JEE Main · 2008 · 27 Apr 2008 · Offline · Gravitation

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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

(A) $1.1 kms^{- 1}$
(B) $11 kms^{- 1}$
(C) $110 kms^{- 1}$ ✓ Correct
(D) $0.11 kms^{- 1}$

Answer: (C)

Solution

Escape velocity from a body of mass

M

and radius

R

v = \frac{2 GM}{R}

. The planet is ten times as massive as the Earth and ten times smaller in radius, so the factor

\frac{M}{R}

under the root is multiplied by

10 \times 10 = 100

. The escape velocity is therefore

100 = 10

times the Earth's value, and with the Earth's figure of

11 km s^{- 1}

this gives

10 \times 11 = 110 km s^{- 1}

JEE Main · 2008 · 27 Apr 2008 · Offline · Single correct option

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