**Question 5: Calculate the escape velocity of a body and show that it is equal to 11.2 kms**ANSWER Calculation of escape velocity From our daily life experience we know that when a body is projected up, it falls down after attaining a certain height. If the initial velocity of projection is increased, it goes to a higher distance above in the atmosphere. This indicates that if we increase the velocity of projection further and further, a stage will come when the body will go out of the influence of gravitational pull and will never come back!

^{-1}.*The initial velocity which a projectile must have at the surface of earth in order to go out of earth’s gravitational field is called escape velocity.*In order to calculate the escape velocity, we know that the work done required to lift a body from earth’s surface to a far off distance where it is not influenced by the force of gravity is

Where m = mass of the object
M_{e} = mass of the earth
G = Gravitational constant.

_{esc}is the escape velocity at earth’s surface, then We know that, Put this value of GM

_{e}in equation (A), Equation (B) gives the initial velocity at earth’s surface that will enable the body to escape from the earth’s gravitational field and, hence, called escape velocity.

**To show that escape velocity is equal to 11.2 km/s**We know that g = 9.8 m/s

^{2}and R

_{e}= 6.4 x 10

^{6}m. Put these values in equation (B),

This proves the result.

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