Gravitational Potential Energy of Square System
Example 8.3: Four masses at vertices of a square
Example
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.
Solution:
Consider four masses each of mass \(m\) at the corners of a square of side \(l\).
There are four mass pairs at distance \(l\) and two diagonal pairs at distance \(\sqrt{2}\,l\).
Hence,
\[
W(r) = -4\frac{Gm^2}{l} - 2\frac{Gm^2}{\sqrt{2}l}
\]
\[
= -\frac{2Gm^2}{l}\left(2+\frac{1}{\sqrt{2}}\right) = -5.41\frac{Gm^2}{l}
\]
The gravitational potential at the centre of the square \(\left( r = \sqrt{2} \frac{l}{2} \right)\) is:
\[
U(r) = -4\sqrt{2}\frac{Gm}{l}
\]
Results:
System's gravitational potential energy: -5.41 Gm²/l
Gravitational potential at center: -4√2 Gm/l
Numerical values:
Total PE: -5.41 Gm²/l
Center potential: -5.66 Gm/l
Physics Explanation:
This simulation demonstrates the gravitational potential energy of four equal masses arranged at the corners of a square.
System's Gravitational Potential Energy:
For four masses (each mass = m) at the corners of a square (side length = l):
- 4 pairs at distance l (sides of the square)
- 2 pairs at distance √2 l (diagonals of the square)
Gravitational Potential at Center:
The distance from center to any corner is \( r = \frac{\sqrt{2}}{2} l \):



