The figure shows a spherical hollow inside a lead sphere of radius R = 3.4 m; th

Question

The figure shows a spherical hollow inside a lead sphere of radius R = 3.4 m; the surface of the hollow passes through the center of the sphere and "touches" the right side of the sphere. The mass of the sphere before hollowing was M = 400 kg. With what gravitational force does the hollowed-out lead sphere attract a small sphere of mass m = 38 kg that lies at a distance d = 11 m from the center of the lead sphere, on the straight line connecting the centers of the spheres and of the hollow? Number Units the tolerance is +/-1 in the 2nd significant digit

Explanation / Answer

This is a job for negative mass. We take the force the solid sphere would exert, then subtract the force of the material which makes up the hollow. In effect, we treat the hollow as negative mass. F = GMm/d^2 G(M/8)m/(d R/2)^2 = 8.29 × 10^9 N – 1.45 × 10^9 N = 6.84 × 10^9 N

The factor of 0.125 is because the hollow has 1/8 the volume of the full sphere. The center of the hollow is at a distance of d R/2 = 9.3 m from the smaller mass.

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