There are several previous questions concerning concentric shells. I won't reference them here because this is different.

I understand that there is no gravitational effect inside a concentric shell*. But what about a non-concentric one?

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I've looked online and found nothing. Maybe I'm just using the wrong search terms?


I'd like to have a cavity inside a small planet. Assuming perfect spheres and uniform density, is there a general equation for the gravitational field inside the cavity, taking into account:

  • The radius of the solid sphere
  • The radius of the hollow spherical cavity
  • The offset between centres.


If no exact solution exists, is there an approximate formula that will let me play with the variables to get a rough idea of the effects?

*Shell Theorem

In classical mechanics, the shell theorem gives gravitational simplifications that can be applied to objects inside or outside a spherically symmetrical body. This theorem has particular application to astronomy ... A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre. If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell.

  • $\begingroup$ You have been amazingly prolific lately. 1) How do you come up with these questions? and, 2) Are these questions 'real' or just a way to give us something to do at work? 8^D $\endgroup$ Commented Nov 30, 2018 at 17:00
  • $\begingroup$ No time for a real answer, but Gauss’ law is a good thing to research here. $\endgroup$
    – Joe Bloggs
    Commented Nov 30, 2018 at 17:14
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    $\begingroup$ @Tracy Cramer - I'm one of those people who comes up with loads of ideas but can never be bothered to turn them into stories. I'd make a terrible author. This my outlet for an over fertile imagination! P.S. If anyone wants to use any of my ideas, they're welcome to :-) P.P.S. I've actually got a backlog, I've just got to formulate them properly. $\endgroup$ Commented Nov 30, 2018 at 20:16
  • $\begingroup$ As a bit of a simplifcation for the fabulous answers you're getting. Take a compass and draw a circle from the center of the cave with a radius from that point to the closest outer surface. Remove all of that. Per the shell theorem, it all cancels out. What's left is the gravitational force you might feel (enter the complex mathematics). $\endgroup$
    – JBH
    Commented Nov 30, 2018 at 20:43

2 Answers 2


The solution isn't actually too bad; I had this as a problem in AP physics in high school. It doesn't have as much symmetry to exploit as a concentric shell, but it still has a lot of symmetry--as long as the cave and the enclosing body are both spheres, and the enclosing body has uniform density, you can treat everything as rotationally symmetric around the line connecting their centers.

From there, you can treat the cave as a body of negative mass whose gravity is added to that of the enclosing body. The somewhat surprising end result is that gravity is constant inside a spherical cave, and antiparallel to the offset of the cave center from the enclosing body's center.

Because of the rotational symmetry, we can reduce the problem to two dimensions to show that the field is in fact constant everywhere in the cave.

Let the enclosing body have radius $R$ and density $\rho$, the offset between centers be $d < R$, and the cave have radius $r < R - d$. The gravitational field inside a body of uniform density is $g \propto \rho x$. Expanded to 2 dimensions, the total gravitational force is $g \propto \rho \sqrt{x^2+y^2}$, but broken down into vector components we get $g_x \propto \rho \sqrt{x^2+y^2}\cos\theta = \rho \sqrt{x^2+y^2} \frac{x}{\sqrt{x^2+y^2}} = \rho x$, and similarly $g_y \propto \rho y$. The gravity due to the negative body that creates the cave when added to the enclosing body is $g_x \propto -\rho (x-d)$ and $g_y \propto -\rho y$. Adding these together, we get the summed components for the net gravity vector to be $g_x \propto \rho x - \rho (x - d) = \rho (x - (x - d)) = \rho d$ (i.e., a non-zero constant), and $g_y \propto \rho y - \rho y = 0$.

So, the total gravity is axis-aligned, constant, and dependent only on the density of the enclosing sphere and the eccentric distance. This is in fact the only way I know of to get an exact constant, uniform gravitational field with finite amounts of material (the infinite option being the space above an infinite plane).

To get the actual force (not just a factor it is proportional to), add a factor of $\frac{4 \pi G}{3}$ to get $g = \frac{4 \pi G}{3}\rho d$.

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    $\begingroup$ Pretty sure this is right, got a headache in the process, so +1 and + resentment for throbbing sensation. Glad the maths was for a sphere, not oblate spheroid, or cavity "arbitrary amount off axis of spin" - noone needs that migrane. $\endgroup$ Commented Nov 30, 2018 at 20:16
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    $\begingroup$ I couldn't handle reading all those $o$'s, so I changed $o$ to $d$. You can roll back if you disapprove. $\endgroup$
    – kingledion
    Commented Nov 30, 2018 at 20:22
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    $\begingroup$ @Duckisaduckisaduck next up I'll need you to calculate this for a squircle. $\endgroup$ Commented Nov 30, 2018 at 22:03
  • $\begingroup$ @Lord Farquaad Cuberoid surely? Damn, the throbbing's worse. Especially since someone will mention "no extended conversations" shortly, no doubt. $\endgroup$ Commented Nov 30, 2018 at 22:11
  • $\begingroup$ So $x$ and $y$ are in a coordinate system where the origin is the center of the enclosing sphere, I guess? And the cave's center is on the $x$ axis? $\endgroup$
    – LarsH
    Commented Nov 30, 2018 at 22:24

This is a classic problem in electrostatics - that is, in an analogous situation where we care about calculating the electric force on an object inside some cavity. The same solution technique applies for Newtonian gravity, and it relies on something called superposition. Effectively, the cavity is like a region of space inside a sphere of mass density $\rho$ centered at a point $\mathbf{p}$, inside which you've placed a smaller sphere of mass density $-\rho$ centered at a point $\mathbf{p}'$. In the region of intersection, the two densities cancel out, leaving you with a net density of $0$.

