# Gravitational inverse-square law

I was looking at the gravitational inverse-square law: $$F_G = G \frac{Mm}{r^2}$$

This law comes from some experimental data? Why it is an exact inverse-square law? Could it be $$F_G = G \frac{Mm}{r^{2.00000000000000001}}$$ or there is a mathematical method to find exactly this law?

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Usually fractional exponents come from empirical data, and integer exponents from theory. –  ja72 Sep 2 '12 at 14:12
Possible duplicate: physics.stackexchange.com/q/22010/2451 –  Qmechanic Feb 1 '13 at 16:33
It is all because of the fact that we live in a 3-spatial ddimensional universe (and not a 3.00000000000000001-dimensional universe!. :) ). See physics.stackexchange.com/questions/32779/…. –  Dimensio1n0 Jun 19 '13 at 6:56

To answer your question of whether or not there is experimental data, here is one of what I am sure are many papers regarding the classical definition of the gravitational force: http://www.physics.uci.edu/~glab/papers/HoskinsPaper.pdf

Nothing in physics is ever exact, but considered an approximation to the reality of the event under consideration. To test whether or not this approximation is valid, experimental data must be obtained and the standard deviation between the expected and measured results must be analysed. If a measured and expected quantity agree to within some deviation of each other over numerous iterations, the approximation is considered "good enough" to be true.

The gravitational inverse square law is a classical approximation, not taking into account relativistic effects. Even so, the standard deviation of the gravitational inverse square law is a close enough approximation that it will in most cases suffice as an accurate description of the system under study.

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It is exact in the Newtonian limit, i.e., for speeds slow compared to speed of light, away from strong gravitational fields or horizons, and for big enough objects that quantum effects can be ignored. The 1/r^2 comes from solving Poisson's equation $\nabla^2\phi=4\pi G \rho$. If there would be corrections to this law it would come from additional higher derivative terms. Such higher derivatives would not affect the 1/r^2 behavior at large r. So, in the appropriate limit, yes, it is exact.