The binding energy of a uniform sphere that interacts through a force that follows the inverse square law is proportional to $a\frac{C^2}{R}$, where $C$ is whatever charge determines the strength of the force on the particle, such as mass or electric charge, $R$ is the radius, and $a$. For gravity, this looks like $\frac{3GM^2}{5R}$. What would the equation be for a Yukawa potential though? I thought the equation might be $a\frac{C^2e^{-bR}}R$, where $a$ and $b$ are constants, but when I predicted the binding energy by using an integral to calculate potential for a given radius and plotting the results, I couldn't get $a\frac{C^2e^{-bR}}R$ to fit the data.
The equation I used to predict the binding energy of the sphere for a given radius was this: $$\frac{\int_0^R4\pi r^2\left(\int_{-R}^R\left(\int_0^\sqrt{1-x^2}\left(2\pi y \frac{e^{-\sqrt{(x-r)^2+y^2}}}{\sqrt{(x-r)^2+y^2}}\right)dy\right)dx\right)dr}{\left(\frac43\pi R^3\right)^2}$$ The outermost integral variable, $r$, represents a radius from the center of the sphere that is less than the radius $R$. The $4\pi r^2$ bit is because each radius $r$ also represents an equipotential shell. The rest of the equation treats $r$ like a distance to a certain point, and determines the potential at that point. $\int_{-R}^R\int_0^\sqrt{1-x^2}2\pi y...dydx$ maps each point in the sphere and calculates each point's "contribution" to the potential at point $r$. I divide the whole thing by the volume squared because I want the total "charge" to remain constant.