IIT JEE Physics (1978 to 2018: 41 Years) Topic-wise Complete Solutions

# Acceleration due to Gravity

## Problems from IIT JEE

Problem (IIT JEE 2010): 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,

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} \label{vda:eqn:1} 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} \label{vda:eqn:2} 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} \label{vda:eqn:3} v_p=\sqrt{2GM_p/R_p}=\sqrt{2g_p R_p}=\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}}=\sqrt{\frac{6/121}{2/3}}=\frac{3}{11}, \end{align} which gives $v_p={3}\;\mathrm{km/s}$.