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/?
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Anonymous2009-07-02 17:28
Correction: F_0 = G \rho^2 d^4
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Anonymous2009-07-04 14:43
I think the center of mass 'approximation' is not an approximation, but an actual result. I think this follows from superposition.
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Anonymous2009-07-04 16:43
>>3
That is the case with spheres. Not so with other objects.
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.
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Anonymous2009-07-06 19:39
you can only use center of mass approximations if the distance between the cubes is much greater than the length of the cube
>>9
You're missing two terms. Each sphere attracts all three other spheres, not just the two that aren't attached to it.
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Anonymous2009-07-10 6:20
>>10
Also, your diagram is horrible and makes no sense.
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Anonymous2009-07-10 6:33
>>10
The effective force on the centre of mass is equal to the sum of the external forces on the constitutent parts of the body in a rigid frame. The internal forces don't make a difference.
O-OO-O
A B
A is a dumbell, B is a dumbell. The diameter of the spheres and the length of light bars connecting them are both of unit length. Each sphere weighs m.
The mass of each body as a whole is 2m and according to you, the total force between the two bodies is equal to G(2m)^2/(distance between the centres of masses)^2 = Gm^2 * 4/9
But the force between the spheres in the centre ALONE is Gm^2. So the force between A and B must be larger than this.
>>5 No, it applies to all objects.
you sir, are dead wrong.
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Anonymous2009-07-11 22:39
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}.
Your center of mass 'approximation' is not necessarily any less accurate than the result you will get from doing the 6-fold integral.
Consider the logic behind the former method. The answer should be just as exact.