# Gravitational Potential Energy

The gravitational potential energy of a system of two masses $M$ and $m$ separated by a distance $r$ is
\begin{align}
U=-\frac{GMm}{r}.
\end{align}

## Escape Velocity

The escape velocity from a planet of mass $M$ and radius $R$ is
\begin{align}
v_e=\sqrt{\frac{2GM}{R}}.
\end{align}
The escape velocity from the surface of the earth is 11.2 km/s.

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

## Related

- Law of Gravitation
- Gravitational Potential and Field
- Acceleration due to Gravity
- Motion of Planets and Satellites in Circular Orbits