Question

A bullet is fired vertically upwards with a velocity $v$ from the surface 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)

Answer 1

We know that at the surface of the earth, value of $g=\frac{GM}{R^2}$
At height $h$ above the Earth's surface, the value of acceleration due to gravity $g^{\prime }=\frac{GM}{(R+h{)}^2}$

So it is given that $g^{\prime }=\frac{g}{4}$

When the bullet reaches maximum height, acceleration due to gravity is $\frac{1}{4}$th of that at planet's surface.
That implies $\frac{GM}{4R^2}=\frac{GM}{(R+h{)}^2}\to h=R$

By conservation of mechanical energy, $\frac{-GMm}{R}+\frac{1}{2}m{v}^2=\frac{-GMm}{h+R}+0$
since velocity is zero at max height. $\Rightarrow \frac{1}{2}m{v}^2=\frac{GMm}{2R}$

$v=\sqrt{\frac{GM}{R}}=\sqrt{\frac{2GM}{2R}}=\frac{1}{\sqrt{2}}\sqrt{\frac{2GM}{R}}=\frac{1}{\sqrt{2}}{v}_{esc}$ $\Rightarrow v_{esc}=\sqrt{2}v$
$\Rightarrow N=2$