The gravitational force acting on a particle, due to a solid sphere of uniform density and radius $R$, at a distance of $3 R$ from the centre of the sphere is $F_1$. A spherical hole of radius $(R / 2)$ is now made in the sphere as shown in the figure. The sphere with hole now exerts a force $F_2$ on the same particle. Ratio of $F_1$ and $F_2$ is

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Aicharya · Accepted Answer

Gravitational force due to solid sphere is $ F_1=\frac{G M m}{(3 R)^2}=\frac{G M m}{9 R^2} $

where, $M$ and $m$ are mass of solid sphere and particle respectively. Gravitational force on particle due to sphere with cavity $ \begin{aligned} F_2 & =\frac{G M m}{9 R^2}-\frac{G\left(\frac{M}{8}\right) m}{(5 R / 2)^2} \\ & =\frac{G M m}{R^2}\left[\frac{1}{9}-\frac{4}{8 \times 25}\right] \\ & =\frac{G M m}{R^2}\left[\frac{41}{50 \times 9}\right] \\ \therefore \quad \frac{F_1}{F_2} & =\frac{50}{41} \end{aligned} $

The gravitational force acting on a particle, due to a solid sphere of uniform density and radius $R$, at a…

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