A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F₁ on a particle placed at
P, distance 2R from the centre O of the sphere. A spherical cavity of radius R/2 is now made in the sphere as shown in figure. The sphere with cavity now applies an gravitational force F₂ on same particle placed at P. The ratio F₂/ F₁ will be

Question

Infinity Learn · Accepted Answer

Gravitational force due to solid sphere,  $$F1=GMm(2R)2$$ where M and mare masses of the solid sphere and particle respectivelyand R is the radius of the sphere. The gravitational force on particle due to sphere with cavity $$=$$ force due to solid sphere creating cavity, assumed to be present above at that position.i, e, $$F2=GMm4R2-G(M/8)m(3R/2)2=736GMmR2So$$, $$F2F1=7GMm36R2/GMm4R2=79$$

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