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A body is projected horizontally from the surface of the Earth (radius = $R$ ) with a velocity equal to $n$ times the escape velocity. Neglect rotational effects of the earth. The maximum height attained by the body from the earth's surface is $R / 2$ . Then, $n$ must be

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$\sqrt{0.6}$

$\frac{\sqrt{3}}{2}$

$\sqrt{0.4}$

$\frac{1}{2}$

answer is A.

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Detailed Solution

$v_{1} = n v_{e} = n \sqrt{2 g R} = n \sqrt{\frac{2 G M}{R}}$

A body is projected horizontally from the surface of the earth (radius = R) with a velocity equal to n times the escape velocity. Neglect rotational effect of the earth. The maximum

From mechanical energy conservative, we have $\frac{1}{2} m {[m \sqrt{\frac{2 G M}{R}}]}^{2} - \frac{G M m}{R} = \frac{1}{2} m v_{2}^{2} - \frac{G M m}{R + \frac{R}{2}} \dots (i)$

From conservation of angular momentum about center of earth

$m [n \sqrt{\frac{2 G M}{R}}] R = m v_{2} (R + \frac{R}{2}) . . . (i i)$

Solving these two equations, we get

$n = \sqrt{0.6}$

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