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 the cavity now applies 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, $F_{1} = \frac{GMm}{{(2 R)}^{2}}$ 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 the particle due to sphere with cavity = force due to a solid sphere - force due to sphere creating a 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 $\frac{F_{2}}{F_{1}} = \frac{\frac{7 GMm}{36 R^{2}}}{(\frac{GMm}{4 R^{2}})} = \frac{7}{9}$

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