The figure below shows a spherical hollow inside a lead sphere of radius R = 4.0

Question

The figure below shows a spherical hollow inside a lead sphere of radius R = 4.00 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 = 3.10 kg.

G = 6.67428 x 10-11 N-m2/kg2

1)

With what gravitational force does the hollowed-out lead sphere attract a small sphere of mass m = 0.430 kg that lies at a distance d = 9.00 cm from the center of the lead sphere, on the straight line connecting the centers of the spheres and of the hollow?

Explanation / Answer

Here,

mass of unaltered sphere

m1 = 3.10kg

mass of altered sphere is
= 3.10 * (1 - (2/4)^3) = 3.10 * (1 - 1/8) = 3.10 * 7/8

mass removed
mx = 3.10 * 1/8

other mass
m2 = 0.430 kg
9 cm from m1 and 7 cm from mx

F = (G*m1*m2 / 0.09^2) - (G*mx*m2 / 0.07^2) = 6.67428 * 10^-11 * 0.430 (3.10/0.09^2 - (3.10/8)/0.07^2)

F = 8.714 * 10^-9 N

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