Gravitation Question 16
Question 16 - 01 February - Shift 1
If earth has a mass nine times and radius twice to the 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
(1) 2
(2) 3
(3) 18
(4) 1
Show Answer
Answer: (1)
Solution:
Formula: Escape velocity
$v _{\text{(escape)plant }}=\sqrt{\frac{2 G M_P}{R_P}}$
$=\sqrt{\frac{2 G(\frac{M_e}{9})}{(\frac{R_e}{2})}}=\frac{v_e \sqrt{2}}{3} \therefore x=2$
