# A point Mass outside a spherical shell

We start by considering a ring on the surface of a shell (Fig. 13.23a), centered on the line from the center of the shell to m. We do this because all of the particles that make up the ring are the same distance s from the point mass m. From Eq. (13.9) the potential energy of interaction between the earth (mass mE) and a point mass m, separated by a distance r, is U = -GmEm>r. From this expression, we see that the potential energy of interaction between the point mass m and a particle of mass mi within the ring is

To find the potential energy dU of interaction between m and the entire ring of mass dM = Eimi, we sum this expression for Ui over all particles in the ring

To proceed, we need to know the mass dM of the ring. We can find this with the aid of a little geometry. The radius of the shell is R, so in terms of the angle f shown in the figure, the radius of the ring is Rsinf, and its circumference is 2pRsinØ. The width of the ring is R dØ, and its area dA is approximately equal to its width times its circumference:

The ratio of the ring mass dM to the total mass M of the shell is equal to the ratio of the area dA of the ring to the total area A = 4pR^{2} of the shell:

Now we solve Eq. (13.19) for dM and substitute the result into Eq. (13.18) to find the potential energy of interaction between point mass m and the ring:

The total potential energy of interaction between the point mass and the shell is the integral of Eq. (13.20) over the whole sphere as f varies from 0 to p (not 2p!) and s varies from r - R to r + R. To carry out the integration, we have to express the integrand in terms of a single variable; we choose s. To express f and df in terms of s, we have to do a little more geometry. Figure 13.23b shows that s is the hypotenuse of a right triangle with sides (r - RcosØ) and RsinØ, so the Pythagorean theorem gives

We take differentials of both sides:

Next we divide this by 2rR and substitute the result into Eq. (13.20):

We can now integrate Eq. (13.22), recalling that s varies from r - R to r + R:

Finally, we have

This is equal to the potential energy of two point masses m and M at a distance r. So we have proved that the gravitational potential energy of spherical shell M and point mass m at any distance r is the same as though they were point masses. Because the force is given by F_{r} = -dU/dr, the force is also the same.

We assumed at the beginning that the point mass m was outside the spherical shell, so our proof is valid only when m is outside a spherically symmetric mass distribution. When m is inside a spherical shell, the geometry is as shown in 1 Fig. 13.24.

The entire analysis goes just as before; Eqs. (13.18) through (13.22) are still valid. But when we get to Eq. (13.23), the limits of integration have to be changed to R - r and R + r. We then have

and the final result is

Compare this result to Eq. (13.24): Instead of having r, the distance between m and the center of M, in the denominator, we have R, the radius of the shell. This means that U in Eq. (13.26) doesn’t depend on r and thus has the same value everywhere inside the shell. When m moves around inside the shell, no work is done on it, so the force on m at any point inside the shell must be zero.

More generally, at any point in the interior of any spherically symmetric mass distribution (not necessarily a shell), at a distance r from its center, the gravitational force on a point mass m is the same as though we removed all the mass at points farther than r from the center and concentrated all the remaining mass at the center.

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