Question

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

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

Answer 1

The correct option is D

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

Step1: Given data.

Mass of the particle, $m=10g=0.01kg$

Mass of the uniform sphere, $M=100kg$

Radius of the uniform sphere, $r=10cm=0.1m$

Gravitation constant, $G=6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}$

Step2: Finding the work to be done against the gravitational force.

Formula used:

$U=\frac{-GMm}{r}$

Where $U$ is the potential energy, $M$ is the mass of the sphere, $m$ is the mass of the particle, and $G$ is the gravitational constant.

Now,

Initial potential energy of the particle,

$U_i=\frac{-GMm}{r}$

where $U_i$ is the initial potential energy of the particle.

$\Rightarrow$$U_i=\frac{-6.67\times {10}^{-11}\times 100\times 0.01}{0.1}$

$\Rightarrow$$U_i=-6.67\times {10}^{-10}J$ ….$\left(i\right)$

Also, we know that.

Work done = Change in potential energy

$W=∆U=U_f-U_i$

Where $U_i$ and $U_f$ is the initial and final potential energy of the particle.

$\Rightarrow$$W=0-\left(-6.67\times {10}^{-10}\right)$ $\left[\because U_f=0\right]$

$\Rightarrow$$W=6.67\times {10}^{-10}J$

Hence, option D is correct.