Suppose we have a planetary system in orbit around a Class O hypergiant star of 225 solar masses (larger than all but R136a1, the largest star we currently know of), 32,000 times as bright as our sun, and with an effective temperature of 40,000 K, and that the star exists for far longer than our understanding of stellar physics says it should because of [historical plot point].

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

Please note: Magic has a strong presence in this setting, but I'm looking for plausible, mundane solutions to the problems presented.

Edit: From HDE's answer, my number of 32,000 times the luminosity of the Sun may be inadequate to the task. HDE suggested 320,000 as a replacement. Feel free to adjust the luminosity if you feel a different value fits better, but please defend your choice.

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    $\begingroup$ Solar wind for O types and planet atmospheres don't mix well. $\endgroup$ Commented Feb 5, 2015 at 23:24
  • $\begingroup$ @MontyWild I've updated the question to better describe the star in question. Yes, I meant capable of supporting liquid water; I'm aware density doesn't impact this factor, but I included it for extra detail should it apply to other parts of the question. $\endgroup$
    – Frostfyre
    Commented Feb 5, 2015 at 23:35
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    $\begingroup$ Do you have the stellar surface temperature? O-class stars have a temperature range of 30000 – 52000 K. It does affect the answer. $\endgroup$
    – Monty Wild
    Commented Feb 5, 2015 at 23:39
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    $\begingroup$ Based on this: news.psu.edu/story/142653/2013/01/29/research/…, the temperature of a star affects the habitable zone. However, this model does not take stars this hot into consideration, probably because they are usually so short-lived for it to be considered effectively impossible to harbour life. The best I could say, then, is that the HZ would likely be rather further out than the usual calculations suggest. $\endgroup$
    – Monty Wild
    Commented Feb 6, 2015 at 1:33
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    $\begingroup$ Huge stars like this are too much unstable, very differently than our quiet and calm Sun. As a result, it is expected that the habitable zone is equally unstable constantly and chaotically moving in and out, which is something really bad for live. You will need to use your [historical plot point] to cover that also. $\endgroup$ Commented Feb 6, 2015 at 2:23

2 Answers 2


Some of your numbers seem to be way off:

  • Mass: The theoretical limit on the size of stars is around $200~M_\odot$. As a star grows larger and larger, it gets brighter and brighter. Eventually it reaches a point where the outer layers of the star are blown off by the radiation pressure. You may be able to get to $300~M_\odot$, but not to $3000~M_\odot$.
  • Luminosity: Your luminosity is a huge underestimate. Most supermassive stars will, for the reasons mentioned above, be close to the Eddington limit: the maximum possible luminosity for a star of a given mass. It is equal to: $$ L_\text{edd} \approx 34\,000\left(\frac M{M_\odot}\right)L_\odot $$ A $225~M_\odot$ star would probably be between $1\,000\,000-10\,000\,000~L_\odot$, several hundred times more than you predict, and a $3000~M_\odot$ star would probably be billions of times brighter than the Sun, exceeding the limit, for the brief period before it collapsed/exploded.

I'll continue, assuming a $225~M_\odot$ star with a luminosity of $3\,200\,000~L_\odot$. (Based off the luminosity of this star, the most massive known currently).

The radiation budget of a planet does not depend on its size, since the amount of solar energy received and amount of thermal energy radiated are both proportional to surface area. However, the surface temperature will depend on the atmosphere's thickness and composition, especially the presence of greenhouse gasses. I will assume an atmosphere similar to Earth's in thickness.

(Note however that, due to the increased gravity, the density and pressure of the atmosphere will necessarily be increased, but it will not reach as far into space.)

Making the above assumptions, we therefore want to place our planet so that the solar constant (incoming solar energy per unit area) is the same as for Earth. Due to the inverse-square law, this disance will scale exactly with the square root of luminosity. The distance is therefore $\sqrt{5\,000\,000}\approx 2200$ times Earth's orbital radius, or $2200~\text{AU}$.

Serban noted that there would be increased stellar wind. Indeed, R136a1 is estimated to be losing mass at a rate of $50~M_\odot/\text{My}$. This only amounts to about $300~\text{kg}/\text{s}$ of wind impacting your (Earth-size) planet, almost 1000 times the wind at Earth. I would suggest giving your super-dense planet a super-strong magnetic field to protect itself (this also nets you super-intense [magical?] aurorae).

