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 ${\mathbf{p}}^{\prime }$. 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, $$g(r)\cdot 4\pi r^2=-4\pi G\rho \frac{4\pi }{3}{r}^3$$ 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: $${\mathbf{g}}_{+}(\mathbf{r})=-\frac{4\pi G\rho }{3}(\mathbf{r}-\mathbf{p}),{\mathbf{g}}_{-}(\mathbf{r})=\frac{4\pi G\rho }{3}(\mathbf{r}-{\mathbf{p}}^{\prime })$$ The total gravitational fields is then $$\mathbf{g}(\mathbf{r})={\mathbf{g}}_{+}(\mathbf{r})+{\mathbf{g}}_{-}(\mathbf{r})=-\frac{4\pi G\rho }{3}(\mathbf{p}-{\mathbf{p}}^{\prime })$$ which is constant, though non-zero. Notice that if the spheres are concentric, $\mathbf{p}-{\mathbf{p}}^{\prime }=\mathbf{0}$ and the field vanishes - the same result as the good old shell theorem.

posted over 6 years ago

HDE 226868‭

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