The ratio of the escape velocity of a planet | vedclass

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(B) The formula for escape velocity is $V_e = \sqrt{\frac{2GM}{R}}$.Let $M_E$ and $R_E$ be the mass and radius of the Earth,and $M_P$ and $R_P$ be the

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$2:1$

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(B) The formula for escape velocity is $V_e = \sqrt{\frac{2GM}{R}}$.Let $M_E$ and $R_E$ be the mass and radius of the Earth,and $M_P$ and $R_P$ be the mass and radius of the planet.Given: $M_P = 16 M_E$ and $R_P = 4 R_E$.The escape velocity of the planet is $V_P = \sqrt{\frac{2GM_P}{R_P}} = \sqrt{\frac{2G(16M_E)}{4R_E}} = \sqrt{4 	imes \frac{2GM_E}{R_E}} = 2 \sqrt{\frac{2GM_E}{R_E}}$.Since $V_E = \sqrt{\frac{2GM_E}{R_E}}$,we have $V_P = 2 V_E$.Therefore,the ratio $\frac{V_P}{V_E} = \frac{2}{1}$,which is $2:1$.

The ratio of the escape velocity of a planet to the escape velocity of the Earth will be: Given: Mass of the planet is $16$ times the mass of the Earth and the radius of the planet is $4$ times the radius of the Earth.

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