The setup and the equation

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

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:

$u$ 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

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 $2s\cdot 2a=4sa$.

Consider that the luminosity of a star is $$L_{\ast }=4\pi \sigma R_{\ast }^2{T}_{\ast }^4$$ where $R_{\ast }$ and $T_{\ast }$ are the star's radius and surface temperature, and $\sigma$ is the Stefan-Boltzmann constant. The flux at the surface of the planet is $$F_{\ast }=\frac{L_{\ast }}{4\pi r^2}=\frac{\sigma R_{\ast }^2{T}_{\ast }^4}{r^2}=x^2\sigma T_{\ast }^4$$ The power received is then $$P_{\text{in}}=4sa{F}_{\ast }=4sa{x}^2\sigma T_{\ast }^4$$ 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 $S=4{\pi }^{2}a(s+a)$. Setting $P_{\text{in}}=P_{\text{out}}$ yields $$4sa{x}^2\sigma T_{\ast }^4=4{\pi }^{2}a(s+a)\sigma T_{eff}^4$$ and so $$T_{eff}={\left(\frac{sa}{{\pi }^{2}a(s+a)}\right)}^{1/4}{x}^{1/2}{T}_{\ast }$$ In your case, this becomes $$T_{eff}=0.56x^{1/2}{T}_{\ast }$$ Now, $x_{\text{crit}}=0.0051$. Say we want $T_{eff}=300\text{ K}$. This means that $T_{\ast }$ must be less than $105000\text{ K}$ for there to still be an eclipse on the far side of the ring - if $T_{\ast }$ was higher, $x\ge x_{\text{crit}}$, 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 $\sim 200000\text{ K}$.

posted over 6 years ago

HDE 226868‭

601 reputation 61 463 80 49