If a rocket is fired with a speed $v = 2 \sqrt{g R}$ near the earth's surface and coasts upwards, its speed in the intersteller space is.

4 \sqrt{g R}

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\sqrt{2 g R}

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\sqrt{g R}

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\sqrt{4 g R}

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Solution

The correct option is B $\sqrt{2 g R}$
In intersteller space, gravitational P.E is zero and let velocity be $v_{1}$ .
Applying conservation of energy at surface and intersteller space-
$\frac{1}{2} m v^{2} + (- \frac{G M m}{R}) = \frac{1}{2} m v_{1}^{2} + 0$
Now using $G M = g R^{2} v = 2 \sqrt{g R}$ (given)
$\frac{1}{2} m (4 g R) - \frac{(g R^{2}) m}{R} = \frac{1}{2} m v_{1}^{2}$
$⟹ v_{1} = \sqrt{2 g R}$

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