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Geometrically if a force radiates uniformly it either A)has infinite density B)drops off as 1/r^2 We know that gravity drops off as 1/r^2 for all cases excep

Geometrically if a force radiates uniformly it either A)has infinite density B)drops off as 1/r^2 We know that gravity drops off as 1/r^2 for all cases except galaxy rotation curves so for it become 1/r for just this specific scale requires that from gravity's perspective the surface area of a sphere usually but doesn't always increase as r^2. Which in turn means it doesn't move through the 3 spacial dimensions that everything else does.

jcranmer 10y ago
The electromagnetic force (specifically, the Coulomb interaction) is 1/r^2. The strong force is actually effectively constant at long ranges, since it becomes energetically favorable to create an antiquark/quark pair. The residual strong force is exp(-r)/r. The weak nuclear force is exp(-r)/r^2 (or exp(-r)/r? I can't really tell with the links on hand--I just know for sure there's an exp(-r) term in it). Gravity being 1/r wouldn't be a surprise given the other known forces.
effie 10y ago
> it either A)has infinite density B)drops off as 1/r^2 How did you arrive at this? Which force has infinite density and where? k/r is finite except at the point where the source is, the formula is invalid there.