The question is a bit vague, so let me provide some more context.
Imagine we have a world like Middle-earth (prior to the end of the Second Age), or any of many fictional Fantasy worlds (Glorantha, the world of the Young Kingdoms in the Eternal Champion series of Moorcock's Elric/Melnibonë books, or even the early Earth of the Pagans) where the surface of the Earth upon which everyone lives was/is "Flat" (this might mean perfectly flat, or it could mean a slight curvature).
But, let us go with "Flat."
Assuming that Gravity works as normal (it is a property of mass), what shape would the world need to be in order for gravity to be constant and in the same direction across the entire "surface" of this world?
Namely, gravity must always point "directly down," and must always be 1g.
Would this same shape work with a slight "warp" or "curvature" so that the surface of the "world" was slightly curved (gave the impression of being a very large sphere, but actually had "edges" you could walk to)?
The idea that it has edges seems to me to make the shape almost impossible, because at the edge, you would have more mass at a diagonal to that edge, leading to gravity "pointing" in a direction other than down.
But…. I would very much like to design a virtual world that operates by normal rules of physics (mechanical, Newtonian physics only at the present – I don't yet know enough Einsteinian Physics) which is "flat" without having to create a kludge for gravity.
Best Answer
There's a cute way to solve this. Consider Newton's equation:
$$F = G \frac{m_{1}m_{2}}{r^{2}}$$
Formally, this is identical to the fundamental law for electric force, Coulomb's law:
$$F= k\frac{q_{1}q_{2}}{r^{2}}$$
Now, it is possible to rewrite Coulomb's law so that it looks like:
$$\oint {\vec E}\cdot d{\vec A} = 4\pi k Q_{inc}$$
where $E$ is the electric field, the integral is over an arbitrary closed surface, and $Q_{inc}$ is the charge enclosed by that surface.
If we consider an infinite sheet of charge with charge per unit area $\sigma$, then we draw a cylinder with ends parallel to the surface, and we know that the field is perpendicular to the ends and parallel to the sides, and our construction gives us $Q_{inc} = \sigma A$, so we have:
$$2EA = 4\pi k \sigma A$$ $$E = 2\pi k \sigma$$
Since our initial equations are identical, we can simply substitute the results for Newton's equation, substituting $k\rightarrow G$ and interpreting $\sigma$ as mass per unit area, and we have:
$$g = 2\pi G \sigma$$
So, if you have an infinite sheet of mass with mass per unit area of 23,000,000,000 $\rm kg/m^{2}$, then the acceleration due to gravity will be 1 g everywhere.