Question

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

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

Answer 1

The correct option is C $V=\sqrt{gR}$

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:
$\Rightarrow \frac{-GMm}{2R}=\frac{1}{2}m{v}^2-\frac{GMm}{R}\Rightarrow \frac{1}{2}{v}^2=\frac{-GM}{2R}+\frac{GM}{R}$
$\Rightarrow \frac{1}{2}{v}^2=\frac{GM}{2R}\Rightarrow \frac{1}{2}{v}^2=\frac{1}{2}gR\Rightarrow v=\sqrt{gR}$