## Escape Velocity

Escape Velocity – 1. If we project a body vertically upwards, it moves up to a certain height then falls back due to gravitational attraction of the earth. The maximum height attained by it depends upon the initial velocity of projection. Higher is initial velocity of projection, higher is the height attained by it. Ultimately, at a certain velocity of projection the body will go out of the gravitational field of the earth and will never return to the earth. Hence, the minimum velocity with which a body must be projected in the atmosphere so as to enable it to just overcome the gravitational attraction of the earth is called as Escape velocity.

2. Let us calculate the energy required by a body of mass m to escape gravitational attraction. We know that gravitational potential energy of a body of mass m placed on earth’s surface is given by (assuming gravitational potential energy zero at infinity).
U= – GMem/Re
Therefore, in order to take a body from the earth’s surface to infinity, the work required is GMem/Re.
Hence it is evident that if we throw a body of mass m with such a velocity that its kinetic energy is GMem/Re, then it will move outside the gravitational field of earth. Hence
Escape energy = + GMem/Re

3. The velocity given to the body is called as the escape velocity, If it is equal to v0, then
(½) mve2=GMem/Re or ve =√(2GMe/Re)
Thus we find that the escape velocity is independent of the mass of the body. If g is the acceleration due to gravity at the earth’s surface, then
G = GMem/Re2 or GMe = gRe2
Putting this value of GMe in eq.(1), we get
ve=√2gRe
We know that g= 9.8m/s2 and Re = 6.4 × 106 m.
Hence
ve = (2×9.8×6.4×106)=11.2×103 m/s = 11.2 km/sec
Therefore, if a body is thrown upwards with a velocity of 11.2 km/sec., then it will never return to earth.

Note:a) In order to send a body to a height of 6.4×106m(i.e., a height equal to the radius of earth) from the earth, we have to project the body with a velocity of 7.92 km/sec; while to send it to infinity, only a little larger velocity 11.2 km/sec. is required.
b) The orbital velocity of a satellite close to the earth is v0 =(gRe), while the escape velocity for a body thrown from the earth’s surface is Ve =( 2gRe)Thus
(v0/ve)=[√(gRe)/(√2gRe)=1/√2 or ve= √2 v0
I.e., if the orbital velocity of a satellite revolving close to the earth happens to increase to 2 times, the satellite would escape.