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

The gravitational force between spherical objects is the same as that between two particles with all masses concentrated at their centers. Therefore,

FM = GMm/d2 = 6.67*10-11*2.56*0.706 / (0.1216m)2 = 8.15*10-9 N

The removed part has radius R/2. Thus its mass is

M’ = (4/3)(R/2)3 = (1/8) × (4/3)R3 = M/8 = 0.32 kg

According to the superposition principle, the gravitational force due to the sphere before hollowing is the sum of the gravitational force due to the removed part and the hollowed out sphere.

FM-M’ + FM’ = FM   FM-M’ = FM - FM’  = GMm/d2 – GM’m/(d-R/2)2

= 8.15*10-9 - [6.67*10-11*0.32*0.706 / (0.1216 - 0.01125)2] = 6.912*10-9  N = 6.912 nN

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