The gravitational force acting on a particle, | vedclass

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(A) Let $M$ be the mass of the solid sphere and $m$ be the mass of the particle. The gravitational force due to the solid sphere on the particle at a

Vedclass · Accepted Answer

$\frac{50}{41}$

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(A) Let $M$ be the mass of the solid sphere and $m$ be the mass of the particle. The gravitational force due to the solid sphere on the particle at a distance $3R$ from its center is given by:$F_1 = \frac{G M m}{(3 R)^2} = \frac{G M m}{9 R^2}$When a spherical hole of radius $r = R/2$ is made,its mass $M'$ is proportional to its volume. Since the density $\rho$ is uniform,$M' = \rho \cdot \frac{4}{3} \pi (R/2)^3 = \rho \cdot \frac{4}{3} \pi R^3 \cdot \frac{1}{8} = \frac{M}{8}$.The center of the hole is at a distance $R/2$ from the center of the original sphere. The particle is at a distance $3R$ from the center of the original sphere,so it is at a distance $(3R - R/2) = 2.5R = 5R/2$ from the center of the hole.The gravitational force $F_2$ exerted by the sphere with the hole is the force of the solid sphere minus the force that would have been exerted by the removed spherical part:$F_2 = F_1 - F_{hole} = \frac{G M m}{9 R^2} - \frac{G (M/8) m}{(5R/2)^2}$$F_2 = \frac{G M m}{R^2} \left[ \frac{1}{9} - \frac{1}{8} \cdot \frac{4}{25} \right] = \frac{G M m}{R^2} \left[ \frac{1}{9} - \frac{1}{50} \right]$$F_2 = \frac{G M m}{R^2} \left[ \frac{50 - 9}{450} \right] = \frac{G M m}{R^2} \left[ \frac{41}{450} \right]$Now,the ratio $F_1 / F_2$ is:$\frac{F_1}{F_2} = \frac{G M m / 9 R^2}{41 G M m / 450 R^2} = \frac{1}{9} \cdot \frac{450}{41} = \frac{50}{41}$

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