Why $1/r^2$ and not another power of $r$ in Newton's law of gravitation? My book introduces the force of gravitation as a non-contact force between two bodies of mass $M_1$ and $M_2$ separated by a distance $r$ . Then it says it is directly proportional to the product of masses $M_1 M_2$ and inversely proportional to $r^2$. Then writes the the force of gravitation as 
$$
F = G\dfrac{M_1M_2}{r^2}.
$$ 
But why does it take square of $r$ and not another power? What is the cause of taking $r^2$? Why not another power of $r$?
 A: A lot of things decrease in intensity as $1/r^2$, such as light intensity, gravity, charge forces, etc. This is because the same force needs to act over a larger spherical area. The further away, the larger the sphere. And you should know that the surface area of a sphere is $SA=4\pi r^2$. Since the area varies as $r^2$, dividing the magnitude of the intensity by the area means it drops as $1/r^2$.
A: One of the reason's I like to cite for this is empirical observation. For years we have observed that the planets in the solar system move in closed orbits (roughly atleast). Mind you, I haven't asked for the orbits to be circular or elliptic, just closed. Now, invoking Bertrand's theorem with regard to the Kepler problem of the motion of a body in a central force, one can show that the only cases for which closed orbits are stable are that of the inverse square law and Hooke's law for the force. I find it kind of remarkable that the requirement of stability of closed orbits is sufficient to constrain the form of the central force to just two cases. Now the Hookean force is rather unphysical as it predicts increasingly large forces for large separations. Hence, by using this simple observation, we have sufficient reason to atleast believe that gravity acts (classically, that is) as an inverse square law force.
