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 $R/2$ 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 $F_{2}/F_{1}$ will be

• A. $\frac{1}{2}$
• B. $\frac{7}{9}$
• C. $3$
• D. $7$

Answer 1

Answer: (B)

$\frac{7}{9}$

Answer 2

Gravitational force due to solid sphere, $\left(\textit{F}\right)_{1}=\frac{\textit{GMm}}{\left(2 \textit{R}\right)^{2}}$ 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., $\quad F_{2}=\frac{G M m}{4 R^{2}}-\frac{G(M / 8) m}{(3 R / 2)^{2}}=\frac{7}{36} \frac{G M m}{R^{2}}$ So $\frac{F_{2}}{F_{1}}=\frac{\frac{7 G M m}{36 R^{2}}}{\left(\frac{G M m}{4 R^{2}}\right)}=\frac{7}{9}$