Now from Kepler's laws we know that the period of an orbit is: $$ T = 2\pi\sqrt{\frac{a^3}\mu} $$ We can again simply compare to the Earth-Sun system by ratios. $\mu$ is 225 times greater, and $a$ is 2200 times greater. Therefore the period is about $\sqrt{2200^3/225}\approx 7000$ times greater: in other words, this planet would have a year as long as agriculture is old on Earth. Assuming a 30 hour day, that's about two million days per year.

A planet's moons are inside its Hill sphere, or sphere of gravitational influence (an exception is our own moon, which is incredibly huge compared to other planets' moons). The size of the Hill sphere is: $$ r\approx a\left(\frac{m}{3M}\right)^\frac 1 3 $$ For your planet, the distance would be about 0.01 times the orbital radius $a$. But remember that $a=1000~\text{AU}$: that gives the Hill sphere a radius of $22~\text{AU}$, twice the size of Saturn's orbit. So you can probably pack in as many moons as you want.

Note however that the moons are likely to be small, like Saturn and Jupiter's moons, which are thousands of times smaller than their parent planet. A large moon would sweep up all the smaller bodies as they formed, resulting in an Earth-Moon-like system.

  • $\begingroup$ Hm. I just reread my source for the size of my star, which specified Canis Majoris as the largest star, but in terms of radius rather than mass, and I confused the two. I've adjusted the size of the star downward, to 225 solar masses. Could you recalculate for this amount? $\endgroup$
    – Frostfyre
    Commented Feb 6, 2015 at 0:58
  • $\begingroup$ I'm rather disappointed that this answer hasn't gotten more recognition. +1; just like an answer I'd write. $\endgroup$
    – HDE 226868
    Commented Feb 6, 2015 at 1:27
  • $\begingroup$ Notes: 1. Hill sphere assumes a 2 body system, whereas there'll actually be other planets, 2. The density affects gravitational force, which affects retained atmosphere depth, which affects temperature. 3. The atmosphere might get blown away by stellar wind. 4. Hill radius seems off, somehow. $\endgroup$ Commented Feb 6, 2015 at 2:51
  • $\begingroup$ @SerbanTanasa The derivation of the Hill sphere radius assumes a two-body system, and adding more planets will perturb moons' orbits. However, the radius of the Hill sphere will not change very much. Even with two massive planets in the same solar system, the Earth manages to hold onto the Moon outside its Hill sphere. $\endgroup$ Commented Feb 6, 2015 at 18:25
  • $\begingroup$ @2012rcampion Thanks for updating your answer. Comparing your results to HDE's, I notice that you place the planet more than twice as far away, probably because of the factor of 10 on luminosity. HDE calculated (rightly or wrongly) that the temperature might be too low, so I'm having trouble reconciling that putting the planet even farther away would make it habitable. $\endgroup$
    – Frostfyre
    Commented Feb 6, 2015 at 23:29

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.

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    $\begingroup$ I thought I might find one of your excellent posts in response to this question. You may or may not have noticed my edit, which scaled the star's mass down and switched the "largest" reference to a different star. I assume you're using Earth-years to measure the orbital period of this planet. Out of curiosity, would it be possible for multiple planets to be in the habitable zone you calculated? $\endgroup$
    – Frostfyre
    Commented Feb 6, 2015 at 1:44
  • $\begingroup$ @Frostfyre I did notice the edit, but forgot to remove the part about the largest star. As to the other things: Yes, I'm using Earth years (I should have clarified that!) and yes, multiple planets could absolutely be in the habitable zone. I'll put all of this in. $\endgroup$
    – HDE 226868
    Commented Feb 6, 2015 at 1:50
  • $\begingroup$ I just did a quick calculation that may be entirely wrong, but it seems the planet would have a temperature of about 2 K. $\endgroup$
    – Frostfyre
    Commented Feb 6, 2015 at 1:58
  • $\begingroup$ @Frostfyre I've thought that mine was wrong for quite some time, so you may very well be right. What numbers did you use? $\endgroup$
    – HDE 226868
    Commented Feb 6, 2015 at 1:59
  • $\begingroup$ Considering the distance between 1 AU and 920 AU, I estimated the 1 AU temperature as the sun's 40,000 K and multiplied it by the 0.000028 number you came up with. I don't have much of a stellar physics background, so I'm sure my estimation is invalid in one way or another. $\endgroup$
    – Frostfyre
    Commented Feb 6, 2015 at 2:06

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