Gravitational Acceleration on the Surface of a Planet is √6/11g...

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Problem: Gravitational acceleration on the surface of a planet is $\frac{\sqrt{6}}{11}g$, where g is the gravitational acceleration on the surface of the earth. The average mass density of the planet is 2/3 times that of the earth. If the escape speed on the surface of the earth is taken to be 11 km/s, the escape velocity on the surface of the planet (in km/s) will be,  (IIT JEE 2010)

Solution: The acceleration due to gravity on the surface of a planet of mass $M_p$ and radius $R_p$ is given by, \begin{align} g_p=GM_p/R_p^2. \end{align} The mass is related to density $\rho_p$ by $M_p=(4/3)\pi R_p^3\rho_p$. Substitute in above equation and simplify to get, \begin{align} R_p=3g_p/(4\pi G\rho_p). \end{align} The escape velocity from the surface of the planet is given by, \begin{align} v_p&=\sqrt{2GM_p/R_p}\nonumber\\ &=\sqrt{2g_p R_p}\nonumber\\ &=\sqrt{3g_p^2/(2\pi G\rho_p)}. \end{align} Substitute values in above equation to get, \begin{align} \frac{v_p}{v_e}&=\sqrt{\frac{g_p^2/g_e^2}{\rho_p/\rho_e}}\nonumber\\ &=\sqrt{\frac{6/121}{2/3}}\nonumber\\ &=\frac{3}{11}, \end{align} which gives $v_p={3}\;\mathrm{km/s}$.

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References

  1. IIT JEE Physics by Jitender Singh and Shraddhesh Chaturvedi
  2. Concepts of Physics Part 1 by HC Verma (Link to Amazon)