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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?

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?

Grade:11

1 Answers

Kevin Nash
askIITians Faculty 332 Points
9 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]
234-523_f19.JPG
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
\rho = M/[(4/3)(\piR3)]
Here, radius of the lead sphere is R.
The spherical sphere has a radius of R/2.
If the density is constant, then the mass of the hole will be the density times the volume of the hollow sphere. Thus, the mass of the hollow sphere is
234-1859_f20.JPG
The hole is closer to the small sphere. This means the center of the hole is d-(R/2) away from the small sphere.
Now, the gravitational force between the hollow sphere and the small sphere is
234-2217_f21.JPG
The force of attraction with which the lead sphere after it is made hollow and the small sphere is equal to the force of the whole lead sphere before it is made hollow minus the force of the hollow lead sphere. The corresponding force is
234-348_f22.JPG

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