Habitable zone around a Class O hypergiant

Oh, good, a planet habitability question. I love these.

What would be the habitable zone (capable of supporting liquid water) for a planet with an atmosphere and size similar to Earth, but several times more dense?

I recently wrote an answer on Worldbuilding referencing an answer I wrote on Earth Science containing formulas from Planetary Biology. In it, I give the formulas for the inner and outer radii of the habitable zone of a star: $$r_i = \sqrt{\frac{L_{\text{star}}}{1.1}}$$ $$r_o = \sqrt{\frac{L_{\text{star}}}{0.53}}$$ Plugging in the luminosity (320,000 times that of the Sun$^1$), I get $$r_i=539.36 \text{ AU}$$ $$r_o=1302.31 \text{ AU}$$

To put this in perspective, the Oort Cloud only extends to 50,000 AU - still substantially larger than this, and very, very far when compared to the orbits of the planets in the Solar System.

Would there need to be any special qualities of the planet itself to protect it and any life on it from the sun's energy output?

Not really, if it's positioned correctly. But the effective temperature might be different. This calculation is relative to the Sun and Earth. The formula is $$T=\left( \frac{L(1-a)}{16 \pi \sigma D^2} \right)^{\frac{1}{4}}$$ I get$^{1,2}$ an effective temperature 0.000028 times that of the planet if it was orbiting the Sun at 1 AU, which seems pretty low. But the habitable zone calculation is done relative to the stellar flux of the star. Still, this is weird. To fix it, the planet needs to have a much lower albedo ($a$), though that still wouldn't make much of a difference, it seems. I'll have to revisit this and see if I made any errors.

Is there a theoretical limit to the number of moons a planet can support and, if so, how many moons can this planet support?

Well, no, in theory. The distance from the star shouldn't affect the stability of a given system of moons around a planet, but something that has to be taken into account is the fact that the circumstellar disk around the star would have had a much lower density this far out, and so there would have been less material nearby for moons to form and be captured. The formula for density is $$\rho(r)=Ce^{-\frac{(r-r_{peak})^2}{2 \sigma ^2}}$$ where $\rho$ is density, $r$ is the distance from the center, $C$ is a constant, $\sigma$ is one standard deviation, and $r_{peak}$ is the radius at which the density is at a maximum. So the density this far out is going to be much less than the density at 1 AU. However, there might be a different $r_{\text{peak}}$ around a class O star.

Assuming one day on the planet is roughly 30 hours, what would be an estimate range for the number of days there are in one complete revolution around the sun?

The rate of rotation shouldn't affect the rate of revolution. As 2012rcampion wrote in his/her excellent answer, we can use Kepler's laws for this. Using the figure of a radius of 920.835, I get a period of 1871.81 Earth years. Have fun with the seasons!

Multiple planets could absolutely exist within the habitable zone, given that this one is so big. The only thing that might make their existence less likely would be the fact that it's unusual for planets to form that far out from the star, and it's unlikely that even one could form that far out, although it could move outwards from a position closer in.

$^1$ As 2012rcampion pointed out, 32,000 $L_{\odot}$ is waaaaay too low for a star like this. I'll multiply that figure by 10, because VY Canis Majoris has a luminosity of 270,000 $L_{\odot}$, and it seems you want to go a bit brighter.

$^2$ From here on out, I'll use an orbital radius of 920.835 AU, the mean of the two radii calculated at the start.

posted almost 10 years ago

HDE 226868‭

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