# The following problem is from the 1946 “Olympic” examination of Moscow State University (see Fig.) A spherical hollow is made in a lead sphere of radius R, such that its surface touches the outside surface of the lead sphere and passes through its center. The mass of the sphere before hollowing was M. With what force, according to the law of universal gravitation, will the hollowed lead sphere attract a small sphere of mass m, which lies at a distance d from the center of the lead sphere on the straight line connecting the centers of the spheres and of the hollow?

Kevin Nash
8 years ago
Question:-

A spherical hollow is made in a lead sphere of radius R, such that its surface touches the outside surface of the lead sphere and passes through its center (see below figure). The mass of the sphere before hollowing was M. With what force, according to the law of universal gravitation, will the hollowed lead sphere attract a small sphere of mass m, which lies at a distance d from the center of the lead sphere on the straight line connecting the centers of the spheres and of the follow? [1946, Olympic, Moscow State University]

Concept:

The force of gravity on the small sphere of mass due to the lead sphere after it is made hollow is equal to the force of gravity from a solid lead sphere minus the force which would have been contributed by the smaller lead sphere which would have filled the hole.
Solution:

From the Newton’s law of universal gravitation, the force of attraction between the lead sphere and the small sphere is
F1 = GMm/d2
Here, gravitational constant is G, total mass of the lead sphere is M, mass of the small sphere is m and distance between the centers of the spheres is d.

So, we need to know about the size and mass of the lead which was removed to make the hole.

The density of the lead is given by
 = M/[(4/3)(R3)]