When we throw an object in the atmosphere, it falls back to the surface of the earth after reaching to a certain height depending on the velocity given to it. If we throw the object with a velocity so that it overcomes the gravitational pull then such velocity is known as escape velocity and the object will never return to the surface of the earth.

*The minimum velocity with which a body is projected in the atmosphere so that it escapes the gravitational field of the earth is known as escape velocity.*

Consider earth to be a uniform sphere of radius $R$ and mass $M$. Suppose an object of mass $m$ is placed at a point $P$ which is $x$ distance apart from the centre of the earth.

The gravitational force of the earth on the object is,

\[F=\frac{GMm}{x^2}\]

Then, the small work done to move the object through infinitesimally small distance $dx$ is,

\[dW=F\;dx\] \[dW=\frac{GMm}{x^2}\; dx\]

Integrating within limits from $R$ to $∞$,

\[W=GMm \int_R^∞ \frac{dx}{x^2}\]\[W=GMm \left[\frac{x^{-1}}{-1}\right]_R^∞\]\[W=-GMm \left[\frac{1}{∞}-\frac{1}{R}\right]\]\[W=\frac{GMm}{R}\]

To escape the gravitational field, the potential energy is converted into kinetic energy.

\[∴K.E.=\frac{GMm}{R}\]

If $v$ is the escape velocity of the object, then,

\[\frac{1}{2} mv^2 = \frac{GMm}{R}\] \[v^2=\frac{2GM}{R}\] \[v=\sqrt{\frac{2GM}{R}}\]

We have, \[gR^2 = GM\] \[∴v=\sqrt{\frac{2gR^2}{R}}\] \[v=\sqrt{2gR}\]

This gives the escape velocity.

For earth, $g = 0.0098\text{ km/s}^2$ and $R=6400\text{ km}$

\[∴v=\sqrt{2×0.0098*6400}\] \[v=11.2 \text{ km/s}\]

Thus, escape velocity of the earth is $11.2\text{ km/s}$.