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A point P lies on the axis of a fixed ring of mass M and radius R at a distance 2R from its center O. A small particle starts from P and reaches O under gravitational attraction only. Its speed at O is

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a

$\sqrt{\frac{2 GM}{R} (1 - \frac{1}{\sqrt{5}})}$

b

$\sqrt{\frac{2 GM}{R} (\sqrt{5 - 1})}$

c

$\sqrt{\frac{2 GM}{R}}$

d

Zero

answer is D.

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$E_{p} = E_{o} \frac{- GM}{\sqrt{R^{2} + 4 R^{2}}} + O = \frac{- GMm}{R} + \frac{1}{2} {mv}^{2} \frac{- GM}{\sqrt{5} R} + \frac{GM}{R} = \frac{1}{2} V^{2} \frac{2 GM}{R} (1 - \frac{1}{\sqrt{5}}) = V^{2} V = \sqrt{\frac{2 GM}{R} (1 - \frac{1}{\sqrt{5}})}$

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