# Why inverse square not inverse cube law? [duplicate]

So as I understand, the inverse-square law which shows up in a variety of physical laws (Newton's universal law of gravitation, Coulomb's law, etc.) is a mathematical consequence of point-like particle emanating a certain physical quantity in all directions in the form of a sphere, and the density of that quantity is inversely proportional the surface area of that sphere on which that physical quantity gets spread out at a certain distance (radius), and since the surface area of a sphere is directly proportional to its radius squared, therefore the density of that physical quantity is inversely proportional to the distance squared.

My question is: consider the specific example of point-like particle with a certain gravitational mass, now if we pictured the gravitational field of that particle through gravitational flux that emanates out of it isotropically, the density of that gravitational lines is inversely proportional to the volume of the sphere at a certain given distance (radius), and since the volume of sphere is directly proportional to the radius cubed.

Therefore the force of gravity (or the electrostatic force or whatever) is inversely proportional to the distance cubed.

What is wrong with this analysis?

## marked as duplicate by Kyle Kanos, ACuriousMind♦, David Hammen, Qmechanic♦May 10 '15 at 21:37

• The strength of the field is given by the areal density for a surface element normal to the field lines. The volume element you describe has no physical significance. – John Rennie May 10 '15 at 16:44
• Stability is another key reason, see Bertrand's theorem. – Kyle Kanos May 10 '15 at 16:58
• – Kyle Kanos May 10 '15 at 17:07
• – Kyle Kanos May 10 '15 at 17:07
• @DanielSank why areal density not volume density that counts when assessing the strength of a field, can you elaborate on that? – Omar Nagib May 11 '15 at 10:45

Flux is proportional to the area of the sphere not the volume of the sphere. It is evident from definition of the flux $\Phi_\mathbf{B}$ of some quantity $\mathbf{B}$ , which is defined in the following way,
$$\Phi_\mathbf{B}= \iint\mathbf{B} \cdot \mathrm d \mathbf{A}$$
Therefore the flux is proportional to the area of the sphere and hence the $1/r^2$ dependency.
Note that the detailed treatment of $1/r^2$ dependence of Coulomb's Law and Newton's law needs Maxwell's theory of EM and GR respectively.