Question

A particle is thrown with escape velocity v from the surface of earth calculate its velocity at height 3r

Answer 1

At the instant , where the particle is at height 3R , following things are taken into concern

The total mechanical energy is conserved , that is

The KE required by the particle to reach height 3R , is equal to the GPE ( Gravitational potential energy) at the point 3R from the surface of earth

Therefore ,

The total height = height from surface + radius of earth

= 3R + R = 4R

Then , we have ;

KE = GPE

½mVA2 = GMm/4R ( VA = Velocity at 3R from surface )

Solving for VA ,

We get : VA = \sqrt{2GM/4R}

= \sqrt{GM/2R} (\sqrt{2} / \sqrt{2}) ( Multiply and divide \sqrt{2} )

=\sqrt{\frac{2GM}{R}} ( \frac{1}{2})

=\frac{_{Ve}}{2} ( Ve = Escape velocity of earth

= \frac{11.2 Km/s}{2}

= 5.6 Km/s