Name: Anonymous 2009-07-02 17:15
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/?
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/?