Gravitation

JEE Main level, single correct, medium

158 of 223

Question 158 of 223

A planet in a distant solar system is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth is 11kms−111 \mathrm{kms}^{-1}, the escape velocity from the surface of the planet would be

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Answer: (C) 110kms−1110 \mathrm{kms}^{-1}

vesc=2GMR=2G×10MR/10=10×11=110 km/sv_{\mathrm{esc}}=\sqrt{\frac{2 \mathrm{GM}}{R}}=\sqrt{\frac{2 \mathrm{G} \times 10 \mathrm{M}}{\mathrm{R} / 10}}=10 \times 11=110 \mathrm{~km} / \mathrm{s}

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