The escape velocity of an object on a planet | vedclass

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Question

(A) The formula for escape velocity is $v_e = \sqrt{2gR}$.Given,$g_p = 9g_e$ and $R_p = 4R_e$,where $g_p$ and $R_p$ are the acceleration due to gravit

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$67.2$

Vedclass · Answer

(A) The formula for escape velocity is $v_e = \sqrt{2gR}$.Given,$g_p = 9g_e$ and $R_p = 4R_e$,where $g_p$ and $R_p$ are the acceleration due to gravity and radius of the planet,and $g_e$ and $R_e$ are those of Earth.The ratio of escape velocities is $\frac{v_p}{v_e} = \sqrt{\frac{g_p}{g_e} 	imes \frac{R_p}{R_e}}$.Substituting the given values: $\frac{v_p}{v_e} = \sqrt{9 	imes 4} = \sqrt{36} = 6$.Since the escape velocity on Earth is approximately $v_e = 11.2 \ km/s$,the escape velocity on the planet is $v_p = 6 	imes 11.2 \ km/s = 67.2 \ km/s$.

The escape velocity of an object on a planet whose $g$ value is $9$ times that of Earth and whose radius is $4$ times that of Earth in $km/s$ is:

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