Question
Find the potential energy of a system of four particles placed at the vertices of a square of side $l$. Also obtain the potential at the centre of the square.
✓
Answer
Consider four masses each of mass $m$ at the corners of a square of side $l$; See Fig. $7.9$.
We have four mass pairs at distance $l$ and two diagonal pairs at distance $\sqrt{2} t$
Hence,
$W(r)=-4 \frac{G m^2}{l}-2 \frac{G m^2}{\sqrt{2} l}$
$=-\frac{2 G m^2}{l}\left(2+\frac{1}{\sqrt{2}}\right)=-5.41 \frac{G m^2}{l}$
The gravitational potential at the centre of the square $(r=\sqrt{2} l / 2)$ is
$
U(r)=-4 \sqrt{2} \frac{ Gm }{l} .
$
We have four mass pairs at distance $l$ and two diagonal pairs at distance $\sqrt{2} t$
Hence,
$W(r)=-4 \frac{G m^2}{l}-2 \frac{G m^2}{\sqrt{2} l}$
$=-\frac{2 G m^2}{l}\left(2+\frac{1}{\sqrt{2}}\right)=-5.41 \frac{G m^2}{l}$
The gravitational potential at the centre of the square $(r=\sqrt{2} l / 2)$ is
$
U(r)=-4 \sqrt{2} \frac{ Gm }{l} .
$
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