Escape Velocity

​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,
$$dW=Fdx$$ $$dW=\frac{GMm}{x^2}dx$$
​Integrating within limits from $R$ to $\infty$,
$$W=GMm{\int }_R^{\infty }\frac{dx}{x^2}$$$$W=GMm{\left[\frac{x^{-1}}{-1}\right]}_R^{\infty }$$$$W=-GMm[\frac{1}{\infty }-\frac{1}{R}]$$$$W=\frac{GMm}{R}$$

To escape the gravitational field, the potential energy is converted into kinetic energy.
$$\therefore K.E.=\frac{GMm}{R}$$
If $v$ is the escape velocity of the object, then,
$$\frac{1}{2}m{v}^2=\frac{GMm}{R}$$ $$v^2=\frac{2GM}{R}$$ $$v=\sqrt{\frac{2GM}{R}}$$
We have, $$g{R}^2=GM$$ $$\therefore v=\sqrt{\frac{2g{R}^2}{R}}$$ $$v=\sqrt{2gR}$$
This gives the escape velocity.
For earth, $g=0.0098{\text{ km/s}}^2$ and $R=6400\text{ km}$
$$\therefore v=\sqrt{2\times 0.0098\ast 6400}$$ $$v=11.2\text{ km/s}$$
​Thus, escape velocity of the earth is $11.2\text{ km/s}$.