# Gravity on a Minecraftian world?

I'm curious what the gravity would be like on an infinite area flat world. The plane of the world would have some finite depth but be infinite in all cardinal directions. This is, in effect, a Minecraftian world. This is expanding on the ideas presented in this question.

How deep would the world have to be in order to have a surface gravity identical to Earth?

• There are 64 blocks below sea level. – KSmarts Mar 23 '15 at 19:45
• @KSmarts Looks like blocks are about 100km to a side then. Good thing there isn't a power of two restriction here. – Samuel Mar 23 '15 at 19:51
• Isn't it a bit weird to add this bounty? Your answer is superb, but I'm not sure if there's a point, given that you'll simply lose 100 rep. – HDE 226868 Jun 18 '15 at 15:11
• @HDE226868 It totally is. It was an experiment with the bounty cap. I targeted the bounty at existing answers in the hope that it wouldn't show up in the featured list. So, unfortunately bounties don't affect the daily cap, but the question was featured anyway and the extra attention awarded more than the 100 I lost, oh well. – Samuel Jun 18 '15 at 15:43
• @PlanetalFlame The middle of the infinite plane? Where do you calculate that to be? – Samuel Jun 18 '15 at 20:17

It would be 6,378km deep. The same depth as the radius of the Earth.

The math

To calculate the gravity of an infinitely flat plane we simply need to integrate the force felt from concentric rings from a radius of zero to infinity.

This is a basic practice in college physics courses and is demonstrated here. It turns out, that the force on a mass above the plane is equal to:

$$F = 2\pi Gm\rho_a h \int_0^\infty {{R}\over{(h^2+R^2)^{3/2}}}dR = 2\pi G \rho_a m$$

Thus, the force does not depend on distance from the surface! This makes sense because the further away one gets from the plane, the more of the gravitational force vectors from the surrounding mass point back toward the plane. That is, the vertical component of all the gravity vectors become larger. The only factors that matter for the force are the gravitational constant, the density of the plane, and the mass of the object. Lets leave the gravitational constant alone and fiddle with the density of the plane. The mass of the object doesn't matter here, but it can be set to one for simplicity.

The density

In order for the gravitational force to be the same as Earth gravity, we need to set $2\pi G \rho_a$ term equal to $9.8{{m}\over{s^2}}$. This comes from $F = ma$, where $m$ is the mass of the object and $a$ is the acceleration due to gravity.

Thus, $\rho_a$, the mass density per unit area needs to be $2.34 * 10^{10} {kg\over{m^2}}$. That's really high, but let's see where it goes.

The depth

This plane is infinitely thin, so the mass density is per unit area, not unit volume. This not actually an issue, it's trivial to see that adding another infinite plane below this first one will simply add gravitational force. That is, it doesn't actually matter if the plane is infinitely thin or a kilometer thick, it only depends on the mass density of a slice of that plane.

From volume

We know the density of the earth (as in dirt, rocks, etc. - the material of the plane) in terms of volume, but what about area? The conversion is $\rho_v = {{\rho_a}\over D}$ where $\rho_v$ and $\rho_a$ are mass density for volume and area respectively and $D$ is the depth of the plane $^1$. So, if we're going to use volume density in place of the area density in the equation, it needs to be multiplied by the depth. Don't scoff yet, this works.

Earth's average density is $5540 {kg\over m^3}$, but the crust where we live has a density of about $2700 {kg\over m^3}$. Let's mix it up a bit and make the world a little iron rich and say the density is $3668 {kg\over m^3}$.

If the surface density is $2.34 * 10^{10} {kg\over{m^2}}$ and the volume density is $3668 {kg\over m^3}$ then the depth must be 6,378km. That's radius of the Earth.

Ok, so I fudged the density to get that answer. Big deal. Select any density you like and change the depth to work.

From force

Want to double check that? We can agree that, given a surface density of $3668{kg\over m^2}$, the gravity above the plane from the first equation would be $1.538 * 10^{-6} {m\over{s^2}}$. A very small amount of gravity.

But then how thick is the plane? I say one meter, see footnote 1. Though any thickness works, but our units are already in meters, so why not.

Now, we can also agree that if an identical plane was anywhere below this first one, its gravitational force would add with the first. Just like how all the dirt below us adds to the total gravity we feel. Recall that the distance from the infinite plane doesn't matter, the gravity is the same all the way up.

