7
$\begingroup$

A common, matter-efficient science-fiction habitat is a hollow cylinder or ring in space that is spun to simulate the pull of gravity on its interior surface. These habitats have been imagined as small as a spaceship, mere meters in radius, up to a ringworld, 1 AU in radius.

This question pertains to a ring somewhere in the middle of these two extremes, placed in orbit around a star. This ring rotates around 2 axes. The first and faster rotation generates the centrifugal force responsible for simulating gravity. This can be visualized as the spin of a wheel. The second rotation is slower and occurs in an axis perpendicular to the first rotation. This can be visualized as a coin spinning in place on a countertop. The first spin results in a day/night cycle as the inner side of the ring exposed to the sun and the side hidden from the sun are constantly swapped by the rotation of the ring. The second spin’s effect is hard to visualize but it creates something akin to “seasons” where the contrast between day and night waxes and wanes. Here is a short gif I made in Unity that should help with visualization. Here the bottom view is from the perspective of the rotating camera in the top view.

Ring Simulation

While this is all interesting my question is about a very specific moment in this dynamic system. Inevitably the ring will spin to a point where it is edge on to the sun. At this moment the sunward-facing portion of the ring blocks the light from reaching the other side of the ring. In this way the ring eclipses itself. This can be seen towards the end of the gif. In an eclipse, the term “umbra” refers to the area totally eclipsed from the sun. What I would like to know is how to compute the size of this umbra and secondly how I can go about maximizing the size of this umbra, because let’s face it, the darker, larger, and longer an eclipse is the cooler it is. Since I already have a design in mind for the ring, maximizing the umbra will have to be done with the star and the distance the ring will orbit it. Importantly, I need the ring to still be habitable so the amount of heat coming from the star needs to be equivalent to what we receive on Earth.

I have prepared a schematic to describe the eclipsed ring and what I think are the relevant variables needed to describe the size of the umbra. Ideally, an answer would provide an equation to calculate the size of the umbra given any ring dimensions but the dimensions of this ring are a total radius of 10,000 km and a thickness of 100 km. Umbra schematic

$\endgroup$

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.

  • $\begingroup$ Are you sure that that sort of rotation can take place? $\endgroup$ – L.Dutch Aug 17 '18 at 4:18
  • $\begingroup$ @L.Dutch I think so, but if you have a reason why you think it can't I'd be interested in hearing it. $\endgroup$ – Mike Nichols Aug 17 '18 at 4:22
  • 1
    $\begingroup$ This sort of rotation can occur, but the possibilities are limited. If the "eclipsing" rotation is synchronised to the orbital rotation period -the ring will always face in the same direction and angular momentum is conserved. But if the rotation period is significantly shorter or longer then much energy will need to be provided to overcome the gyroscopic effects. For example if the rotation were once every 24 hours, once every 24 hours almost all of the angular momentum needs to be completely reversed. It would be like trying to turn a huge gyroscope upside down without any gimbals. $\endgroup$ – Slarty Aug 17 '18 at 12:54
  • 1
    $\begingroup$ @Slarty Thank you for your input, this is not very intuitive for me. If I’m understanding you correctly then this image would describe a stable orbit and spin that would produce seasons and eclipses as I desire. Here the absolute orientation of the spinning ring is unchanging but because it orbits the star the relative angles between the two bodies are changing. So there is no spin perpendicular to the gyroscopic (wheel) spin but it appears there is due to the orbital motion. If this gets any more complex I'll probably make another question. $\endgroup$ – Mike Nichols Aug 17 '18 at 17:34
  • 2
    $\begingroup$ Yes correct, your image shows the exact case where this would be stable. Gyroscopes do not like being turned over due to the conservation of angular momentum. That's why Earth's poles alway point in the same dircetion throughout the year $\endgroup$ – Slarty Aug 17 '18 at 17:43
8
$\begingroup$

