Question

A body is projected up with a velocity equal to ${\left(\frac{3}{4}\right)}^{th}$ to the escape velocity from the surface of the earth. Find the height above earth's surface up to which it reaches. Given R is the radius of the earth.

Answer 1

The velocity of projection, $v=\frac{3}{4}\times v_e$
$=\frac{3}{4}\times \sqrt{\frac{2GM}{R}}$ ------ (i)
Total energy of the body when it is projected
$E=KE+PE=v=\frac{1}{2}\times m{v}^2-\frac{GMm}{R}$ ----(ii)
From (i) ,$v^2=\frac{9}{16}\times \frac{2GM}{R}$
Putting it in (ii)
$E=\frac{1}{2}m\times \frac{9}{16}\times \frac{2GM}{R}-\frac{GMm}{R}$
$=-\frac{7}{16}\frac{GMm}{R}$
Suppose, h= maximum height attained by the body.
So, distance of the body from the center of earth, r = R+h
At this height, KE = 0
So from $E^{\prime }=0-\frac{GMm}{r}=-\frac{GMm}{(R+h)}$
Using the principle of conservation of energy.
$-\frac{7}{16}\frac{GMm}{R}=-\frac{GMm}{(R+h)}$
$\Rightarrow 7h=16R-7R$
$\Rightarrow h=\frac{9}{7}R$