# 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 $r$?

• You've asked an intellectually sophisticated question, and you deserve an intellectually sophisticated answer. The trouble is that there is no uniquely defined answer to this type of "why" question. If there's an answer that you will find satisfactory, it will be an answer that appeals to something you consider more fundamental. But we don't know what you'd say was more fundamental than this law. That's a matter of taste. Empirical observations? General relativity? Heuristic arguments about the commonness of inverse-square laws, because we live in three dimensions and area goes like $r^2$? – Ben Crowell Sep 29 '14 at 3:40
• Related: physics.stackexchange.com/q/22010/2451 and links therein. – Qmechanic Sep 29 '14 at 8:52

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 $1/r^2$.