Question

A body starts from rest from a point a >radius of earth from center .The velocity acquired by it when it reaches the earth surface

Answer 1

When the body is at a distance a, the potential energy of the Earth-body system is:

$U_{i} = -\frac{GM_{e}M_{b}}{a}$

At the surface of Earth, the potential energy for the same system shall be:

$U_{f} = -\frac{GM_{e}M_{b}}{R_{e}}$

The change in internal energy is mostly transferred to the kinetic energy of the body, due to the large mass of Earth.

So, $KE_{f} - KE_{i} = U_{i} - U_{f}$

Note: If this isn't clear, try to see the above equation by rearranging the terms, to obtain the conservation of mechanical energy: for the Earth-body system, due to the absence of external forces, KE + U = constant.

Now, the initial kinetic energy was zero, since the body and Earth start from rest.

So, $KE_{f} = U_{i} - U_{f} = -\frac{GM_{e}M_{b}}{a} + \frac{GM_{e}M_{b}}{R_{e}} = \frac{GM_{e}M_{b}}{aR_{e}} * (a - R_{e})$

Finally, the kinetic energy of a body is given by:

$KE = \frac{1}{2}mv^2$

So...

$\frac{GM_{e}M_{b}}{aR_{e}} * (a - R_{e}) = \frac{1}{2}M_{b}v^2$

And: $v = \sqrt{\frac{2GM_{e}M_{b}}{aR_{e}} * (a - R_{e})}$