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Consider a planet of mass M and radius R as shown in the figure. A particle is projected from the surface of the planet at an angle $θ$ from radial direction and with velocity equal to the orbital velocity of circular orbit at that point. Then choose the correct statement(s):

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Time taken by the particle to return to the surface of planet is $\sqrt{\frac{R^{3}}{G M}}$

Maximum distance of particle from centre of the planet is $(R + R \cos θ)$

Time taken by the particle to reach the surface of planet is equal to $π \sqrt{\frac{R^{3}}{G M}}$

Path of the particle is parabola

answer is A, D.

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Detailed Solution

From energy conservation,

$V_{0} = \sqrt{\frac{G M}{R}} - \frac{G M m}{R} + \frac{1}{2} m {v_{0}}^{2} = - \frac{G M m}{R + h} + \frac{1}{2} m v^{2} \begin{array}{l} m v_{0} \sin θ R = m v (R + h) \\ F r o m (i) a n d (i i) h = R \cos θ \end{array}$

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