7
$\begingroup$

Based on this question about an impossible barbell planet, how hot would the bar be if it extended from halfway in the Earth's mantle to an altitude of 2r of Earth's radius? The diameter of the bar itself is 20% of the radius of Earth. The rod is made of mild steel (we are handwaving the structural integrity of molten steel and ignoring that a structure this high would instantly collapse.) Also, the rod is permanently affixed at halfway through the mantle. The orientation and distance between Earth's core and the bar are fixed points.

Barbell Planet, Broo

We are going to ignore orbital mechanics, structure strength of megastructures, and structural strength of the Earth's crust surrounding this megastructure.

Given the Earth's core temperature and the conductivity of mild steel, how high would a person have to go on the rod in order to touch the rod with their bare hands and not get burned?

Bonus Question(s): What would happen to the weather near the base of the rod where it intersects with earth? If you want to talk about how fast Earth's core would completely cool off, that's cool too.

Extra Bonus Question(s): Account for orbital mechanics, interaction with the magnetosphere, and compression heating of the rod against the earth's core and friction with the crust.

Remember, this is a question. Equations, official government sources, and journal references are most appreciated. A minimum of handwavium, please.

$\endgroup$

This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.

  • 1
    $\begingroup$ This question would be more suitable for physics.SE.😃 $\endgroup$ – user6760 Jul 29 '15 at 4:32
  • $\begingroup$ @user6760 I think it's fine here. $\endgroup$ – HDE 226868 Jul 29 '15 at 14:23
  • 1
    $\begingroup$ I suppose you're ignoring the melting point of steel and the fact that the core can't rotate with a rod sticking out of it up through the crust? $\endgroup$ – Samuel Jul 29 '15 at 17:13
  • 1
    $\begingroup$ @Green Yeah, I started to calculate the heat transfer and temperature in terms of distance from the core already. It is complex, the rod is too thick to be considered isothermal for a given cross-section and the shift from rock, to soil+water, to air, to vacuum (and back!) is a lot to deal with. Hard-science is the perfect description for this and I have abandoned my answer (no time this week, not that much!). $\endgroup$ – Samuel Jul 29 '15 at 19:05
  • 3
    $\begingroup$ "We're going to hand wave all these things. Oh yeah, and hard science, a minimum of handwavium, please." $\endgroup$ – AndyD273 Aug 3 '15 at 17:35
9
+100
$\begingroup$

As a first approximation, we treat the rod as a one-dimensional rod of length $L$. On each end is a heat source (the mantles of each of the two "Earths". We can begin to model the system as following the one-dimensional heat equation: $$\frac{\partial T}{\partial t}=k\frac{\partial^2T}{\partial x^2}\tag{1}$$ where $T$ represents temperature, $x$ is the distance from one end of the rod, $t$ is time and $k$ is a constant, the thermal diffusivity. Consider a case with the following boundary conditions: $$T(0,t)=T_1,\quad T(L,t)=T_2,\quad T(x,0)=f(x)$$ The first condition is because both heat sources are identical. We can use the method of separation of variables to arrive at our solution: $$T(x,t)=T_1+\frac{T_2-T_1}{L}x+\sum_{n=1}^{\infty}B_n\sin\left(\frac{n\pi x}{L}\right)e^{-k\left(\frac{n\pi}{L}\right)^2t}$$ where $$B_n=\frac{2}{L}\int_0^L(f(x)-u_E(x))\sin\left(\frac{n\pi x}{L}\right)dx,\quad u_E(x)=T_1+\frac{T_2-T_1}{L}x$$ In our case, $T_1=T_2$ and $f(x)=T_0$ is uniform. Then, if $\Delta T=T_0-T_1$, $$B_n=\frac{2\Delta T(1-\cos(\pi n))}{\pi n}$$ and $$T(x,t)=T_1+2\Delta T\sum_{n=1}^{\infty}\left(\frac{1-\cos(\pi n)}{\pi n}\right)\sin\left(\frac{n\pi x}{L}\right)e^{-k\left(\frac{n\pi}{L}\right)^2t}$$ Wikipedia cites steel as having roughly $k=2\times10^{-5}\text{ m}^2\text{ s}^{-1}$. Let's say $L=4R_{\oplus}$, $T_1=T_2=6000\text{ K}$ (according to estimates), and $T_0=300\text{ K}$.

It turns out that the bridge between the cores heats up really slowly. I computed the first 100 terms of $T(x,t)$ at a variety of times, and plotted them.

Temperature plot

Second temperature plot

The strong oscillations at the ends are just examples of the Gibbs phenomenon, and don't have physical significance.

It becomes dangerous to walk on the rod after timescales longer than about 100 million to 1 billion years, from what I can tell. The center should remain habitable.

There are a couple of things we haven't considered:

  • Radiative cooling, which I think will be important. This also might mean that the ends of the rods - at least, the sections immediately protruding from the surfaces of the planets - could appear quite bright. After all, 6000 K is approximately the temperature of the surface of the Sun!
  • The fact that the rod is a cylinder, not one-dimensional. I don't think this is a major factor when it comes to the relevant timescales, though, especially if the rod is thin.
  • The ends of the rod are embedded inside planets, and heat will diffuse into the outer layers of the planets. There should also be a non-zero gradient at the cores, at $t=0$.
  • Atmospheres have an impact on temperature circulation; the (cooler) sections of the atmosphere around the two planets should take in heat from the (hotter) sections around the rod.

That said, I think this simple 1-D model can give us an order-of-magnitude estimate.

$\endgroup$
  • $\begingroup$ I'm not physicist or mechanical engineer but I think your approach is sound. $\endgroup$ – Green Jul 30 '15 at 20:51
  • 1
    $\begingroup$ On the one hand, you're ignoring radiative cooling once the rod leaves the crust. On the other hand, you're misapplying the geothermal gradient, since heat will diffuse away from the rod into the crust as soon as the rod temperature gets above the gradient temp. On the gripping hand, atmospheric temperatures aren't enough: you also need to factor in both the changing density and the added cooling caused by the massive convective currents which will accompany the increased temperatures. $\endgroup$ – WhatRoughBeast Jul 31 '15 at 3:25
  • 2
    $\begingroup$ It's definitely heading in the right direction though :) $\endgroup$ – Tim B Jul 31 '15 at 8:50
  • $\begingroup$ I already mentioned in the chat, but for those who don't go there, you are missing a few points (so far), like day-night variations, climate influence and seasonal variations. All those will make some differences. But, and maybe more importantly, the rod will influence its environment. Intuitively, it might start to cool the Earth, and affect the climate. At first close by and with butterfly effect on the whole surface of the Earth. Those will affect any values you actually put in your model. So as Tim has it, yes it a good starting point. The question might be a bit demanding, though... $\endgroup$ – bilbo_pingouin Jul 31 '15 at 9:46
  • $\begingroup$ When I started to calculate this I found this document incredibly helpful. You can treat the rod like a pin fin in a fluid (except in space you stop subtracting the Q lost to convection out the sides of the rod). $\endgroup$ – Samuel Jul 31 '15 at 16:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.