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 $11 {kms}^{- 1}$ , the escape velocity from the surface of the planet would be

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Escape-xA0-velocity-x221A-2Gmrvnewvearth-x221A-mnewrearthmearthrnew-xE001-xE000-xE000-xE000-x23B7-10110-10vnew-110-xA0-kms-x2212-1

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 $11 k m$ $s^{- 1}$ , the escape velocity from the surface ofthe planet would be

$E s c a p e v e l o c i t y = \sqrt{\frac{2 G m}{r}} \frac{v_{n e w}}{v_{e a r t h}} = \sqrt{\frac{m_{n e w} r_{e a r t h}}{m_{e a r t h} r_{n e w}}} = \sqrt{\frac{10}{\frac{1}{10}}} = 10 v_{n e w} = 110 k m s^{- 1}$

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 11km s−1, the escape velocity from the surface ofthe planet would be

Escape velocity=√2Gmrvnewvearth=√mnewrearthmearthrnew=  ⎷10110=10vnew=110 kms−1

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 $11 k m$ $s^{- 1}$ , the escape velocity from the surface ofthe planet would be

$E s c a p e v e l o c i t y = \sqrt{\frac{2 G m}{r}} \frac{v_{n e w}}{v_{e a r t h}} = \sqrt{\frac{m_{n e w} r_{e a r t h}}{m_{e a r t h} r_{n e w}}} = \sqrt{\frac{10}{\frac{1}{10}}} = 10 v_{n e w} = 110 k m s^{- 1}$