Question

Two particles $\text{A}$ and $\text{B}$ each of mass $m$ are separated by a distance $r$. Another particle $\text{C}$ of mass $M$ is placed at the midpoint of $\text{A}$ and $\text{B}$. Find the work done in taking $\text{C}$ to a point at a distance $r$ from both $\text{A}$ and $\text{B}$ without acceleration.
$(G=$ Gravitational constant and only gravitational interaction between $\text{A, B}$ and $\text{C}$ is considered$)$

• A. $\frac{GMm}{r}$
• B. $\frac{2GMm}{r}$
• C. $\frac{3GMm}{r}$
• D. $\frac{4GMm}{r}$

Answer 1

The correct option is B $\frac{2GMm}{r}$

Since, particle $\text{C}$ is moved without any acceleration,

$\therefore \Delta K.E=0$

And we know that work done by external agent will be equal in magnitude and opposite in sign to work done by gravitational force.

$\therefore$ Work done by external agent$=-W_g$

$\Rightarrow W_{ext}=-(-\Delta U)=\Delta U=U_f-U_{in}$

So, initial potential energy of the system will be

$U_i=-\frac{GMm}{r/2}-\frac{GMm}{r/2}=-\frac{4GMm}{r}$

Similarly, final potential energy of system is given by

$U_f=-\frac{GMm}{r}-\frac{GMm}{r}=-\frac{2GMm}{r}$

$\therefore W_{ext}=U_f-U_i=-\frac{2GMm}{r}+\frac{4GMm}{r}$

$\Rightarrow W_{ext}=\frac{2GMm}{r}$

Hence, option $\left(b\right)$ is correct.

Key Concept - For no change in kinetic energy, $W_{ext}=\Delta U$ Why this question: To make students understand the application of work- energy theorem in gravitational systems.