Question

A BODY DROPPED FROM A HEIGHT EQUAL TO HALF OF THE RADIUS OF EARTH. IF Vé IS THE ESCAPE VELOCITY AND AIR RESISTANCE IS NEGLECTED.IT WILL STRIKE THE SURFACE OF EARTH WITH SPEED????

Answer 1

R=radius of Earth

g=acceleration due to gravity at Earth's surface

h=height of the body above Earth's surface=half radius of earth (given)

Let the escape velocity for earth be $v_{e} = \sqrt{2gR}$

First we find acceleration due to gravity at height 'h' by the formula ,as height is too large so we can not take g=9.8 m$s^{-2}$   :-   $g_{h} = g(\frac{R^{2} }{(R+h)^{2} })$

⇒ $g_{h} = g(\frac{R^{2} }{(\frac{3R}{2})^{2} } ) \\g_{h} = \frac{4}{9} g$

The mechanical energy at the height 'h' = $mg_{h} h$(only PE as KE is zero due to rest)

The mechanical energy just before collision=$\frac{1}{2} mv^{2}$(v is the final velocity,to be found)

The mechanical energy just before collision = The mechanical energy at the height h =$\frac{1}{2} mv^{2} =mg_{h} h$$=mg_{h} \frac{R}{2}$  (as h=$\frac{R}{2}$)

Equating it, we get :-

                         ⇒       $v^{2} = g_{h}R\\v=\sqrt{g_{h}R}$

$as,v_{e}=\sqrt{2g_{h}R}\\ so,v=\frac{v_{e}}{\sqrt{2} }$(ANSWER)