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 ${\nu }_{esc}=\nu \sqrt{N},$ then the value of N is (ignore energy loss due to atmosphere)

Answer 1

Answer: 2

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