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Lesson 3, Topic 5
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# Escape Velocity

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This is the minimum velocity with which an object would just escape from the earth gravitational force influence.

If a body is projected upward, it moves against the gravitational influence of the earth. The maximum height attained depends on the velocity of projection of the body, with enough velocity it can escape the earthâ€™s gravitational attraction.

Considering a rocket of mass â€˜mâ€™ escape from earthâ€™s gravitational influence with velocity â€˜vâ€™ the work done by the rocket is

V= $$– \frac{GMm} {r}$$

The kinetic energy of the rocket is equal to the work done against the gravity

k.e=work done against gravity = potential energy

Therefore,

Â $$\frac{1}{2} \scriptsize MV^2 = \frac{GMm} {r}$$Â  = MgR

Â V 2 = 2gr

Or Â

V = $$\scriptsize \sqrt{2gr}$$

R = radius of the earth

g = acceleration due to gravity

The escape velocity depends on gravity and distance or height from where it is launched. It is independent of mass of a body.

If R = 6.4 x 10 6 m , g = 9.8 Â x 10 6 ms-1

Then, the magnitude of velocity that a body must possess to escape from the earthâ€™s gravitation pull is

V = $$\scriptsize \sqrt{2\; \times \; 9.8 \times 10^6 \; \times \; 6.4 \times 10^6}$$

=1.13 x 10 4 ms-1

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