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JEE Advanced 2010Paper 1
Exam
JEE Advanced
Year
2010
Paper
Paper 1
Subject
Physics
Chapter
Gravitation

JEE Advanced 2010 · Paper 1 · Physics · Gravitation

Gravitational acceleration on the surface of a planet is 6 11 ~g ,...

Gravitational acceleration on the surface of a planet is 611 g\frac{\sqrt{6}}{11} \mathrm{~g}, where gg is the gravitational acceleration on the surface of the earth. The average mass density of the planet is 23\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\mathrm{kms}^{-1}, the escape speed on the surface of the planet in kms−1\mathrm{kms}^{-1} will be

Correct answer

3

Step-by-step solution

g=GMR2=G(43πR3)ρR2g=\frac{G M}{R^2}=\frac{G\left(\frac{4}{3} \pi R^3\right) \rho}{R^2} or g∝ρR\quad g \propto \rho R or R∝gρ\quad R \propto \frac{g}{\rho} Now escape velocity, ve=2gRv_e=\sqrt{2 g R} or ve∝gR\quad v_e \propto \sqrt{g R} or ve∝g×gρ∝g2ρ∴(ve)planet =(11km−s−1)6121×32∴\quad v_e \propto \sqrt{g \times \frac{g}{\rho}} \propto \sqrt{\frac{g^2}{\rho}} \therefore\left(v_e\right)_{\text{planet }}=\left(11 \mathrm{ km}-\mathrm{s}^{-1}\right) \sqrt{\frac{6}{121} \times \frac{3}{2}} \therefore The correct answer is 3 .
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