A body is projected vertically from earth's surface of radius R with velocity equal to half the escape velocity. The maximum height reached by the body is

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TestHub · Accepted Answer

Escape velocity from the surface of the earth is given by $v_{e} = \frac{2 GM}{R}$

Initial velocity of the body is $v = \frac{v _{e}}{2} = \frac{GM}{2 R}$

Applying conservation of mechanical energy at the surface of the earth and at maximum height $h$ :

$K_{i} + U_{i} = K_{f} + U_{f}$

$\frac{1}{2} m v^{2} - \frac{GM m}{R} = 0 - \frac{GM m}{R + h}$

Substituting the value of $v^{2}$ :

$\frac{1}{2} m (\frac{GM}{2 R}) - \frac{GM m}{R} = - \frac{GM m}{R + h}$

$\frac{GM m}{4 R} - \frac{GM m}{R} = - \frac{GM m}{R + h}$

$- \frac{3 GM m}{4 R} = - \frac{GM m}{R + h}$

$\frac{3}{4 R} = \frac{1}{R + h}$

$3 R + 3 h = 4 R$

$3 h = R \Rightarrow h = \frac{R}{3}$

Answer: $\frac{R}{3}$

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