Question

What is the maximum height attained by a body projected with a velocity equal to one-third of the escape velocity from the surface of the earth? (Radius of the earth = R) [AIIMS 2017]

Moderate

Solution

SHORT CUT : WHEN V= $\frac{V_{E}}{N} t h e n t h e h e i g h t r e a c h e d b y t h e b o d y h = \frac{R}{N^{2} - 1}$
From conservation of energy,
$\frac{1}{2} m v_{1}^{2} - \frac{G M_{e} m}{R} = \frac{1}{2} m v_{2}^{2} - \frac{G M_{e} m}{R + H}$
Here,v₂ = 0 (At maximum height)
$∴ \frac{1}{2} m v_{1}^{2} - \frac{G M_{e} m}{R} = - \frac{G M_{e} m}{R + H}$
Solving, we get
$H = \frac{R}{\frac{2 g R}{v_{1}^{2}} - 1} = \frac{R}{\frac{v_{e}^{2}}{v_{1}^{2}} - 1} (∵ v_{e} = \sqrt{2 g R})$
where, v_e is escape velocity.
Here, $v_{1} = \frac{1}{3} v_{e} \Rightarrow \frac{v_{e}}{v_{1}} = 3$
$∴ H = \frac{R}{9 - 1} = \frac{R}{8}$

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