A solid sphere of uniform density and radius r applies a gravitational force of attraction equal to F 1 on a particle placed at P , distance 2 R from the centre O of the sphere. A spherical cavity of

Question

A solid sphere of uniform density and radius $r$ applies a gravitational force of attraction equal to $F_{1}$ on a particle placed at $P$ , distance $2 R$ from the centre $O$ of the sphere. A spherical cavity of the radius $\frac{R}{2}$ is now made in the sphere as shown in the figure. The sphere with cavity now applied a gravitational force $F_{2}$ on the same particle placed at $P$ . The ratio $\frac{F_{2}}{F_{1}}$ will be

TestHub · Accepted Answer

Gravitational force due to solid sphere, $F_{1} = \frac{G M m}{{(2 R)}^{2}},$
where $M$ and $m$ are mass of the solid sphere and particle respectively and $R$ is the radius of the sphere. The gravitational force on particle due to sphere with cavity = force due to solid sphere creating cavity, assumed to be present above at that position

i.e., $F_{2} = \frac{GMm}{4 R^{2}} = \frac{G (M / 8) m}{{(3 R / 2)}^{2}} = \frac{7}{36} \frac{GMm}{R^{2}}$

So, $\frac{F_{2}}{F_{1}} = \frac{7 GMm}{36 R^{2}} / (\frac{GMm}{4 R^{2}}) = \frac{7}{9}$

Physics - Gravitation Question with Solution | TestHub

Options:

Doubts & Discussion