Question

A body is released from height equal to r what is its final velocity

Answer 1

Since gravitational field is conservative, we can solve this problem by considering conservation of mechanical energy.

At height R above the surface of the earth, mechanical energy,

E1=kinetic energy + potential energy

=0 -GmM/2R………(1).

M is mass of the earth, m is mass of the body, R is radius of the earth and G is universal constant of gravitation.

At the surface of the earth,

E2=(1/2)mv^2-GMm/R………..(2).

Now, E1=E2 ( conservation of energy). Therefore ,

-GMm/2R=(1/2)mv^2-GMm/R

OR (1/2)mv^2=GMm/2R, therefore ,

v=( GM/R)^1/2

But GM/R^2=g, therefore,

v=(gR)^1/2. Here, g is acceleration due to gravity at the surface of the earth.

We have neglected the resistive forces and relativistic effect.

I hope its help you