation for two different radii? That would be the easiest way to tell if they were different. But as you said you can write $M=M(r)=\rho 4 \pi r^2 \Delta r$ ($\Delta r$ being the thickness of the shell which is very small in comparison to the radii). $\endgroup$
    – Triatticus
    Jul 16, 2022 at 3:02
  • $\begingroup$ You answered the question in the question. Note that if it gravity got smaller (larger), then in the limit $r\rightarrow \infty$, you would expect and infinite slab to have a zero (infinite) field....which makes no sense. $\endgroup$
    – JEB
    Jul 16, 2022 at 4:32
  • $\begingroup$ Hint: Since the shell has uniform density and thickness, it is spherically symmetric and you can use Newton's Shell Theorem (en.wikipedia.org/wiki/Shell_theorem). $\endgroup$ Jul 16, 2022 at 20:06


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