Question

A body of mass $m\text{kg}$ starts falling from a point $2R$ above the earth's surface. Its kinetic energy when it has fallen to a point $R$ above the earth's surface will be

• A. $\frac{1}{2}\frac{GmM}{R}$
• B. $\frac{1}{6}\frac{GmM}{R}$
• C. $\frac{2}{3}\frac{GmM}{R}$
• D. $\frac{1}{3}\frac{GmM}{R}$

Answer 1

The correct option is B $\frac{1}{6}\frac{GmM}{R}$


As there is no external force present on the system of $M\&m$. So energy of the system is conserved.

Initial energy of the system,

$E_i=U_i+K_i=-\frac{Gm.M}{(2R+R)}+0=-\frac{GmM}{3R}$

Final energy of the system will be,

$E_f=U_f+K_f=\frac{-Gm.M}{(R+R)}+\frac{1}{2}m{v}^2$

where $v$ is the velocity of the mass $m$ when it falls down by distance $R$

As energy is conserved

$\therefore E_i=E_f$

$\frac{-GmM}{3R}=\frac{1}{2}m{v}^2-\frac{GmM}{2R}$

$\Rightarrow KE=\frac{Gmm}{R}[\frac{1}{2}-\frac{1}{3}]=\frac{GmM}{6R}$

Hence, option $\left(b\right)$ is correct.