# 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,
$dW=Fdx$ $dW=\frac{GMm}{x^2}\cdot dx$
​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 = 0.0098km/s² and R=6400 km
$∴v=\sqrt{2×0.0098*6400}$ $v=11.2 \text{ km/s}$
​Thus, escape velocity of the earth is 11.2km/s.

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