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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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Yes, this is correct. Formally, you could write for circles around the point mass $C_1$ and $C_2$: $$\int_{C_1}\vec F\left(r_1\right)\cdot \mbox{d}\vec s=\int_{C_2}\vec F\left(r_2\right)\cdot\mbox{d}\vec s$$

By rotational symmetry, this can be written as:

$$2\pi r_1 F_1=2\pi r_2 F_2 \Rightarrow F_2r_2=F_1r_1$$

See also.

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  • $\begingroup$ I was wondering, isn't there something weird happening in 2 dimensions given $\int_1^\infty \frac{1}{r} dr = \infty$ ? $\endgroup$
    – reuns
    Commented Jul 27, 2018 at 22:40

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