Question

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

Answer 1

Applying conservation of mechanical energy
$\frac{1}{2}mν^{2}+\frac{-GMm}{R+h} = 0 - \frac{GMm}{R + h}$
$\frac{1}{2}mν^{2} = GMm\left[\frac{1}{R}-\frac{1}{R+h}\right]$
$\frac{1}{2}ν^{2} = GM \left[\frac{h}{R\times\left(R+h\right)}\right]\quad\quad\dots\dots.\left(i\right)$
We have
$gh = g\left(1+\frac{h}{R}\right)^{-2}$
$\frac{g}{4} = g \left(1+\frac{h}{R}\right)^{-2}$
$g\left(2\right)^{-2} = g\left(1+\frac{h}{R}\right)^{-2}$
$h = R\quad\quad..\dots\dots\left(ii\right)$
From equation (i) & (ii)
$\frac{1}{2}ν^{2} = GM\left[\frac{h}{2R^{2}}\right] = \frac{GMR}{2R^{2}} = \frac{GM}{2R}$
$ν^{2} = \frac{GM}{R}$
$ν = \sqrt{\frac{GM}{R}}$
Escape velocity at surface of planet
$ν_{esc} = \sqrt{\frac{2GM}{R}}$
$ν_{sec} = \sqrt{2}ν\quad\quad$[Given $ν_{sec} = ν\sqrt{N}$]
$N = 2$