Question

A body is thrown with a velocity equal to $n$ times the escape velocity $\left(v_e\right)$. Velocity of the body at a large distance away will be

• A. $v_e\sqrt{n^2-1}$
  
• B. $v_e\sqrt{n^2+1}$
• C. $v_e\sqrt{1-n^2}$
  
• D. None of these

Answer 1

The correct option is A $v_e\sqrt{n^2-1}$


At large distance, $\text{potential energy}=0$

Let, the velocity of body at very large distance be $v$.

Applying the conservation of mechanical energy at surface of earth and at a large distance from earth,

${\text{K.E}}_{surface}+{\text{P.E}}_{surface}={\text{K.E}}_{largedistance}+{\text{P.E}}_{largedistance}$

$\Rightarrow \frac{1}{2}m{\left(n{v}_e\right)}^2-\frac{GMm}{R}=\frac{1}{2}m{v}^2+0$

As, $v_e=\sqrt{\frac{2GM}{R}}$

$\Rightarrow \frac{2GMm}{R}(n^2-1)=m{v}^2$

$\Rightarrow v=\sqrt{\frac{2GM}{R}}{(n^2-1)}^{\frac{1}{2}}$

$\Rightarrow v=v_e\sqrt{n^2-1}$

Hence, option (a) is the correct answer.

Key Concept: The escape velocity for a planet is given by $$v_e=\frac{\sqrt{2GM}}{R}$$Why this question: To make students apply conservation of mechanical energy in a planet-satellite system.