Question

Gravitational acceleration on the surface of a planet is $\frac{ \sqrt{6}}{11}$ where g is the gravitational acceleration on the surface of the earth. The average mass density of the planet is 1/3 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.

• A. $\frac{-GM}{R}$
• B. $\frac{-GM}{2R}$
• C. $\frac{-2GM}{3R}$
• D. $\frac{-2GM}{R}$

Answer 1

Answer: (C)

$\frac{-2GM}{3R}$

Answer 2

g = $\frac{GM}{R^2} = \frac{G (\frac{4}{3} \pi R^3) \rho}{R^2}$ or $\, \, \, \, \, \, \, \, \, \, g \propto \rho R \, or \, R \propto \frac{g}{\rho}$ or $\, \, \, \, \, \, \, \, \, \, \, \, \, v_e \propto \sqrt{gR}$ or $\, \, \, \, \, \, \, \, \, \, \, \, \, v_e \rho \sqrt{gR}$ or $\, \, \, \, \, \, \, \, \, v_e \propto \sqrt{g \times \frac{g}{\rho} \propto \sqrt{\frac{g^2}{\rho}}}$ $\therefore \, \, \, \, \, \, \, \, \, (v_e)_{plant} = (11kms^{-1}) \sqrt{\frac{6}{121} \times \frac{3}{2}}= 3 kms^{-1}$ $\therefore$ The correct answer is 3