A solid sphere of radius r/2 is cut out of a solid sphere of radius r such that the spherical cavity so formed touches the surface on one side and the centre of the sphere on the other side, as shown . The initial mass of the solid sphere was m. If a particle of mass m is placed at a distance 2.5r from the centre of the cavity, then what is the gravitational attraction on the mass m ?
Answers
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, (see the diagram)
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
