Question

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 ?

Answer 1

Distance of the particle from center of the cavity = 2.5r

=>  Distance of the particle from center of the sphere = 2.5r - 0.5r = 2r

mass of the whole sphere = m

=> mass of the cavity, m' = m x (4π(r/2)³/3) ÷ 4πr³/3

=> m' = m/8

using the principle of superposition gravitational attraction on the particle is equal to gravitational attraction due to whole sphere minus the attraction due to the cavity.

=> F = $\frac{Gmm}{(2r)^2 }$ - $\frac{Gmm/8}{(2.5r)^2}$

=> F = Gm²/4r² - Gm²/ 50

= 23Gm²/100r²

Hence the gravitational attraction on the mass m will be $\frac{23Gm^2}{100r^2}$

Answer 2

Answer:

let us consider mass of gravity =M'

our problem is to find mass of this gravity 4/3 3.14 R^3 =M

1= 4/3 3.14R^3/4/3 3.14 R^3/8

M'=M/8

now,

1. F=GMm/2R^2-GM/8(2.5)R^2×m
2. F=GM/4R^2 -GM^2/50
3. =23Gm^2/100R^2