A body is projected up with a velocity equal to ${(\frac{3}{4})}^{t h}$ 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.

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Solution

The velocity of projection, $v = \frac{3}{4} \times v_{e}$
$= \frac{3}{4} \times \sqrt{\frac{2 G M}{R}}$ ------ (i)
Total energy of the body when it is projected
$E = K E + P E = v = \frac{1}{2} \times m v^{2} - \frac{G M m}{R}$ ----(ii)
From (i) , $v^{2} = \frac{9}{16} \times \frac{2 G M}{R}$
Putting it in (ii)
$E = \frac{1}{2} m \times \frac{9}{16} \times \frac{2 G M}{R} - \frac{G M m}{R}$
$= - \frac{7}{16} \frac{G M m}{R}$
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 $E^{'} = 0 - \frac{G M m}{r} = - \frac{G M m}{(R + h)}$
Using the principle of conservation of energy.
$- \frac{7}{16} \frac{G M m}{R} = - \frac{G M m}{(R + h)}$
$\Rightarrow 7 h = 16 R - 7 R$
$\Rightarrow h = \frac{9}{7} R$

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