# 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 $r^2$?
– user4552
Sep 29, 2014 at 3:40
• Related: physics.stackexchange.com/q/22010/2451 and links therein. Sep 29, 2014 at 8:52

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.

• Is this correct? I seem to remember a lecture by Feynman (on Youtube) where he explains that a force always pointing to one of the foci is a sufficient condition for an elliptical orbit; conditions concerning the magnitude of the force were not necessary. It's always possible that my memory is not correct though. May 20, 2019 at 13:45

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 $1/r^2$.

• This is fine as a heuristic argument, but as a matter of taste, one may or may not like it. Gravity isn't a physical substance that spreads out over an area like peanut butter.
– user4552
Sep 29, 2014 at 3:43
• @BenCrowell Neither is the force between two charges, but the electric field generated by one of the charges acts that way. I guess it is all how one looks at the situation. BTW, light intensity also acts like peanut butter.
– LDC3
Sep 29, 2014 at 3:48
• It should be noted that not all the forces between point particles are governed by inverse square law. See e.g. Yukawa potential. Sep 29, 2014 at 5:08
• @Ruslan - but all forces mediated by massless bosons are. Sep 29, 2014 at 5:32