Question 122: Gravitation

Question

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

Accepted Answer

$$
\begin{aligned}
& M_E=9 M_P \\
& R_E=2 R_P
\end{aligned}
$$
Escape velocity $=\sqrt{\frac{2GM}{R}}$
For earth $v_e=\sqrt{\frac{2 G M_E}{R_E}}$
$$
\begin{aligned}
& 	ext { For } P, v_e=\sqrt{\frac{\frac{2 G M_E}{9}}{\frac{R_E}{2}}}=\sqrt{\frac{2 G M_E}{R_E} 	imes \frac{2}{9}} \\
& =\frac{v_e \sqrt{2}}{3}
\end{aligned}
$$

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Gravitation

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

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