The gravitational force acting on a particle, | vedclass

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Question

(A) The gravitational force due to a solid sphere of mass $M$ on a particle of mass $m$ at a distance $3R$ from its center is given by:$F_1 = \frac{G

Vedclass · Accepted Answer

$\frac{50}{41}$

Vedclass · Answer

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

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