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A particle is projected from point A, that is at a distance 4R from the centre of the earth, with speed $V_{1}$ in a direction making 30° with the line joining the centre of the earth and point A, as shown. Consider gravitational interaction only between these two. ( $Use \frac{GM}{R} = 6.4 \times 10^{7} m^{2} / s^{2}$ ). The speed $V_{1}$ if particle passes grazing the surface ofthe earth is

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$4 \sqrt{2} \times 10^{3} m / s$

$4 \times 10^{3} m / s$

$4 \sqrt{3} \times 10^{3} m / s$

$2 \sqrt{2} \times 10^{3} m / s$

answer is B.

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Detailed Solution

Conserving angular momentum

$m . (V_{1} \cos 60 °) . 4 R = m . V_{2} . R; \frac{V_{2}}{V_{1}} = 2 conserving energy of the system$

$- \frac{G Mm}{4 R} + \frac{1}{2} {mV}_{1}^{2} = - \frac{GMm}{R} + \frac{1}{2} {mV}_{2}^{2}$

$\frac{1}{2} {V_{2}}^{2} - \frac{1}{2} {V^{2}}_{1} = \frac{3}{4} \frac{GM}{R}$

${V^{2}}_{1} = \frac{1}{2} \frac{GM}{R}$

$V_{1} = \frac{1}{\sqrt{2}} \sqrt{64 \times 10^{6}} = \frac{8000}{\sqrt{2}} m / s$

= $4 \sqrt{2} \times 10^{3} m / s$

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