Confusions on The Gravitational Energy of a Point P in a Cube

Question

I have been working, quite tirelessly, to try and find an answer to a question that has been bothering me for some time now.

I have been working over some proofs, in the Newtonian Mechanics world, to help myself see and verify that these given equations and numbers work. One of the big ones I was working on was the gravitational potential of a point P inside a uniform mass distribution of a cube. For context, this cube would have a corner at (0,0,0) in the cartesian plane, and have side length values of $L$, and uniform mass density $ \frac{M}{L^3} $. That being said, the position of P would then be $r = \sqrt{x^2 + y^2 + z^2} $.

I was curious enough to ask ChatGPT about this, and the answer that it gave me was the potential of a point inside a cube is ${-3GM\over 2L}$. This was supposedly the "well-established answer", but I haven't found any sources that directly support this. So, I went ahead and tried to do the integration out to actually see if this was a valid answer.

My math is as follows:

1. Defining the proper variables/setup parameters.

   • This setup included noting the position of 'P', as shown above, as well as the fact that the potential was equivalent to $ \frac{G \ dV \ \rho}{r} $ where $ \rho $ is the mass density.

2. Setting up the integral.

   • Now, I can set up an integral, I, which would be equivalent to $ -G* \iiint_ \frac{\rho}{(x^2 + y^2 + z^2)^{1/2}} \ dx \ dy \ dz $ where each bound would be denoted by $ {-\frac{L}{2}} to {\frac{L}{2}} $. This would effectively become $ -\frac{G M}{L^3} \iiint_{-\frac{L}{2}}^{\frac{L}{2}} \frac{1}{(x^2 + y^2 + z^2)^{1/2}} \, dx \, dy \, dz $.

3. Solving the integral.

   • Solving this integral was not exactly trivial, given the denominator of my integrand. So, I wanted to introduce 'dimensionless variables' that would help reduce this integral to one of a tabular form. I did so by noting that x, y, and z would equal $ \frac{L}{2} u $, $ \frac{L}{2} v $, and $ \frac{L}{2} w $ respectively. This would imply that my new dx, dy, and dz would become $ \left( \frac{L}{2} \right)^3 \ du \ dv \ dw $. This would change each of my bounds to be from -1 to 1 and would leave me an integral that looks like $ -\frac{GM}{L^3} \left( \frac{L}{2} \right)^3 \iiint_{-1}^{1} \frac{du \ dv \ dw}{\left[ \left( \frac{L}{2} u \right)^2 + \left( \frac{L}{2} v \right)^2 + \left( \frac{L}{2} w \right)^2 \right]^{1/2}} $.
   • I solve this integral via an integral table, which results in a value of $ \ln(1 + \sqrt{2}) - \frac{1}{2} $, which I multiply by 8 for the entire cube, and ultimately arrive at an answer of $ -\frac{GM}{L} \cdot \frac{1}{4} \left[ 8 \left( \ln(1 + \sqrt{2}) - \frac{1}{2} \right) \right] $.

4. My final conclusion.

   • So, my final conclusion is that for a point P in the center of a cube, you get a gravitational potential value of $ -\frac{GMm}{L} \left( 2 \ln(1 + \sqrt{2}) - 1 \right) $. This is clearly different than what the supposed "standard" was, and worried me a bit.

I would now like to present this question here to see if my math, intuition, setup, etc. was wrong. Is there an actual standard for a question like this?

Answer 1

ChatGPT hallucinated the answer. The potential it came up with is at the center of a sphere of constant density with radius $L$ and mass $M$. There is no standard answer because there aren't many cubes massive enough to have an appreciable gravitational field.

Your math setup does look correct. I found one paper from 2012 on the arXiv that calculated the field and potential of a cube. The result looks much more like your result, though I didnt actually calculate the potential at the origin. The paper might have a mistake because the limits of integration go from $-L$ to $+L$ in each dimension, leading to a possible factor of 8 error.