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When Newton tested his theory of gravity by comparing the orbital

acceleration of the moon to the acceleration of a falling apple on earth, he

assumed he could compute the earth’s force on the apple using the distance

from the apple to the earth’s center. Was he wrong. After all, it isn’t just the

earth’s center attracting the apple, it’s the whole earth. A kilogram of dirt a

few feet under his backyard in England would have a much greater force on

the apple than a kilogram of molten rock deep under Australia, thousands

of miles away. There’s really no obvious reason why the force should come

out right if you just pretend that the earth’s whole mass is concentrated at

its center. Also, we know that the earth has some parts that are more dense,

and some parts that are less dense. The solid crust, on which we live, is

considerably less dense than the molten rock on which it floats. By all

rights, the computation of the vector sum of all the forces exerted by all the

earth’s parts should be a horrendous mess.

Actually, Newton had sound mathematical reasons for treating the

earth’s mass as if it was concentrated at its center. First, although Newton no

doubt suspected the earth’s density was nonuniform, he knew that the

direction of its total gravitational force was very nearly toward the earth’s

center. That was strong evidence that the distribution of mass was very

symmetric, so that we can think of the earth as being made of many layers,

like an onion, with each layer having constant density throughout. (Today

there is further evidence for symmetry based on measurements of how the

vibrations from earthquakes and nuclear explosions travel through the

earth.) Newton then concentrated on the gravitational forces exerted by a

single such thin shell, and proved the following mathematical theorem,

known as the shell theorem:

If an object lies outside a thin, uniform shell of mass, then the

vector sum of all the gravitational forces exerted by all the parts of

the shell is the same as if all the shell’s mass was concentrated at its

center. If the object lies inside the shell, then all the gravitational

forces cancel out exactly.

For terrestrial gravity, each shell acts as though its mass was concentrated at

the earth’s center, so the final result is the same as if the earth’s whole mass

was concentrated at its center.

The second part of the shell theorem, about the gravitational forces

canceling inside the shell, is a little surprising. Obviously the forces would

all cancel out if you were at the exact center of a shell, but why should they

still cancel out perfectly if you are inside the shell but off-center. The

whole idea might seem academic, since we don’t know of any hollow planets

in our solar system that astronauts could hope to visit, but actually it’s a

useful result for understanding gravity within the earth, which is an impor-

tant issue in geology. It doesn’t matter that the earth is not actually hollow.

In a mine shaft at a depth of, say, 2 km, we can use the shell theorem to tell

us that the outermost 2 km of the earth has no net gravitational effect, and

the gravitational force is the same as what would be produced if the remain-

ing, deeper, parts of the earth were all concentrated at its center.

An object outside a spherical shell of

mass will feel gravitational forces from

every part of the shell — stronger

forces from the closer parts and

weaker ones from the parts farther

away. The shell theorem states that

the vector sum of all the forces is the

same as if all the mass had been con-

centrated at the center of the shell.

Chapter 10Gravity