4 6 Escape Velocity – Class 11TH Physics

4.6 Escape Velocity

It is our daily life experience that an object projected upward comes back to the ground after rising to a certain height. This is due to the force of gravity acting downward. With increased initial velocity, the object rises to the greater height before coming back. If we go on increasing the initial velocity of the object, a stage comes when it will not return to the ground. It will escape from the influence of gravity.

The initial velocity of an object with which it goes out of the Earth's gravitational field, is known as escape velocity.

Some Escape speeds ( $\mathrm{km\,s^{-1}}$ )

Heavenly body	Escape speed
Moon	$2.4$
Mercury	$4.3$
Mars	$5.0$
Venus	$10.4$
Earth	$11.2$
Uranus	$22.4$
Neptune	$25.4$
Saturn	$37.0$
Jupiter	$61$

The escape velocity corresponds to the initial kinetic energy gained by the body, which carries it to an infinite distance from the surface of the Earth.

\text{Initial K.E.} = \tfrac{1}{2}mv_{\mathrm{esc}}^{2}

We know that the work done in lifting a body from Earth's surface to an infinite distance is equal to the increase in potential energy.

\text{Increase in P.E.} = 0 - \left(-\frac{GMm}{R}\right) = \frac{GMm}{R}

where $M$ and $R$ are the mass and radius of the Earth respectively. The body will escape out of the gravitational field if the initial K.E. of the body is equal to increase in P.E. Then:

\tfrac{1}{2}mv_{\mathrm{esc}}^{2} = \frac{GMm}{R}

v_{\mathrm{esc}} = \sqrt{\frac{2GM}{R}} \quad \ldots (4.14)

As $g = GM/R^{2}$ or $gR = GM/R$

Hence

v_{\mathrm{esc}} = \sqrt{2gR} \quad \ldots (4.15)

The value of $v$ comes out to be approximately $11\,\mathrm{km\,s^{-1}}$ .

4.6 Escape Velocity

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