The figure below shows a spherical hollow inside a lead sphere ofradius R = 4.00

Question

The figure below shows a spherical hollow inside a lead sphere ofradius R = 4.00 cm. The surface of the hollow passesthrough the center of the sphere and "touches" the right side ofthe sphere. The mass of the sphere before hollowing was M= 2.60 kg. With what gravitational forcedoes the hollowed-out lead sphere attract a small sphere of massm = 0.483 kg that lies at adistance d = 9.00 cm from the center of the lead sphere,on the straight line connecting the centers of the spheres and ofthe hollow?

Explanation / Answer

    in the given problem if the sphere wee notallowed the magnitude of the force it exerts on m would be     F1 = G M m / d2     part of this force is due to material that isremoved     we calculate the force exerted on m by a spherethat just fills the cavity, at the position of cavity, and subtractfrom the     force of the solid sphere     the cavity has a radius r = R / 2     the material that fills it has the same densityas the solid sphere that is     Mc / r3 = M /R3     where Mc is the mass that fills thecavity     so we get     Mc = (r3 / R3)M           =(R3 / 8R3) M           = M / 8     the center of the cavity is     d - r = d - (R / 2)     so the force it exerts on m will be     F2 = G (M / 8) m / [d - (R /2)]2     the force of the hollowed sphere on m willbe     F = F1 -F2       = (G M m / d2) [1 - (1 / 8(1 - (R / 2d))2)]        = ....... N

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