A small point of mass m is placed at a distance 2R from the centre 'O' of a big uniform solid sphere of mass M and radius R . The gravitational force on 'm' due to M is $$F_1$$. A spherical part of radius R/3 is removed from the big sphere as shown in the figure and the gravitational force on m due to the remaining part of M is found to be $$F_2$$ . The value of ratio $$F_1 : F_2$$ is

For a point mass outside a uniform solid sphere, the entire mass $$M$$ acts as if it is concentrated at the center $$O$$.

$$F_1 = \frac{GMm}{(2R)^2} = \frac{GMm}{4R^2}$$

The cavity has a radius $$r = R/3$$. Since mass is proportional to volume ($$M \propto R^3$$),

Mass of cavity ($$m_c$$): $$M \cdot (\frac{R/3}{R})^3 = \frac{M}{27}$$

Distance to $$P$$ ($$d_c$$): The center of the cavity $$O'$$ is at a distance $$R - R/3 = 2R/3$$ from $$O$$. Thus, its distance from point $$P$$ is $$2R - \frac{2R}{3} = \frac{4R}{3}$$.

The force exerted by this removed mass is $$F_{cav} = \frac{G(M/27)m}{(4R/3)^2} = \frac{GMm}{27} \cdot \frac{9}{16R^2} = \frac{GMm}{48R^2}$$

The force from the remaining part ($$F_2$$) is $$F_2 = F_1 - F_{cav} = \frac{GMm}{4R^2} - \frac{GMm}{48R^2}$$

$$F_2 = \frac{12GMm - 1GMm}{48R^2} = \frac{11GMm}{48R^2}$$

$$\frac{F_1}{F_2} = \frac{\frac{GMm}{4R^2}}{\frac{11GMm}{48R^2}} = \frac{1}{4} \times \frac{48}{11} = \frac{12}{11}$$