Question

A body is dropped from a height equal to radius of the earth. The velocity acquired by it before touching the ground is

• A. $V=\sqrt{2gR}$
• B. V=3gR
• C. $V=\sqrt{gR}$
• D. V=2gR

Answer 1

Answer: (C)

To find the velocity, we conserve the energy between the point of release and at Earth's surface. Since the distance from the Earth's surface to the body $(R)$ is large, we shouldn't just take $g$ while calculating the potential energy. Total energy before dropping: $\frac{−GMm}{R+R}$ Total energy as the body hits Earth's surface: $\frac{1}{2}m{v}^2−\frac{GMm}{R}$ Now these two are equal: $⇒\frac{−GMm}{2R}=\frac{1}{2}m{v}^2−\frac{GMm}{R}⇒\frac{1}{2}{v}^2=\frac{−GM}{2R}+\frac{GM}{R}$ $⇒\frac{1}{2}{v}^2=\frac{GM}{2R}⇒\frac{1}{2}{v}^2=\frac{1}{2}gR⇒v=\sqrt{gR}$