Question

A body is released from a point of distance $R^{\prime }$ from the centre of earth. Its velocity at the time of striking the earth will be $(R^{\prime }>R_e)$

• A. $\sqrt{2g{R}_e}$
• B. $\sqrt{R_{e}g}$
• C. $\sqrt{2g(R^{\prime }-R_e)}$
• D. $\sqrt{2g{R}_e(1-\frac{R_e}{R^{\prime }})}$

Answer 1

Answer: (D)

$\sqrt{2g{R}_e(1-\frac{R_e}{R^{\prime }})}$

$\text{Initial energy of the body}=-\frac{GMm}{R^{\prime }}$
$\text{Applying conservation of energy}-\frac{GMm}{R^{\prime }}=\frac{1}{2}m{v}^2-\frac{GMm}{R_e}$
$\text{Here}v\text{is the required velocity or}\frac{v^2}{2}=\frac{GM}{R_e}-\frac{GM}{R^{\prime }}=GM\left(\frac{R^{\prime }-R_e}{R^{\prime }{R}_e}\right)$
$\therefore v=\sqrt{\frac{2GM(R^{\prime }-R_e)}{R^{\prime }{R}_e}}$
But $GM=g{R_e}^2$
$v=\sqrt{\frac{2g{R_e}^2(R^{\prime }-R_e)}{R^{\prime }{R}_e}}$
$\therefore v=\sqrt{2g{R}_e(1-\frac{R_e}{R^{\prime }})}$