Question

A particle placed at a distance $2R$ from the centre of a solid sphere of uniform density and radius $R$ experiences a gravitational force of attraction equal to $F_1$ . A spherical cavity of radius $\frac{R}{2}$ is now made in the sphere as shown in figure and the same particle now experiences a gravitational force $F_2$ . If the ratio $\frac{F_1}{F_2}$ is $\frac{x}{7}$ then find $x$ .

Answer 1

Answer: 9

Mass of the cavity $=\frac{1}{8}$ of mass of sphere
$\Rightarrow M_c=\frac{M_s}{8}$
$F_1=\frac{GMm}{(2R}=\frac{GMm}{4R^2}$
$M_c=\frac{M_s}{8}$
$F_2=\frac{GMm}{4R^2}-\frac{\frac{GMm}{8}}{{\left(\frac{3}{2}R\right)}^2}$
$=\frac{GMm}{R^2}\left[\frac{1}{4}-\frac{1}{18}\right]$
$=\frac{\text{ GMm }}{R^2}\left[\frac{18-4}{72}\right]=\frac{7}{36}\frac{GMm}{R^2}$
$\frac{F_1}{F_2}=\frac{GMm}{\frac{7}{36}\frac{GMm}{R^2}}=\frac{9}{7}=1.29$