Show that the escape velocity of a body from earth surface is √2Vc where Vc is critical velocity of body orbitingvery close to earth surface

Answer 1

To escape Earth's gravity, an object's kinetic energy must be greater than or equal to its gravitational potential energy:

(1/2)mv² = GMm/R

1. M = the object's mass
2. v = the object's velocity (escape velocity)
3. G = gravitational constant
4. M = Earth's mass
5. R = Earth's radius

Solving for v:

v² = 2GM/R

v = √(2GM/R)

Now for an object orbiting right at the Earths surface, its centripetal force (mv²/R) will exactly equal the gravitational force:

mv_c²/R = GMm/R²

v_c² = GM/R

So the escape velocity is:

v = √(2GM/R) = √(2v_c²) = √(2)·v_c