Gravitation

JEE Main level, single correct, medium

119 of 223

Question 119 of 223

The ratio of escape velocity of a planet to the escape velocity of earth will be:-
Given: Mass of the planet is 1616 times mass of earth and radius of the planet is 44 times the radius of earth.

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Answer: (D) 2:12:1

The formula to calculate the escape velocity is given by
ve=2GMR...(1){v}_{e}=\sqrt{\dfrac{2GM}{R}}...\left(1\right)
For the Earth, it is given by
veE=2GMERE...(2){{v}_{e}}^{E}=\sqrt{\dfrac{2G{M}_{E}}{{R}_{E}}}...\left(2\right)
and, for the planet, it is given by
veP=2GMPRP...(3){{v}_{e}}^{P}=\sqrt{\dfrac{2G{M}_{P}}{{R}_{P}}}...\left(3\right)
Divide equation (2) by equation (3) and simplify to obtain the required ratio.
veEveP=MEMP×RPRE...(4)\dfrac{{{v}_{e}}^{E}}{{{v}_{e}}^{P}}=\sqrt{\dfrac{{M}_{E}}{{M}_{P}}}\times \sqrt{\dfrac{{R}_{P}}{{R}_{E}}}...\left(4\right)
Substitute the values of the known parameters into equation (4) to obtain the required ratio.
veEveP=ME16ME×4RERE=12⇒veP:veE=2:1\begin{matrix}\dfrac{{{v}_{e}}^{E}}{{{v}_{e}}^{P}} & = & \sqrt{\dfrac{{M}_{E}}{16{M}_{E}}}\times \sqrt{\dfrac{4{R}_{E}}{{R}_{E}}} \\ & = & \dfrac{1}{2} \\ & \Rightarrow & {{v}_{e}}^{P}:{{v}_{e}}^{E}=2:1\end{matrix}

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