Question

A ball of small size is released from height $R_E$. Find the speed of the ball when it reaches the surface of earth. Assume $R_E$ is the radius of earth and $g$ is acceleration due to gravity on earth's surface.

• A. $\sqrt{2g{R}_E}$
• B. $\sqrt{g{R}_E}$
• C. $\sqrt{\frac{g{R}_E}{2}}$
• D. None

Answer 1

The correct option is B $\sqrt{g{R}_E}$

From law of conservation of mechanical energy,
$\text{KE + PE= constant}$
$\Rightarrow 0-\frac{GMm}{2R_E}=\frac{1}{2}m{v}^2-\frac{GMm}{R_E}$
On solving this, we get,
$v=\sqrt{g{R}_E}$