Question

expression for escape velocity earth

Answer 1

It is define as the minimum velocity with which a body must be projected upward so that it escapes away from the gravitational pull of earth
V =
$\sqrt{2gr} \\ \sqrt{2 \times 9.8 \times 6.4 \times {10}^{6} } = 11.2km \: per \: sec$

Answer 2

hell... ooo!!

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

At the ground, PE = GMm/R

And the KE = ½ mv2

To overcome earth’s gravitational field,

KE>PE

=> ½ mv2 > GMm/R

=> v > (2GM/R)1/2

We know,

g = GM/R2

Therefore,

v > [2(GM/R2)R]1/2

=> v > [2gR]1/2

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

ve = [2gR]1/2

hope this helps you