So how many of these one meter thick planes need to be below the top one to sum to a force of $9.8{{m}\over{s^2}}$? You guessed it, 6,378,000, or 6,378km deep if stacked one on top of the other. $$1.538 * 10^{-6} {m\over{s^2}} * 6,378,000 = 9.8 {m\over{s^2}}$$

Wait, what about 1/2 meter planes? Ok, well 1/2 the density and two times the number of planes then. Same depth, same gravity.

Wait, what about 1/x meter planes? Ok, well 1/x the density and x the number of planes then. Same depth, same gravity.

Fun fact

The gravity extends infinitely in all directions. This means that both sides of this world will have Earth surface gravity. The people that live on the other side are called Australians.

Footnote

1: The mass of one of the slices is equal to the mass density per unit area times the area, $M = \rho_a A$. The mass density of a volume is mass per unit volume, $\rho_v = {M\over V}$. Volume can be expressed as area times depth, $V = A*D$. So we can say the mass density of the volume is $\rho_v = {M\over {A*D}}$ The volume density of the plane would then be, $\rho_v = {{\rho_a A}\over {A*D}}$ or $\rho_v = {{\rho_a}\over D}$. For a depth of 1 meter then, mass density of the volume would equal the mass density of the area.

• Nice. I'll admit that I was quite surprised that surface gravity doesn't depend on shape (in this case). You don't want to delve into the general relativity . . . ? :-) – HDE 226868 Mar 20 '15 at 23:46
• @HDE226868 Nope, this is a Minecraftian world :) – Samuel Mar 21 '15 at 0:08
• People on the other side aren't going to be people without something seriously weird going on; and establishing contact would be quite the undertaking. – John_H Mar 23 '15 at 21:48
• @John_H nothing weird necessary, they just spawn there instead of here. (It is a Minecraftian world after all) – Mr.Mindor Jun 17 '15 at 20:59
• "The people that live on the other side are called Australians." -- this Australian approves. – Amziraro Jun 21 '15 at 18:06

If I may, I'll make a slightly simpler (I think) derivation of the answer.

We start with Gauss's law for gravity. In its integral form, it is $$\oint_{\partial V}\mathbf{g}\cdot \mathrm{d}\mathbf{A}=-4\pi GM$$ Here, $\mathbf{g}$ and $\mathrm{d}\mathbf{A}$ are vector quantities of the gravitational acceleration and an infinitesimal vector change in area. $\partial V$ is part of the surface area encompassing a volume - in this case, a Gaussian pillbox (see Example 4.2). Since the surface is presumably of uniform surface density, $\mathbf{g}$ must be the same throughout any region, and therefore $\mathbf{g}$ is independent of $\mathbf{A}$1. Using this, we can write the integral as1 $$\int_{\partial V}||\mathbf{g}||\text{ }||\mathrm{d}\mathbf{A}||\cos\theta=-4\pi GM$$ because $\theta=0$, and so $\cos\theta$ goes to $1$. We then get $$\int_{\partial V}||\mathbf{g}||\text{ }||\mathrm{d}\mathbf{A}||=||\mathbf{g}||\int_{\partial V}\mathrm{d}\mathbf{A}=g(2\pi r^2)=-4\pi GM$$ where $g=||\mathbf{g}||$. The factor of $2$ came about from the use of the Gaussian pillbox for the area. We then have $$g=-\frac{4\pi GM}{2\pi r^2}=-2\pi G\frac{M}{\pi r^2}=-2\pi G\sigma$$ where $\sigma$ is the surface mass density, Samuel's $\rho_{\alpha}$. We've gotten there by slightly different means than Samuel's answer, bypassing that integration (which really isn't too bad). We've also used the postulate that $\mathbf{g}$ (and thus $g$) is independent of location, which makes perfect sense on an infinite plane.

From here, it's easy to find the height of the plane, using $$\rho=\frac{\sigma}{h}\to h=\frac{\sigma}{\rho}\approx R_{\text{Earth}}$$

This is a direct analog of finding the charge density on an infinite plane - or doing the reverse, finding the electric field in a point on an infinite plane due to a charge density. Likewise, we can give the infinite plane a density, and then find the gravitational acceleration at its surface.

1 For two vectors $\mathbf{a}$ and $\mathbf{b}$, the dot product is equivalent to $$||\mathbf{a}||\text{ }||\mathbf{b}||\cos\theta$$ where $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$, and $||\text{ }||$ denotes the magnitude of a vector.