Question

Define escape velocity of a body and hence obtain the expression for it

Answer 1

The minimum velocity given to a body on the surface of the Earth so that it escaped the Earths gravitational field and reached an infinite distance from the surface of the Earth

OR

The velocity with which the body goes out from the Earth gravitational field.

Initial K.E = 1/2mv^2

When object reaches infinity K.E = 0

So change in K.E = 1/2mv^2

P.E will be increased when body is lifted from surface of the Earth so

Increase in P.E = 0- ( -GMm/R ) = GMm/R

-GMm/R is the work done by gravitational force on a body in moving it from the surface of the Earth to infinity where the force if gravity becomes 0.

1/2mv^2 = GmM/R

Vesc = √(2GM/R)...(I)

Gravitational field strength = Fg/m

g = GMm/r^2 x 1/m

Gravitate field strength, g= GM/R^2

gR^2 = GM....(ii)

Vesc = √2gR