The radius of a planet is frac{1}{4} of earth | vedclass

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(A) The formula for escape velocity is $v = \sqrt{2gR}$.Let $g_e$ and $R_e$ be the acceleration due to gravity and radius of the earth,and $g_p$ and $

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$\frac{1}{\sqrt{2}}$

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(A) The formula for escape velocity is $v = \sqrt{2gR}$.Let $g_e$ and $R_e$ be the acceleration due to gravity and radius of the earth,and $g_p$ and $R_p$ be the acceleration due to gravity and radius of the planet.Given: $R_p = \frac{1}{4} R_e$ and $g_p = 2g_e$.The ratio of escape velocity on the planet $(v_p)$ to the escape velocity on earth $(v_e)$ is:$\frac{v_p}{v_e} = \sqrt{\frac{g_p}{g_e} 	imes \frac{R_p}{R_e}}$Substituting the given values:$\frac{v_p}{v_e} = \sqrt{2 	imes \frac{1}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$Therefore,the escape velocity on the planet is $\frac{1}{\sqrt{2}}$ times the escape velocity on earth.

The radius of a planet is $\frac{1}{4}$ of earth's radius and its acceleration due to gravity is double that of earth's acceleration due to gravity. How many times will the escape velocity at the planet's surface be as compared to its value on earth's surface?

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