Question

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 planets 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)

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

Answer: (A)

Given situation is shown in the figure.

Let acceleration due to gravity at the surface of the planet be $g$. At height $h$ above planets surface $v=0$.
According to question, acceleration due to gravity of the planet at height $h$ above its surface becomes $g/4$.
$g_h=\frac{g}{4}=\frac{g}{{\left(1+\frac{h}{R}\right)}^2}$
$4={\left(1+\frac{h}{R}\right)}^2$
$\Rightarrow 1+\frac{h}{R}=2$
$\frac{h}{R}=1$
$\Rightarrow h=R$.
So, velocity of the bullet becomes zero at $h=R$.
Also $v_{esc}=v\sqrt{N}$
$\Rightarrow \sqrt{\frac{2GM}{R}}=v\sqrt{N}\dots \left(i\right)$
Applying energy conservation principle,
Energy of bullet at surface of earth
$=$ Energy of bullet at highest point
$\frac{-GMm}{R}+\frac{1}{2}m{v}^2=\frac{-GMm}{2R}$
$\frac{1}{2}m{v}^2=\frac{GMm}{2R}$
$\therefore v=\sqrt{\frac{GM}{R}}$
Putting this value in eqn. $\left(i\right)$, we get
$\sqrt{\frac{2GM}{R}}=\sqrt{\frac{NGM}{R}}$
$\therefore N=2$