Question

A solid sphere of radius $R/2$ is cut out of a solid sphere of radius $R$ such that the spherical cavity so formed touches the surface on one side and the centre of the sphere on other side, as shown. The initial mass of the solid sphere was $M$. If a particle of mass $m$ is placed at a distance $2.5R$ from the centre of the cavity, then what is the gravitational attraction on the mass $m$?

• A. $\frac{GMm}{R^2}$
• B. $\frac{GMm}{{2R}^2}$
• C. $\frac{GMm}{{8R}^2}$
• D. $\frac{23}{100}\frac{GMm}{R^2}$

Answer 1

The correct option is D $\frac{23}{100}\frac{GMm}{R^2}$

The system is equivalent to an attracting sphere of radius R and mass M, and a repelling sphere of radius R/2 and mass equal to that of a sphere with mass density as that of sphere of mass M.

Mass is proportional to volume which is proportional to cube of radius. The radius of the cavity is half of that of bigger sphere, hence the mass of the the repelling sphere is M/8.

Hence the mass of the repelling sphere=$M/8$.

Attractive force by the attracting sphere=$\frac{GMm}{(2R{)}^2}$

Repelling force by repelling sphere=$\frac{G\left(\frac{M}{8}\right)m}{(2.5R{)}^2}$

Hence net force=$\frac{GMm}{(2R{)}^2}$ $-$ $\frac{G\left(\frac{M}{8}\right)m}{(2.5R{)}^2}$=$\frac{23}{100}\frac{GMm}{R^2}$.