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.

  • $\begingroup$ See my answer here: physics.stackexchange.com/q/26427 $\endgroup$
    – JamalS
    Feb 19, 2016 at 18:07
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    $\begingroup$ youtube.com/watch?v=VNqNnUJVcVs $\endgroup$ Feb 19, 2016 at 18:11
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    $\begingroup$ If you can tolerate slight curvature, I would think a conical wedge would work. Imagine a dunce cap, with us living on the nearly-flat spot where the head would be, and most of the mass concentrated in high-density matter near the point. $\endgroup$ Feb 19, 2016 at 21:06
  • $\begingroup$ Infinite plane would have constant gravity! $\endgroup$
    – Slereah
    Feb 20, 2016 at 12:32

3 Answers 3


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.

  • $\begingroup$ I suppose it could be interesting to show that there can be no rotating and/or variable-density (finite) discs that have the same effect on their inhabitants. (Ignoring discs that exert no gravity but simply accelerate upwards.) $\endgroup$
    – Řídící
    Feb 19, 2016 at 19:25
  • $\begingroup$ I came to a similar solution here. Note that you can have normal density for earth if you like. However, this requires that there are no edges, where the OP asked for them. $\endgroup$
    – Samuel
    Feb 20, 2016 at 16:54
  • $\begingroup$ @Samuel I looked over your analysis in the other thread you referenced where you obtained the thickness of the slab equal to the radius of the earth using 2/3 the average density. This is consistent with what I obtained in my answer below where I used the average density and obtained a slab thickness equal to 2/3 the earth's radius. $\endgroup$ Feb 20, 2016 at 20:51
  • $\begingroup$ The volumnetric density will depend on the thickness of the slab, which you can adjust to be whatever you want without affecting the value of $g$, since the final expression is independent of the distance to the sheet, with $\rho = \sigma / t$, and where $t$ is arbitrary. $\endgroup$ Feb 21, 2016 at 1:40
  • $\begingroup$ @Samuel: Also note that any solution is necessarily edgless, just by the nature of the symmetry of a gravitational acceleration vector that is always pointing downward. This means that the underlying geometry will have to be planar translation invariant, which then says that your matter distribution will have to be infinite in extent. $\endgroup$ Feb 21, 2016 at 1:41

If you are talking about Newtonian gravity, then one crucial piece of information you need is that gravity is a central force. That means between any two masses the force acts along the line connecting the two masses.

The flat constant gravity that we usually assume in real world is because the Earth is a large enough sphere that at sufficiently small space, it looks uniform. But all gravitational force vectors point to the centre of the Earth. So if you look at the flat slice, it looks like it is all pointing perpendicular to the plane. If you curve it a little (which is equivalent to taking a larger piece of land large enough to detect curvature), then every vector will still point perpendicular to the surface. This is how it should be done.

If you naively just take a sheet of masses and ask how gravitational field looks like, then notice that such a sheet is unstable: gravitational force will force the whole sheet to curl up, eventually stabilizing into a sphere, precisely because it is a central force. This is the reason why we do not see "cuboidal planets", or any other not closely spherical.

Short answer: if you are mimicking surface of the Earth, then your gravitational force simply has to be just perpendicular to the surface at every point, the same way as how gravitational force points towards center of the Earth. The "edge" of your world will have to be dealt somewhat artificially since if you follow conventional physics, your curved sheet is not stable unless you can keep them rigid by other forces (e.g. maybe the world has steel scaffolding or stuff). You will need to really do numerical calculations when that happens to account for e.g. forces on one edge due to the other edges.


The characteristics you are looking for would be exhibited by an infinitely wide flat slab having the same average density as the earth, and having a thickness equal to 2/3 the radius of the earth. The gravitational acceleration would be 1g at every point on the surface. And, even more interesting, it would also be 1g at any distance above the surface. Thus, the gravitational acceleration would not vary with altitude, no matter how far away you got. Unfortunately, you could not walk to the edge of this "earth" because it would be infinitely wide in all directions.

These results are consistent with what @Jerry Schirmer found in his analysis using Gauss' Law if I take the product of slab thickness and density equal to his parameter $\sigma$, representing the mass per unit area of his surface. However, in the analysis that I carried out, I integrated the actual gravitational inverse-square force equation over the slab, both laterally and through the thickness of the slab for an arbitrary test mass located above the slab surface.


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