What is an escape velocity? Obtain an expression for it.

Hint: On the surface of the Earth, the escape velocity is near $11.2km{{s}^{-1}}$, which is approximately 33 times the speed of sound and so many times the muzzle velocity of a rifle bullet . However, at $9000km$ altitude in space, it is slightly lesser than $7.1km{{s}^{-1}}$.

Complete step-by-step answer:
Escape velocity is described as the minimum velocity with which an object is needed to leave a planet or moon. The formula for escape velocity consists of a constant, \[g\] which we mention as the acceleration due to gravity.
The formula of the escape velocity is given as,
\[v=\sqrt{2gR}\]
Where R is the distance between earth and the object. Now let us derive the formula for escape velocity.
Suppose the body of mass $m$ which is residing on the earth surface. Earth is having a total mass of $M$ and radius of \[R\] respectively. Let the force experienced over the small mass will be written as,
\[F=\dfrac{GMm}{{{R}^{2}}}\]
Where \[G\] the universal gravitational constant, $M$ is the mass of earth, $m$ is the mass of the small body on earth and \[R\] is the distance from the earth’s surface.
The gravitational potential energy is given by,
\[w=F\cdot R\]
\[w=\dfrac{GMm}{R}\]
In order to make a body escape from the surface of earth should be given a kinetic energy equivalent to its potential energy.
That is,
Kinetic energy=potential energy
\[\dfrac{1}{2}m{{v}^{2}}=\dfrac{GMm}{R}\]
From this equation after rearranging the terms, the escape velocity is given as,
\[v=\sqrt{\dfrac{2GM}{R}}\]
We know,
$gR=\dfrac{GM}{R}$
Substituting this,
\[v=\sqrt{2gR}\]
Hence proved.

Note: Basically, it is meant by the escape from the land without any chance of falling back. Therefore, any object or body which acquires escape velocity on the surface of the earth can completely escape the gravitational field of the earth in addition by avoiding the losses due to the atmosphere.