Question

what is escape speed ? derive an expression for it on the surface of earth .

Answer 1

An object can be thrown up with a certain minimum initial velocity so that, the object goes beyond the earth's gravitational field and escape from earth, this velocity is known as escape velocity of the earth. So, escape velocity is defined as the minimum initial velocity that will take a body away above the surface of a planet when it's projected vertically upwards.

On throwing the object upwards, work has to be done against the gravity. If a certain minimum velocity is given to an object, such that the work done against gravity from the planet's surface to infinity(outside gravitational field of planet) is equal to the kinetic energy of the object, then it will not return back to the planet.

Derivation of Escape Velocity
Suppose that the planet be a perfect sphere of radius R having mass M. Let a body of mass m is to be projected from point A on the earth's surface as shown in the figure. Join OA and produce it further. Let us take two points P and Q which are at distances x and (x + dx) from the center of the earth.

To calculate the escape velocity of the earth, let the minimum velocity to escape from the earth's surface be ve. Then, kinetic energy of the object of maVelocity
Ke=1/2 mve sq.
When the projected object is at point P which is at a distance x from the center of the earth, the force of gravity between the object and earth is
F=gmn/r
Work done in taking the body against gravitational attraction from P to Q is given by The total amount of work done in taking the body against gravitational attraction from surface of the earth to infinity can be calculated by integrating the above equation within the limits x = R to x = ∞. Hence, total work done is
Dw=fdx=gmn/x sq.
For the object to escape from the earth's surface, kinetic energy given must be equal to the work done against gravity going from the earth's surface to infinity, hence

K.E. of Object should be Equal to Magnitude of P.E.
W=gmn/r
Since,
K=w or 1/2 mve sq. =gmn/r
Ve=root 2gmn/r
The relation shows that the escape velocity of an object does not depend on the mass of the projected object but only on the mass and radius of the planet from which it is projected.
G=gm/r sq.
Escape Velocity of Earth
For the earth, g = 9.8 m/s2 and R = 6.4 X 106 m, then
2*9.8*6.4*10 power 6 hole root
11.2*10 power 3 m/s
11.2 km/s


That means, a spacecraft leaving earth surface should have 11.2 km/sec or 7 miles/sec initial velocity to escape from earth's gravitational field. 

Answer 2

Escape Speed(Ve) :

• It is minimum speed with which a body is thrown it will go beyond coverage of earth ie never returned back.

Derivation:

Gravitational potential energy at a height h above earth surface is

u = -GMem/(Re+h)

where ;

• G= universal gravitational constant
• Me= mass of the earth
• m = mass of object
• Re= radius of the earth
• h= height of the object

Now,

Body will go beyond coverage of earth if we give minimum energy:

+GMem/(Re+h)

$\longrightarrow$ K.E = 1/2mVe²

$\longrightarrow$ 1/2mVe²= +GMem/(Re+h)

$\longrightarrow$Ve²= +2GMem/(Re+h)m

$\longrightarrow$Ve²= +2GMe/Re+h

$\longrightarrow$ Ve=√(2GMe/Re+h)