Question

What is the work done by the gravitational field in placing three identical particles each of mass $m$ at the vertices of an equilateral triangle of side $l$?

Answer 1

We know that the gravitational field to conservative.
$W=-(△PE{)}_{system}.....(1)$
$△PE=(P{E}_{final}-P{E}_{Initial})$
Since initially the particles $A,B$ and $C$ were at infinity, their initial gravitational potential energy was zero
$\Rightarrow △PE=P{E}_{final}$,
The gravitational potential energy of the configuration of the particles $=PE$ between $A$ and $B+PE$ between $B$ and $C+PE$ between $C$ and $A$
$\Rightarrow △PE=|(-\frac{G{m}_A{m}_B}{l})+(-\frac{G{m}_B{m}_c}{l})+(-\frac{G{m}_C{m}_A}{l})|$
Putting $m_A=m_B=m_C=m$, we obtaine,
$△PE=\frac{3G{m}^2}{l}.....\left(2\right)$
Using $\left(1\right)$ and $\left(2\right)$, we obtain, $W=3G{m}^2/l$.