# Newtonian gravity equation in a 2 dimensional world [duplicate]

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

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
–  Dimensio1n0 Sep 29 '13 at 10:25

Yes, this is correct.

In 2 dimensions,

So:

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

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.

So,

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

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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