A solid sphere of uniform density and radius $R$ exerts a gravitational force of attraction $F_{1}$ on the particle $P$ , distant $2 R$ from the centre of the sphere. A spherical cavity of radius $R / 2$ is now formed in the sphere as shown in figure. The sphere with cavity now applies a gravitational force $F_{2}$ on the same particle $P$ . Find the ratio $F_{2} / F_{1}$ .

Open in App

Solution

Here $F_{1} = \frac{G M m}{{(2 R)}^{2}} = \frac{G M m}{4 R^{2}} = 0.25 \frac{G M m}{r^{2}}$
We assume that density of sphere is constant.
Mass of cut out sphere ,
$\frac{M}{R^{3}} = \frac{M^{'}}{{\frac{R}{2}}^{3}}$
$M^{'} = \frac{M}{8}$
$F_{2} = F_{1} -$ (force experineced by point mass due to spheer of mass M/8 and at a distance of (R+R/2) from its center)
$F_{2} = F_{1} - \frac{G (\frac{M}{8}) m}{{(\frac{3 R}{2})}^{2}} = \frac{G M m}{4 R^{2}} - \frac{G M m}{18 R^{2}}$
$= (0.1944) \frac{G M m}{R^{2}} \frac{F_{2}}{F_{1}} = \frac{0.1944}{0.25} = 0.7777 \frac{F_{2}}{F_{1}} = \frac{7}{9}$

Suggest Corrections

Similar questions

Q. 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 2R from the centre O of the sphere. A spherical cavity of radius R/2 is now made in the sphere as shown in the figure. The sphere with cavity now applies a gravitational force

F_{2}

on same particle placed at P. The ratio

F_{2} / F_{1}

will be

A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to $F_{1}$ on a particle placed at a distance 3R from the centre of the sphere. A spherical cavity of radius $\frac{R}{2}$ is now made in the sphere as shown in the figure. The sphere with cavity now applies a gravitational force $F_{2}$ on the same particle. The ratio $\frac{F_{2}}{F_{1}}$ is:

Join BYJU'S Learning Program