The setup and the equation

Let's look at the geometry involved here. I created two diagrams:

enter image description here

On the left, we have the star of radius $R$. On the right, we have a cross-section of the ring. The center of the ring is a distance $r$ from the star, and the ring has a diameter of $2s$ and a cross-sectional radius of $a$. We can calculate $\theta$ using trigonometry: $$\theta\approx\tan\theta=\frac{R-a}{r-s}\approx\frac{R}{r}$$ We make the assumption that $r\gg s$ and $R\gg a$, and use the small-angle approximation to say that $\tan\theta\approx\theta$. Now, let's look at a slightly more in-depth diagram:

https://i.stack.imgur.com/PFlqt.png

$u$ is the radius of the umbra as projected onto the opposing side of the ring. Again, using the small-angle approximation, $$\theta\approx\tan\theta=\frac{a-u}{2s-a}$$ Setting both equations equal, $$\frac{R}{r}=\frac{a-u}{2s-a}$$ and $$u=a-(2s-a)\frac{R}{r}$$ Yesterday in chat, we talked a bit about striking a balance when it comes to the star the ring is orbiting. To maximize the size of the umbra, you need a small star, but it should also be relatively far away. On the other hand, you also want the ring to be in the habitable zone.

A red dwarf comes to mind, but red dwarfs are dim. A red dwarf of $0.1R_{\odot}$ would have a luminosity of about $0.01L_{\odot}$, meaning you would need to orbit at $0.1\text{ AU}$ to receive the same flux the Earth does. Plugging in the relevant numbers, at $R=0.1R_{\odot}$, $2s=9900\text{ km}$, $a=50\text{ km}$, and $r=0.1\text{AU}$, I get $u=4.2\text{ km}$. That's small.

Now, a white dwarf - a stellar remnant, sure - could have a radius of perhaps $10000\text{ km}$, maybe even less by a factor of two. The hottest white dwarfs come in at $0.5L_{\odot}$, meaning the ring could orbit at $0.71\text{ AU}$. Plugging these values in, we get $u=49.07\text{ km}$, which would essentially cover the opposite side of the ring (which has an inner radius of $50\text{ km}$).

Limits on stellar temperatures

As an interesting aside, we can find the point at which the umbra disappears by setting $u=0$, and getting $$\frac{R}{r}=\frac{a}{2s-a}=0.0051$$ This can yield some useful information. Define $x\equiv R/r$, and $x_{\text{crit}}=0.0051$. The umbra only covers part of the opposite side of the ring for $x<x_{\text{crit}}$.

Let's derive the equilibrium temperature of this ring. Assume that it's positioned face-on towards the star, as in the diagrams. Then the cross-sectional area facing the star is $2s\cdot2a=4sa$.

Consider that the luminosity of a star is $$L_*=4\pi\sigma R_*^2T_*^4$$ where $R_*$ and $T_*$ are the star's radius and surface temperature, and $\sigma$ is the Stefan-Boltzmann constant. The flux at the surface of the planet is $$F_*=\frac{L_*}{4\pi r^2}=\frac{\sigma R_*^2T_*^4}{r^2}=x^2\sigma T_*^4$$ The power received is then $$P_{\text{in}}=4saF_*=4sax^2\sigma T_*^4$$ If the planet is a black body, the emitted power $P_{\text{out}}$ is then its surface area multiplied by $\sigma T_{eff}^4$, where $T_{eff}$ is the planetary equilibrium temperature. The surface area is the surface area of a torus, or $S=4\pi^2a(s+a)$. Setting $P_{\text{in}}=P_{\text{out}}$ yields $$4sax^2\sigma T_*^4=4\pi^2a(s+a)\sigma T_{eff}^4$$ and so $$T_{eff}=\left(\frac{sa}{\pi^2a(s+a)}\right)^{1/4}x^{1/2}T_*$$ In your case, this becomes $$T_{eff}=0.56x^{1/2}T_*$$ Now, $x_{\text{crit}}=0.0051$. Say we want $T_{eff}=300\text{ K}$. This means that $T_*$ must be less than $105000\text{ K}$ for there to still be an eclipse on the far side of the ring - if $T_*$ was higher, $x\geq x_{\text{crit}}$, and the planetary equilibrium temperature would be greater than $300\text{ K}$. In fact, at the far end of the habitable zone, $T_{eff}=373\text{ K}$ - hotter than this an water boils. This sets a firm limit for a habitable and eclipse-producing star at $131000\text{ K}$. This is much hotter than the vast majority of stars, but it does rule out a number of hot white dwarfs, which may have temperatures of $\sim200000\text{ K}$.

$\endgroup$

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.