The figure shows a spherical hollow inside a lead sphere of radius R = 2.25 cm;

Question

The figure shows a spherical hollow inside a lead sphere of radius R = 2.25 cm; 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 = 2.56 kg. With what gravitational force does the hollowed-out lead sphere attract a small sphere of mass m=0.706 kg that lies at a distance d= 12.16 cm from the center of the lead sphere, on the straight line connecting the centers of the spheres and of the hollow? Enter your answer in nanonewtons (nN) and round your answer to 3 significant figures.

Explanation / Answer

Given Data

radius R = 2.25 cm = 0.0225 m

M = 2.56 kg.

m=0.706 kg

distance d= 12.16 cm = 0.1216 m

Solution:-

Required force F = force from whole lead sphere F1 - force from the hollow F2

where F1 = GMm/d2

the radius of the hollow = R/2, volume = 4(R/2)3/3= volume of the lead sphere/8
so the mass = M/8

and F2 = G(M/8)m/(d - R/2)2

so F = GMm[1/d2 - 1/(8(d - R/2)2)]

F = (6067*10^-11)*(2.56)*(0.706)*[1/0.12162 - 1/(8(0.1216 - 0.0225/2)2)]

      = 6.9*10-9 N

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