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Q&A

Gravity on a Minecraftian world?

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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?

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This post was sourced from https://worldbuilding.stackexchange.com/q/12442. It is licensed under CC BY-SA 3.0.

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1 answer

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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.

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