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How inverse square law is derived from area of sphere

  1. Aug 23, 2015 #1
    I almost understand how the inverse square law is derived from the area of sphere equation, 4πr2, but I'm not quite clear on what happens to the 4π. I found one equation that seemed to say that the intensity is equal to the area of the sphere of the source point times the amount of whatever coming from that source point, divided by the area of the sphere at the distance one is interested in. So that way the 4π cancels out since it's common to the numerator and denominator. Is that right? If so, does that mean that the radius of the source point is always assumed to be 1?
  2. jcsd
  3. Aug 23, 2015 #2
    What context are you speaking of? For instance, in electromagnetism, it was historically postulated that the electrostatic force obeys an inverse-square law, and this was based off of experiment. We can then draw assumptions from that. For instance, I don't know your background, but if you're in college, you may know of Gauss's law for electrostatics: [tex]\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q}{\epsilon_0}[/tex]

    For a spherically symmetric force, the left side of the equation can just be written as [itex] \oint \mathbf{E} \cdot d\mathbf{A} = EA = E(4 \pi r^2)[/itex] so [tex]E = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2}.[/tex]

    Thus the electric field falls off as an inverse square. But Iike I said, inverse square laws aren't, as far as I know, derived. Physicists performed experiments and then assumed an inverse square law, and all the other math came out of that. In fact, any inverse square law can be written in the form above, even the gravitational force.
  4. Aug 23, 2015 #3
    Hi axmls,

    I was thinking of the gravitational formula in particular:

    I'm just wondering what happens to the 4π part. Why isn't it written as something like this:

  5. Aug 23, 2015 #4
    G is defined with the 1/4π already applied. Its just a constant.

    Remember, they could'a made G anyway they liked.

    - stp
  6. Aug 23, 2015 #5
    I also found this link from the Georgia State University site where they depict the gravitational force formula being something like this:


    Please check out the link:

    it seems to suggest that the 4π cancels out, but I don't know why it appears in the numerator. My best guess was that it means to represent the virtual source point from which the gravity comes from.
  7. Aug 23, 2015 #6
    What Steve Pitcher said. It's just a constant. In the case of electromagnetism, the term [itex]\frac{1}{4 \pi \epsilon_0}[/itex] is just written [itex]k[/itex]. It's a constant, so the [itex]\pi[/itex] terms are already taken into account.

    In case you're curious, there's a corresponding equation for gravity: Gauss's law for gravity--[tex]\oint \mathbf{g} \cdot d\mathbf{A} = -4 \pi G M[/tex] In this case, [itex]\mathbf{g}[/itex] is the gravitational field produced by the mass [itex]M[/itex], and assuming a spherically symmetric field produced by the mass, the left side is [tex]\oint \mathbf{g} \cdot d\mathbf{A} = gA = g(4 \pi r^2)[/tex] so we have [tex]g(4 \pi r^2) = -4 \pi G M \implies g = -\frac{GM}{r^2}[/tex] So it's very similar to the case with electromagnetism. But again, it's assumed that gravity falls off as an inverse square--that's why we can write it in that form.
  8. Aug 23, 2015 #7


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    Staff: Mentor

    What is the area of the sphere with radius ##r_E##? It is ##4\pi{r}_E^2##.
    What is the area of the sphere with radius ##2r_E##? It is ##4\pi(2{r}_E)^2##.

    The ratio of those two quantities will be the ratio between the strength at ##r_E## and ##2r_E##. What is that ratio?
  9. Aug 24, 2015 #8
    I am assuming here that you mean the inverse square law according to which the intensity of a point source of light reduces as the inverse square of the distance from the source. The reasoning for that is the following:
    Assume a point source of light emitting power (Energy per unit time) Ps equally in all directions. At a distance r from the source, the intensity of light is defined as the power per unit area. How do we calculate this? Well, the power Ps coming from the source is spread out over a total surface area 4πr2. The intensity at that distance is therefore Ps / 4πr2. So the intensity falls off according to the inverse square of the distance. In words, it is simply because as you go further, the same power is spread out over larger and larger spherical surface areas.
  10. Aug 25, 2015 #9
    Hi Chandra,

    I'm interested in gravity in particular, because in the Newton formula for gravitational force the 4π is not there. To me, it makes it harder to explain in laymen terms the purpose of the inverse square aspect of the formula if the entire area of the sphere formula is not present.


    Thank you for posting the info about Gauss' law. It's over my head, but I think that must be what's depicted here:

    axmls and Steve Pitcher,

    I can't find any information that demonstrates that the 1/4π is already incorporated into the Universal Gravitational constant. Do you know where I could find more information on this?

    Thank you
    Last edited: Aug 25, 2015
  11. Aug 25, 2015 #10
    Yes, it all has to do with the fact that the surface area of a sphere increases proportionally to the square of the radius. Therefore, the strength of gravity is in effect spread out inversely as the square of the radius.

    Also, the gravitational constant is measured, not derived from other laws, so when we say that the Pi term is already included, we don't mean that we cancelled out Pi's and ended up with G. We literally mean that since G is just a constant, we could easily come up with another constant, like J for instance, such that [itex]G = J/(4 \pi)[/itex]. It makes no difference, so we go with the easy choice. Remember, it's the same number regardless. We just go with G. It's the same reason that in electromagnetism, we just replace the constant [itex]\frac{1}{4 \pi \epsilon_0}[/itex] with [itex]k[/itex].
  12. Aug 25, 2015 #11


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    As you say, the surface area of a sphere is given by [itex]4\pi r^2[/itex]. So if you have some quantity being "spread" over two spheres (imagine jam being spread on the two spherical pieces of bread!) we would have [itex]A= (4\pi r^2)d[/itex] and [itex]A= (4pi R^2)D[/itex] where "A" is the amount being spread over the two sphere, one having radius "r" and density (of whatever is being spread) "d", the other radius "R" and density "D". Since A is the same for both, [itex]4\pi r^2 d= 4\pi R^2 D[/itex] so [itex]\frac{r^2}{R^2}= \frac{D}{d}[/itex], the "[itex]4\pi[/itex]" terms cancel. That is why anything that is "conserved" (the total amount stays the same), as it spreads out, obeys an "inverse square" law.
  13. Aug 25, 2015 #12
    Thank you for the clarification axmls. I think I get it now. Let me test out my understanding. Since the 1/4π is a constant, the original Newton formula simplifies the equation by factoring it out and making it a part of G. And so when Cavendish did his experiment to find the gravitational constant, since he was using Newton's formula, then that meant the number he came up with already included 1/4π.
    Last edited: Aug 25, 2015
  14. Aug 25, 2015 #13
    Thanks HallsofIvy for the explanation. It looks like the 4π would only cancel out if it were being compared to another sphere and both spheres had the same source or two different sources of the same amount.
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