Newtonian gravity between point masses follows an inverse-square law. For extended bodies, the gravitational field is the vector sum of the contributions from all mass elements. For spherically symmetric mass distributions, that sum has especially simple consequences known as Newton's shell theorem.

Thin spherical shell

For a thin spherical shell of total mass $M$ and radius $R$:

• at any point outside the shell, the gravitational field is exactly the same as if all the shell's mass were concentrated at its center;
• at any point inside the shell, the net gravitational field from the shell is zero.

Thus, for $r>R$,

$$\boxed{\mathbf g(r)=-\frac{GM}{r^2}\hat{\mathbf r}},$$

while for $r<R$,

$$\boxed{\mathbf g(r)=\mathbf0}.$$

The zero interior field is not because every nearby mass element is equally distant. Nearby pieces pull more strongly, but they occupy a smaller solid angle than the more distant pieces on the opposite side. For the inverse-square law, these effects cancel exactly.

Any spherically symmetric body

A spherically symmetric body can be imagined as many concentric thin shells. At radius $r$ inside the body:

• shells lying at radii larger than $r$ contribute zero net field at that point;
• the mass enclosed within radius $r$ acts gravitationally as if it were concentrated at the center.

If $M(r)$ is the mass enclosed within radius $r$, then

$$\boxed{\mathbf g(r)=-\frac{G M(r)}{r^2}\hat{\mathbf r}}.$$

Outside the entire body, $M(r)=M_{\rm total}$, so

$$\mathbf g(r)=-\frac{GM_{\rm total}}{r^2}\hat{\mathbf r}.$$

This is why a spherically symmetric planet or star produces the same external gravitational field as a point mass with the same total mass located at its center.

Uniform solid sphere

Consider a sphere of radius $R$, total mass $M$, and uniform density. The density is

$$\rho=\frac{M}{\frac43\pi R^3}.$$

At radius $r<R$, the enclosed mass is

$$M(r)=\rho\frac43\pi r^3 =M\frac{r^3}{R^3}.$$

Therefore the field magnitude inside is

$$g(r)=\frac{G M(r)}{r^2} =\boxed{\frac{GM}{R^3}r}.$$

Inside a uniform sphere, gravity increases linearly with distance from the center. At the center,

$$g(0)=0,$$

and at the surface,

$$g(R)=\frac{GM}{R^2},$$

which matches the external inverse-square expression continuously.

Worked example

For an ideal uniform sphere, compare the gravitational field halfway from the center to the surface with the surface field.

At

$$r=\frac R2,$$

$$g\left(\frac R2\right) =\frac{GM}{R^3}\frac R2 =\frac12\frac{GM}{R^2}.$$

Thus

$$\boxed{g(R/2)=\frac12g(R)}.$$

A naive point-mass formula using the entire mass at $r=R/2$ would instead predict four times the surface field, which is wrong because the outer shells make no net contribution inside them and only the enclosed mass contributes.

Newton's shell theorem is therefore the key bridge between the inverse-square point-mass law and gravitational fields of spherical planets, stars, shells, and other symmetric bodies.