what is escape velocity?obtain the expression for the escape velocity on earth ?find escape velocity on the surface of the earth.

Dear Student!!

The minimum velocity with which a projectile has to be projected to escape the earth’s gravitational field is called escape velocity.

Let a body of mass ‘m’ be projected with velocity ‘v’.

At the ground, PE = GMm/R

And the KE = ½ mv^{2}

To overcome earth’s gravitational field,

KE>PE

=> ½ mv^{2} > GMm/R

=> v > (2GM/R)^{1/2}

We know,

g = GM/R^{2}

Therefore,

v > [2(GM/R^{2})R]^{1/2}

=> v > [2gR]^{1/2}

Thus, to escape the earth’s gravitational field, the minimum velocity must be,

v_{e} = [2gR]^{1/2}

For the last part please refer the answer posted by Ashley.

@ Ashley: Good effort keep it up!!

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