A uniform solid sphere of mass M and radius R exerts a gravitational force `F_(g)` on a particle of mass m placed at a point P at a distance of 4 R from centre of sphere. Suppose a spherical cavity of radius `(R )/(2)` is cut into the sphere.

The sphere with cavity exerts a force `F_(c )` on the same particle, placed at P. Find the ratio `F_(g)//F_(c )`. Also find the force of attraction due to remaining part of bigger mass. Let force due to remaining part is F'.

Explanation

The force on mass m due to solid sphere.
`F_(g) = (GMm)/((4R)^(2)) = (GMm)/(16R^(2))` …(i)
Radius of cavity `= (R )/(2)`
`:.` Mass of cavity `= (M)/((4)/(3)piR^(3)) xx (4)/(3)pi((R )/(3))^(3) = (M)/(8)`
`F. + F_(c ) = F_(g)`
`F. = F_(g) - F_(c ) = (GMm)/(16R^(2)) - (GMm)/(98R^(2))`
`= (GMm)/(R^(2)) ((1)/(16) - (1)/(98))`
Force on mass m due to cavity `= F_(c )`
`= (g((M)/(8)).m)/((4R - (R )/(2))^(2)) = (GMm)/(98R^(2))` ...(ii)
From equations (i) and (ii)
`(F_(g))/(F_(c )) = (GMm98 R^(3))/(16R^(2).GMm) = (98)/(16)`