How would Newtonian gravity work in a 1-dimensional universe? I have come across the idea of gravity in different dimensional space. From the standard formula for gravity $F=\frac{GMm}{r^2}$ I have found that the $1/r^2$ term is a result of a three dimensional space which a gravitational field permeates. In general, for an $n$ dimensional space the force of gravity is $F\propto \frac{1}{r^{n-1}}$.
In 1 dimensional space this leads to an odd conclusion of $F\propto 1$, the force doesn't decrease as the distance between objects increases. 
This is quite counter-intuitive. Is there a false assumption which I am making?
If this conclusion is true then is there a simple explanation for why the gravitational field doesn't weaken over distance? We are used to thinking of gravity as just a force but really it is a field, perhaps if this was explained in terms of fields it would make sense.
 A: On way to think about it is the gravitational flux: In 3D, when you are further away from an object, you receive a lesser portion of overall flux, whereas in 1D, the flux does not decrease with the distance. This also explains why an infinite plane in 3D also yields a constant gravitational field, as the flux in this case also does not change with distance.
A: A gravitational force that is distance independent and constant is nothing new. This is possible even in three dimensions. If we imagine an infinite plane in $\mathbb R^3$,

then solving Poisson's equation for gravity, $\nabla^2\phi=2\pi G \rho \delta(z)$ and computing the force $F = -\nabla \phi$ will show that the gravitational field is a constant and always towards the plane as shown.
A: The fact $F \propto r^{1-d}$ follows from Gauss' law for gravity. In more detail, one has that,
$$\iint_S g \cdot dS \sim M$$
where $M$ is the mass enclosed by the closed surface $S$. If we take a point mass, enclosed by a sphere, then because of the symmetry of the problem, we can take $g$ to only depend radially. Then,
$$g \mathrm{Vol}(S^{d-1}) \sim M$$
and since $\mathrm{Vol}(S^{d-1}) = 2\pi^{d/2} r^{d-1}/\Gamma(d/2)$, one has,
$$g \sim \frac{M}{r^{d-1}}.$$
