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A particle of mass ‘m’ is kept at rest at a height 3R from the surface of earth, where ‘R’ is radius of earth and ‘M’ is mass of earth. The minimum speed with which it should be projected, so that it does not return back, is (g is acceleration due to gravity on the surface of earth)

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${(\frac{2 g}{R})}^{1 / 2}$

${(\frac{G M}{2 R})}^{1 / 2}$

${(\frac{g R}{4})}^{1 / 2}$

${(\frac{G M}{R})}^{1 / 2}$

answer is A.

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Detailed Solution

To escape from gravitational field
PE = KE.
At a height 3R,

$PE = - \frac{GMm}{R + 3 R} = - \frac{GMm}{4 R}$

$So, KE = \frac{1}{2} m v^{2} = \frac{GMm}{4 R}$

$v^{2} = \frac{G M}{2 R} = \frac{G M R}{2 R^{2}} = \frac{g R}{2} (\frac{G M}{R^{2}} = g)$

Where M is the mass of the planet

m is the mass of the body

and R is the radius of the planet

$v = {(\frac{Gm}{2 R})}^{1 / 2} or {(\frac{gR}{2})}^{1 / 2}$

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