If earth has a mass nine times and radius twice to that of a planet P. Then $\frac{v_{e}}{3} \sqrt{x} \mathrm{~ms}^{-1}$ will be the minimum velocity required by a rocket to pull out of gravitational force of $\mathrm{P}$, where $v_{e}$ is escape velocity on earth. The value of $x$ is
Solution
$$
\begin{aligned}
& M_E=9 M_P \\\\
& R_E=2 R_P
\end{aligned}
$$
<br/><br/>Escape velocity $=\sqrt{\frac{2GM}{R}}$
<br/><br/>For earth $v_e=\sqrt{\frac{2 G M_E}{R_E}}$
<br/><br/>$$
\begin{aligned}
& \text { For } P, v_e=\sqrt{\frac{\frac{2 G M_E}{9}}{\frac{R_E}{2}}}=\sqrt{\frac{2 G M_E}{R_E} \times \frac{2}{9}} \\\\
& =\frac{v_e \sqrt{2}}{3}
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Gravitation · Topic: Escape Velocity
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