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  4. If earth has a mass nine times and radius twice to that of a planet P. Then (ve/3) √x ms -1 will be the minimum velocity required by a rocket to pull out of gravitational force of P, where ve is escape velocity on earth. The value of x is

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Q. 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

JEE MainJEE Main 2023Gravitation

Solution:

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