Question

A stationary object is released from a point $\text{P}$ at a distance $3R$ from the center of the moon which has radius $R$ and mass $M$. Which of the following are the correct expression for speed of the object on hitting the moon?

• A. ${\left(\frac{2GM}{3R}\right)}^{\frac{1}{2}}$
• B. ${\left(\frac{4GM}{3R}\right)}^{\frac{1}{2}}$
• C. ${\left(\frac{GM}{3R}\right)}^{\frac{1}{2}}$
• D. ${\left(\frac{GM}{R}\right)}^{\frac{1}{2}}$

Answer 1

The correct option is B ${\left(\frac{4GM}{3R}\right)}^{\frac{1}{2}}$

Since, there are no external or non-conservative forces acting on the system (moon + object), work done by them would be zero and hence mechanical energy of the system will be conserved.


Conserving mechanical energy between points $\text{P}$ and $\text{S}$,

$P.E_P+K.E_P=P.E_S+K.E_S$
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$\Rightarrow -\frac{GMm}{3R}+0=\frac{1}{2}m{v}^2-\frac{GMm}{R}$

Where, $v$ is the speed of the object on hitting the moon.

$\Rightarrow \frac{1}{2}m{v}^2=-\frac{GMm}{3R}+\frac{GMm}{R}$

$\Rightarrow \frac{1}{2}m{v}^2=\frac{GMm}{R}(\frac{-1}{3}+1)$

$\Rightarrow \frac{1}{2}m{v}^2=\frac{2GMm}{3R}$

$\Rightarrow v=\sqrt{\frac{4GM}{3R}}$

Hence, option (b) is correct answer.

Why this question? To familiarize students with the application of conservation of mechanical energy, which is a core concept of this chapter.