How to calculate the gravitational binding energy of a uniform cube of length $L$ and mass $M$?

The functional form is known already (as attached). But what is the solution for this integral?

• In the 21st century, go numerically. Or google it, probably somebody in the 19th century has done it analitycally – patta May 13 at 15:25
• I did a Monte-Carlo evaluation of $Q$ and found that it is between -0.941 and -0.942. – G. Smith May 13 at 23:09
• The three integrals over either $\mathbf{x}$ or $\mathbf{y}$ can be done analytically to get the gravitational potential of the cube. See arxiv.org/abs/1206.3857. I haven't been able to then integrate this potential over the cube to get the binding energy, but I was able to numerically integrate it and again get -0.941... . – G. Smith May 14 at 0:11
• More accurately, $Q=-0.941156...$. – G. Smith May 14 at 2:50

1 Answer

The integral can apparently be done exactly and the answer is

$$Q=\frac{2\sqrt{3}-\sqrt{2}-1}{5}+\frac{\pi}{3}+\ln{[(\sqrt{2}-1)(2-\sqrt{3})]}=-0.94115632219483008005...,$$

which is consistent with the numerical evaluation that I mentioned doing in my comments on the question.

The value was apparently first derived in 2000 by Skeidov and Skvirsky in this paper:

https://arxiv.org/abs/astro-ph/0002496

As an explanation of their evaluation, they unfortunately have only this to say: "After some lengthy interactive session with Mathematica, we get...". I personally am unable to get Mathematica to produce this result, but I have no doubt that it is correct.