Gravity on a Minecraftian world?

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 $${\int }_{\partial V}||\mathbf{g}||||\mathrm{d}\mathbf{A}||\cos \theta =-4\pi GM$$ because $\theta =0$, and so $\cos \theta$ goes to $1$. We then get $${\int }_{\partial V}||\mathbf{g}||||\mathrm{d}\mathbf{A}||=||\mathbf{g}||{\int }_{\partial V}\mathrm{d}\mathbf{A}=g(2\pi r^2)=-4\pi GM$$ 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 $$||\mathbf{a}||||\mathbf{b}||\cos \theta$$ where $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$, and $||||$ denotes the magnitude of a vector.

posted about 9 years ago

HDE 226868‭

601 reputation 61 463 80 49