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  4. 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's surface is 11 kms -1, the escape velocity from the surface of the planet would be

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Q. 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's surface is $11\, kms ^{-1}$, the escape velocity from the surface of the planet would be

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Solution:

$\frac{\left(v_{e}\right)_{p}}{\left(v_{e}\right)_{e}}=\frac{\sqrt{\frac{2 G M_{p}}{R_{p}}}}{\sqrt{\frac{2 G M_{e}}{R_{e}}}}=\sqrt{\frac{M_{p}}{M_{e}} \times \frac{R_{e}}{R_{p}}}$
$=\sqrt{\frac{10 M_{e}}{M_{e}} \times \frac{R_{e}}{R_{e} / 10}}=10$
$\therefore \left(v_{e}\right)_{p}=10 \times\left(v_{e}\right)_{e}$
$=10 \times 11=110\, km / s$