If earth has a mass nine times and radius twice to of a planet P . Then frac{vₑ}{3} √{x} ms ^{-1

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7.Gravitation

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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

A

$2$

B

$3$

C

$18$

D

$1$

(JEE MAIN-2023)

Solution

$v_{\text {(escape) plant }}=\sqrt{\frac{2 G M_P}{R_P}}$

$=\sqrt{\frac{2 G\left(\frac{M_e}{9}\right)}{\left(\frac{R_e}{2}\right)}}=\frac{v_e \sqrt{2}}{3} \therefore x=2$

Standard 11

Physics

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