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Question

A solid sphere of uniform density and radius R exerts a gravitational force of attraction F1 on the particle P, distant 2R from the centre of the sphere. A spherical cavity of radius R/2 is now formed in the sphere as shown in figure. The sphere with cavity now applies a gravitational force F2 on the same particle P. Find the ratio F2/F1.
794555_0c81127cc98946d0bfe31b6c404c6184.PNG

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Solution

Here F1=GMm(2R)2=GMm4R2=0.25GMmr2
We assume that density of sphere is constant.
Mass of cut out sphere ,
MR3=MR23
M=M8
F2=F1(force experineced by point mass due to spheer of mass M/8 and at a distance of (R+R/2) from its center)
F2=F1G(M8)m(3R2)2=GMm4R2GMm18R2
=(0.1944)GMmR2F2F1=0.19440.25=0.7777F2F1=79

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