An asteroid of mass m (<< M) was fast approaching the earth of mass ‘M’. Scientists fired a racket which hit the asteroid at a distance 5R from the centre of the earth, where R is radius of earth. Immediately after the hit the asteroids velocity $V_{0}$ was making a angle of $θ = 30^{0}$ with the line joining the centre of the earth to the asteroid. The asteroid just grazed past the surface of the earth. Find its speed when it grazes the earth.

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$\sqrt{\frac{16 G M}{21 R}}$

$\sqrt{\frac{32 G M}{105 R}}$

$\sqrt{\frac{64 G M}{105 R}}$

$\sqrt{\frac{40 G M}{21 R}}$

answer is D.

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Detailed Solution

Angular momentum conservation

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$m v_{0} \sin 30 = m v R ......... (1)$

Energy conservation

$\frac{1}{2} m v_{0}^{2} - \frac{G M m}{5 R} = \frac{1}{2} m v^{2} - \frac{G M m}{R} ......... (2)$

$\Rightarrow v_{0}^{} = \sqrt{\frac{32 G M}{105 R}} & v = \sqrt{\frac{40 G M}{21 R}}$

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