Gravity between two cubes

Two identical cubes with diameters of d = 1 m, made of lead, are standing on a hard floor. One of their faces are fully in contact so that their vertices are aligned.

How strong is the gravitational force between them?


First approximation: Consider the case where all of the cubes' masses are concentrated in their centers. Then the gravitational force between them will be F_0 = G \rho^2 d^{-4}.

Now if one was to calculate the force exactly, a sixfold integration would have to be evaluated (basically (x1-x2)/r^3 has to be integrated over x1,x2,y1,y2,z1,z2). This can be simplified to a threefold integral with variable substitution.

However, this integral doesn't seem to be explicitly solved, at least I could not do it.

Numerical evaluation gives F = 0.926... F_0 (that is, the force between the actual cubes is 0.926... times smaller than the case where all mass is concentrated in two point masses).

Any insight from /sci/?

>>4

No, it applies to all objects. Although I suspect you messed up the integral somehow, because that should have also given the right answer. Basically, the integral for calculating the force on one cube is the same integral as the one you would use to calculate the center of mass.