Question

A particle of mass $10\text{g}$ is kept on the surface of a uniform sphere of mass $100\text{kg}$ and radius $10\text{cm}$, find the work to be done against the gravitational force between them to take the particle for away from the sphere:-
$(G=6.67\times {10}^{-11}\frac{N{m}^2}{k{g}^2})$

• A. $6.67\times {10}^{-9}\text{J}$
• B. $6.67\times {10}^{-10}\text{J}$
• C. $13.34\times {10}^{-10}\text{J}$
• D. $3.33\times {10}^{-10}\text{J}$

Answer 1

The correct option is B $6.67\times {10}^{-10}\text{J}$

Potential energy of system of two masses is,

$U=\frac{-G{m}_1{m}_2}{R}=-\frac{6.67\times {10}^{-11}\times 100\times 10\times {10}^{-3}}{10\times {10}^{-2}}$

$=-6.67\times {10}^{-10}\text{J}$

So the amount of work done against the gravity is,

$=-U=6.67\times {10}^{-10}\text{J}$

Why this question? The potential energy stored in the object is same as work done against the gravity to move to a given height.