Question

A uniform sphere (mass $M$ , radius $R$ ) exerts a force $F$ on a small mass $m$ placed at distance $2R$ from the centre of the sphere. A sphere of radius $\frac{R}{2}$ is excavated from the sphere as shown in the figure. Find the force applied by the remaining part of the sphere on the point mass.

• A. $\frac{F}{3}$
• B. $\frac{4F}{9}$
• C. $\frac{2F}{3}$
• D. $\frac{7F}{9}$

Answer 1

Answer: (D)

The mass of the sphere $M=\frac{4}{3}\pi R^3\rho$ where $\rho$ is the density.
The force of attraction between the sphere and the mass m is
$F=\frac{GMm}{{\left(2R\right)}^2}=\frac{GMm}{4R^2}$ ... (1)
Mass of the cut out sphere is $m\prime =\frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3\rho =\frac{1}{8}\left[\frac{4}{3}\pi R^3\rho \right]=\frac{1}{8}\left(M\right)$
The force of attraction between the cut out portion and 'm'.
$f=\frac{Gm\prime m}{{\left(\frac{3R}{2}\right)}^2}$ $\because$ $O\prime P=2R-\frac{R}{2}=\frac{3R}{2}$
$=\frac{G\left(\frac{M}{8}\right)\cdot m}{\frac{9R^2}{4}}=\frac{GMm}{2\times 9R^2}$
$=\left(\frac{GMm}{4R^2}\right)\times \frac{2}{9}=\frac{2}{9}F$ ... from (1)
$\therefore$ The force of attraction between the remaining portion and $m=F-\frac{2}{9}F=\frac{7}{9}F$ .