# 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 $v_p={3}\;\mathrm{km/s}$.