A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to `F_(1)` on a particle placed at P, distance 2R from the centre of the sphere. A spherical cavity of radius R/2 is now made in te sphere as shown in figure the sphere with cavity now applies a gravitational force `F_(2)` on the same particle placed at P. find the ratio of `F_(2)//F_(1)`

Explanation

`F_(1)=(GMm)/(4R^(2)), F_(2)=` force due to whole sphere - force due to the sphere forming the cavity
`=(GMm)/(4R^(2))-(GMm)/(18 R^(2))implies (7 GMm)/(36 R^(2))" ":. F_(2)/F_(1)=7/9`