A bullet is fired vertically upwards with velocity V from the surface of a spherical planet. When it reaches its maximum height, its acceleration due to the planet’s gravity is 1/4 th of its value at the surface of the planet. If the escape velocity from the planet is $V_{esc} = V \sqrt{N}$ , then the value of N is ______. (Ignore energy loss due to atmosphere).

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answer is 2.

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Detailed Solution

For a particle at a distance r form the earth

force is given by, $F = \frac{{Gm}_{1} m_{2}}{r^{2}}$

Force becomes one-fourth, when r = 2R

(R = radius of earth)

Escape velocity, $V_{e} = \sqrt{\frac{2 GM}{R}}$

Using conservation of energy for the given particle,

$\frac{1}{2} {mV}^{2} - \frac{GmM}{R} = \frac{- GmM}{2 R}$

$This gives, V = \sqrt{\frac{GM}{R}}$

$\Rightarrow V_{e} = V \sqrt{2}$

Hence, n = 2

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