Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Is there an intuitive explanation why it is plausible that the gravitational force which acts between two point masses is proportional to the inverse square of the distance $r$ between the masses (and not only to the inverse of $r$)?

share|improve this question
2  
Possible duplicate: physics.stackexchange.com/q/47084/2451 –  Qmechanic Dec 19 '12 at 14:20
    
This is essentially the same question, with charge replaced by mass: physics.stackexchange.com/q/32948 –  DJBunk Dec 19 '12 at 14:23
    
In this context there is Bertrand's theorem: books.google.de/…, but I don't know if there is an intuitive proof for this (which would lead to another answer to your question) –  student Nov 14 '13 at 17:29
add comment

5 Answers

Richard Feynman explains it wonderfully in his lectures on the "Character of Physical Law". You could have a look at this video.

One way to see it is that the inverse square law just is. There were experiments done to determine how one mass affects the other and then the result was that the force between them varies as the inverse square of the distance between them. But clearly that is neither intuitive nor obtained from "first principles". In other words, once we have the experimental data, who is there to explain it?

You can see that it has to be an inverse square law by symmetry. The other arguments for this question involve gravitons, but the argument works just as well even for fields. Conservation of flux

From the image a general idea of the argument can be seen. The argument goes like so:

If you consider a point mass in space, and then claim that it has a "gravitational field" around it, then from the law of conservation of energy the energy of the field has to be conserved in all space. But since the source of the field is a point mass, it is also clear that the field has to be spherically symmetric (since from every region around the point mass, the point mass looks the same). So if we consider a spherical region around the point and measure the total field passing through that surface (called the flux of the field) and then consider another spherical region with a larger radius and measure the field passing through that surface, the total field through both spherical surfaces must be equal.

But the surface area of the sphere goes up as $4\pi r^2$. Therefore for the flux through both spherical surfaces to be the same, the field has to go down as $1/r^2$.

share|improve this answer
add comment

If you think of the gravitational force carried by particles like gravitons, the flux of these force-carriers would fall off like $1/r^2$, just like photons from a point source. And that is due to the fact that they are spread out over a sphere with area $4\pi r^2$.

share|improve this answer
add comment

Inverse-square laws are completely geometry-based.

Let us assume a point source. This point source may emit anything such as Gravity, Light, Sound, Radiation, etc. One thing to note in these point sources - They spread their influence equally in all directions (approaching infinity) without limit to their covered range.

For instance let us assume light. It emerges as spherical wave-fronts from this point source. Taking a slice off this sphere and when projected in a perspective view, we'd get a square matrix of order of $\text{your distance}$. If you take $2 m$, then it'd be in a $2\times 2$ matrix which has $4$ elements which basically takes the square. If we look - into with our goggles somewhat, we'd see that these spherical wave-fronts spread in the form of $\text{your distance}$ as their radii $r$.


We know that the actual source strength of gravity is $4\pi GM$ whereas the surface area traversed by the spheres is $4\pi r^2$ and it results in the gravitational field intensity $$I=\frac{GM}{r^2}$$

share|improve this answer
add comment

Two answers come to mind (I assume this explanation is to be provided to non-specialists and non-would-be-specialists, as there is intuitive in the question):

  1. Argue by "heavy body and satellite problem" that other values make no sense (of course this can rule out $r^{-1}$ but not $r^{-1.9992}$). Data for comparison - how long is the Martian year (or year on Pluto).

  2. Say that the body emits gravitons in all directions and the total flux of gravitons has to be preserved which implies $r^{-2}$.

The second "argument" gives -2 power, but also is an "proof" by a unverified conjecture. Nevertheless it requires no calculation, therefore qualifies for intuitive.

share|improve this answer
add comment

No, there is no intuitive explanation why the gravitational force which acts between two point masses is proportional to the inverse square of the distance $r$ between the masses. All geometrical reasonings involve surfaces, not the point-like bodies, so they are too speculative.

share|improve this answer
    
Why is proceeding from point-like bodies to surfaces is "too speculative"? At least in the framework of classical physics. Though, frankly speaking, I see no explanation for the square power in classics, being it geometrical or not. Nevertheless if there some general principal prohibiting transition from points to surfaces, I'd like to know it. –  Yrogirg Dec 20 '12 at 8:19
    
Points do not have surfaces (or their surfaces vanish) whereas the force does not depend on the actual surfaces, but on the distance. –  Vladimir Kalitvianski Dec 20 '12 at 8:28
    
I want to do something with it, could you please join the chat? chat.stackexchange.com/rooms/6805/… –  Yrogirg Dec 20 '12 at 17:01
    
What do you think is more flawed --- transition from points to surfaces or from the force between two bodies to a field? –  Yrogirg Dec 21 '12 at 12:48
    
Transition to a field. Field makes sense only as an "external" force in the equation of motion of another body. –  Vladimir Kalitvianski Dec 21 '12 at 12:52
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.