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.

Escape Velocity

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 $11.2\text{ km/s}$.


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