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I am wondering if my line of thought is correct - and thus the resulting answer to the problem above would be correct.

As we know the gravitational force (of two point masses) is given by $$F = G\frac{m_1m_2}{r^2}.$$

So the gravitational force/vector field reduces with the distance squared. Now this is the formula in 3 spatial dimensions - and I always picture it as a point with gravitational field lines moving outward. Then the "strength" of the field would be the density of the lines. And hence the density drops with the distance squared (as it is inversely proportional to the area of the sphere at that distance).

Now taking this line of thought to other situations we can think of course about a hypothetical 2 dimensional world. Here gravity would also be. And here we can also see the density of the "gravitational field lines". However as they propagate only in 2 spatial dimensions the density would be inversely proportional to the circumference of the circle at a distance $r$. And hence the formula would lose the square and become like:

$$F = G\frac{m_1m_2}{r}$$

(With change $G$, and obviously we can't talk about mass in 2d).

Is this line of thought correct?

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marked as duplicate by Emilio Pisanty, Qmechanic Sep 29 '13 at 19:59

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Possible duplicate: physics.stackexchange.com/q/48447/2451 –  Qmechanic Sep 29 '13 at 10:00
the answer to this question can also be found at Intuitive explanation of the inverse square power $\frac{1}{r^2}$ in Newton's law of gravity –  Dimensio1n0 Sep 29 '13 at 10:24
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2 Answers 2

up vote 3 down vote accepted

Yes, this is correct.

In 2 dimensions,

enter image description here


$$\int_{C_1}\vec F\left(r_1\right)\cdot \mbox{d}\vec s=\int_{C_2}\vec F\left(r_2\right)\cdot\mbox{d}\vec s$$

Now, would it make sense, ; if the gravitational force on two points on the Very same circle, were not the same?

Certainly not! Formally, to say it's the same, we could say it's an $SO(2)$ symmetry.


$$2\pi r_1 F_1=2\pi r_2 F_2$$ $$r_1 F_1=r_2 F_2$$ $$F_2=\frac{r_1F_1}{r_2}$$

So we have it that $F_2$ is inversely proportional to $r_2$. To see how this exact form, $F=G\frac{m_1m_2}{r}$ (in 2 dimensions, not 3) arises, c.f. the near - last part of my answer here, but apply it to 2 - dimensions instead.

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You can never know whether your line of thought is correct or not because you have asked an unscientific question.In science the test of all ideas is experiment and you can never conduct an experiment in your hypothetical 2 dimensional world.

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A circle in 2 dimensional space, would exhibit the same behaviour as a infinitely long cylinder in 3 spatial dimensions. Or almost similar when staying close to a "very long" cylinder - which is something we CAN test. -- Funny: when I actually saw the "gauss law" I knew I was asking a stupid question and was correct. –  paul23 Sep 29 '13 at 13:32
There are hundreds of peer-reviewed papers on 2d and 1d gravity (not speaking about of thousands of papers on string theory and such, most of which also cannot be experimentally verified in the foreseeable future). I think your definition of science needs to be considerably broadened. –  user23660 Sep 29 '13 at 13:55
-1. This is off - topic trolling. –  Dimensio1n0 Sep 29 '13 at 14:10
This answer appears to be off - topic because it is not constructive. –  Dimensio1n0 Sep 29 '13 at 14:10
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