The gravitational force acting on a particle, due to a solid sphere of uniform density and radius $R$, at a distance of $3R$ from the centre of the sphere is $F_1$. A spherical hole of radius $\dfrac{R}{2}$ is now made in the sphere as shown in the figure. The sphere with the hole now exerts a force $F_2$ on the same particle. The ratio $\dfrac{F_1}{F_2}$ is
Answer: (A) $\dfrac{50}{41}$
Let the full sphere have mass $M$. The hole has radius $\dfrac{R}{2}$, so its mass would be $\dfrac{M}{8}$. Its centre is $\dfrac{R}{2}$ from O, towards the particle, so it is $3R - \dfrac{R}{2} = \dfrac{5R}{2}$ from the particle.
$$F_1 = \frac{GMm}{9R^2}$$
$$F_2 = \frac{GMm}{9R^2} - \frac{G(M/8)m}{(5R/2)^2} = \frac{GMm}{R^2}\left(\frac{1}{9} - \frac{1}{50}\right) = \frac{GMm}{R^2}\cdot\frac{41}{450}$$
$$\frac{F_1}{F_2} = \frac{1/9}{41/450} = \frac{50}{41}$$
Solution by Sreeraj P, M.Sc Physics