From a sphere of mass M and radius R, smaller sphere of radius R/2 is carved out such that cavity made in original sphere is between its centre and periphery. For configuration in fig. where distance between centre of original sphere and removed sphere is 3R, gravitational force between 2 sphere is

7GM^2/576R^2.                                                                                                                                                                 .....................................................

Hello Suban Abdul

Given  r = radius of the solid sphere

m = mass of the solid sphere

Taken  d =density of the solid

m' = mass of the removed sphere

M = mass of the remaining body

a = distance between center of cavity and gravitational point of the remaining body.

b = distance between gravitational point of the remaining body & the particle

In first, we have to find values for m' & M.

m = d * (4/3)πr³ = 4dπr³/3

m' = d * (4/3)π(r/2)³ = dπr³/6 ----------- (1)

M = m-m'

=  4dπr³/3 - dπr³/6 = 7dπr³/6 = 7m/8 ------- (2)

Then we can find value of 'a' by taking torque respect to center of gravity of initial sphere,

Mg * (a-r/2) - m'g * (r/2) = 0

M(a-r/2) = m'r/2

(7dπr³/6) * (a-r/2) = (dπr³/6)*(r /2)         [by (1) & (2)]

7 * (a-r/2) = r /2

a = 4r/7 = 0.57r

Then we can write equation for 'b' as below,

b = a + 2.5r

b = 0.57r + 2.5r

b = 3.07r

From the gravitational force formula,

gravitational force = GMm/b² (G =Newtonian constant of gravitation)

= [G * (7m/8 ) *m ] / (3.07r²)

= (6.674 08 x 10-11 * 7m²)/24.56r²

= 1.0902 x 10-11m²/r²

Hope it helps you.