The mass of the moon is 1/144 times the mass | vedclass

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(A) The formula for escape velocity is $V_e = \sqrt{\frac{2GM}{R}}$.Let $M_p$ and $R_p$ be the mass and radius of the planet,and $M_m$ and $R_m$ be th

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$v/3$

Vedclass · Answer

(A) The formula for escape velocity is $V_e = \sqrt{\frac{2GM}{R}}$.Let $M_p$ and $R_p$ be the mass and radius of the planet,and $M_m$ and $R_m$ be the mass and radius of the moon.Given: $M_m = \frac{M_p}{144}$ and $D_m = \frac{D_p}{16}$,which implies $R_m = \frac{R_p}{16}$.Escape velocity on the planet is $V_p = \sqrt{\frac{2GM_p}{R_p}} = v$.Escape velocity on the moon is $V_m = \sqrt{\frac{2GM_m}{R_m}} = \sqrt{\frac{2G(M_p/144)}{(R_p/16)}} = \sqrt{\frac{2GM_p}{R_p} 	imes \frac{16}{144}} = \sqrt{\frac{2GM_p}{R_p} 	imes \frac{1}{9}}$.Substituting $V_p = v$,we get $V_m = \sqrt{\frac{v^2}{9}} = \frac{v}{3}$.

The mass of the moon is $1/144$ times the mass of a planet and its diameter is $1/16$ times the diameter of a planet. If the escape velocity on the planet is $v$,the escape velocity on the moon will be:

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