Question

If earth has a mass nine times and radius twice to that of a planet $P$ Then $\frac{v_e}{3} \sqrt{x} ms ^{-1}$ will be the minimum velocity required by a rocket to pull out of gravitational force of $P$, where $v_e$ is escape velocity on earth The value of $x$ is

• A. 1
• B. 18
• C. 3
• D. 2

Answer 1

The Correct Option is D

$v_{\text{(escape) plant}}=\sqrt{\frac{2G{M}_P}{R_P}}$
$=\sqrt{\frac{2G\left(\frac{M_e}{9}\right)}{\left(\frac{R_e}{2}\right)}}=\frac{v_e\sqrt{2}}{3}$
$\therefore x=2$

Answer 2

Escape velocity for a planet is given by: \[ V_{\text{escape}} = \sqrt{\frac{2GM}{R}} \] For planet P, \[ V_{\text{escape, P}} = \sqrt{\frac{2G \left(\frac{M}{9} \right)}{\frac{R}{2}}} \] \[ = V_e \frac{\sqrt{2}}{3}, \quad x = 2 \]