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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