Question

what is the escape velocity and derive escape velocity

Answer 1

Hey there!!!


=>> 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 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 it helped!!!!☺️

Answer 2

In physics, escape velocity is the minimum speed needed for an object to escape from the gravitational influence of a massive body. The escape velocity from Earth is about 11.186 km/s (6.951 mi/s; 40,270 km/h; 25,020 mph) at the surface