A simpler case

Say we have a body with uniform density $\rho$. We can use something called Gauss's law for gravity. This shouldn't be confused with its cousin, Gauss's law for electrostatics - usually just called "Gauss's law" - or the underlying mathematical theorem behind them both, referred to as the divergence theorem or Gauss's theorem. Regardless of how you want to refer to it, the law goes like this: $$\int_{\mathcal{S}}\mathbf{g}\cdot\mathrm{d}\mathbf{A}=-4\pi G\int_{\mathcal{V}}\rho\mathrm{d}V$$ where $\mathcal{V}$ is a surface with boundary $\mathcal{S}$, $\mathbf{g}$ is the gravitational field and $d\mathbf{A}$ is an area element. Then, in our case of a uniform sphere, $$g(r)\cdot4\pi r^2=-4\pi G\rho\frac{4\pi}{3}r^3$$ and $$g(r)=-\frac{4\pi G\rho}{3}r$$ We know that $\mathrm{d}\mathbf{A}$ points radially outwards, so does $\mathbf{g}$ by spherical symmetry, and so $$\mathbf{g}(\mathbf{r})=-\frac{4\pi G\rho}{3}\mathbf{r}$$ as claimed.

Modeling the cave

The principle of superposition says that to calculate the gravitational field due to two objects, we can simply add the gravitational fields created by each object. Let's call these fields $\mathbf{g}_+$ and $\mathbf{g}_-$, coming from the sphere of density $\rho$ and the sphere of density $-\rho$, respectively. Now we just apply the result from the last section: $$\mathbf{g}_+(\mathbf{r})=-\frac{4\pi G\rho}{3}(\mathbf{r}-\mathbf{p}),\quad\mathbf{g}_-(\mathbf{r})=\frac{4\pi G\rho}{3}(\mathbf{r}-\mathbf{p}')$$ The total gravitational fields is then $$\mathbf{g}(\mathbf{r})=\mathbf{g}_+(\mathbf{r})+\mathbf{g}_-(\mathbf{r})=-\frac{4\pi G\rho}{3}(\mathbf{p}-\mathbf{p}')$$ which is constant, though non-zero. Notice that if the spheres are concentric, $\mathbf{p}-\mathbf{p}'=\mathbf{0}$ and the field vanishes - the same result as the good old shell theorem.

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    $\begingroup$ Terminology alert! You're actually applying divergence theorem (which, according to Wikipdia, is itself sometimes called Gauss's theorem, though I've never heard that) to gravity, whereas Gauss's law is application of the same divergence theorem to electrostatics. Divergence theorem is itself a specialized case of Stokes' theorem, which is how I would have described your method. $\endgroup$
    – kingledion
    Commented Nov 30, 2018 at 20:18
  • $\begingroup$ @kingledion I've always heard this particular usage referred to as Gauss's law, too (note the Wikipedia page I linked too, as well as many notes online on this particular problem) - perhaps with the qualifier that it's Gauss's law for gravity. In my mind, the divergence theorem is the more general case of an arbitrary vector field $\mathbf{F}$, without identifying, say, $\nabla\cdot\mathbf{F}$ or $\mathbf{F}\cdot\mathbf{\hat{n}}\mathrm{d}A$ as another physical quantity of interest. $\endgroup$
    – HDE 226868
    Commented Nov 30, 2018 at 20:23
  • $\begingroup$ Wow, there is a "Gauss's law for gravity." They will name anything after that guy, I swear. That is really a super confusing terminology, given that the mathematical principals have their own names... $\endgroup$
    – kingledion
    Commented Nov 30, 2018 at 20:26
  • $\begingroup$ @kingledion Honestly, it's almost as confusing as the term "Euler's equation", which can apply to half a dozen different identities and laws. . . I'll make some edits to make things maybe a bit clearer, though - thanks for the input. $\endgroup$
    – HDE 226868
    Commented Nov 30, 2018 at 20:28
  • $\begingroup$ @HDE226868: Indeed. There's an only half-joking saying that mathematical discoveries are usually named after the first person to discover them after Leonhard Euler, because otherwise nearly everything would be named "Euler's theorem/equation/whatever". The guy really was a genius, and worked on incredibly many things. And the same goes for Gauss, too. $\endgroup$ Commented Dec 1, 2018 at 14:56

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