JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Gravitation

Gravitational acceleration on the surface of a planet is

\frac{6}{11} g

, where g is the gravitational acceleration on the surface of the earth. The average mass density of the planet is

\frac{2}{3}

times that of the earth. If the escape speed on the surface of the earth is taken to be

11 kms^{- 1}

, the escape speed on the surface of the planet in

kms^{- 1}

will be

Answer: 3

Solution

Given:

\frac{g ^{'}}{g} = \frac{6}{11}

and

\frac{ρ ^{'}}{ρ} = \frac{2}{3}

Since

g = \frac{GM}{R ^{2}} = \frac{4 π GρR}{3}

, we have:

\frac{g ^{'}}{g} = \frac{ρ ^{'} R ^{'}}{ρR}

Therefore:

\frac{R ^{'}}{R} = \frac{g ^{'} / g}{ρ ^{'} / ρ} = \frac{6 /11}{2/3} = \frac{3 6}{22}

The escape velocity is

v_{esc} = 2 g R

, so:

\frac{v _{esc}^{'}}{v _{esc}} = \frac{g ^{'} R ^{'}}{g R} = \frac{6}{11} \cdot \frac{3 6}{22} = \frac{18}{242} = \frac{9}{121} = \frac{3}{11}

Therefore:

v_{esc}^{'} = \frac{3}{11} \times 11 = 3 km/s

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Integer answer