What would life on Earth be like if the Sun was replaced by two blue giants?

Question

Allright, what i'm asking is how life would be like on Earth (which pretends Earth could develop life in the "short" life of a blue giant) if we had two blue giants in the sky instead of the Sun?

Answer 1

Blue supergiants are massive. Really, really massive - up to dozens of times the mass of the Sun. They are also extremely hot, large and luminous. They live much shorter lives that stars like the Sun, and may die spectacularly. All of this means that they aren't great stars to host habitable planets.

The luminosity of the Sun is on the order of $10^{26}$ watts - an incredible amount (think of that compared to a lightbulb)! But that's nothing compared to a blue supergiant. Rigel A has a luminosity 120,000 times that! Let's see if we can replicate the Sun's effects by comparing the stellar flux density. The flux $F$ from a star at a radius $r$ is $$F=\frac{L}{4 \pi r^2}$$ Setting the flux of the Sun equal to that of Rigel, we find that $$F_{\text{Sun}}=F_{\text{Rigel}}$$ $$\frac{L_{\text{Sun}}}{4 \pi r_{\text{Earth}}^2}=\frac{L_{\text{Rigel}}}{4 \pi r^2}$$ Doing some cancellations, and writing Rigel's luminosity in terms of the Sun's luminosity, $$r^2=\frac{120,000 L_{\text{Sun}}}{L_{\text{Rigel}}}r_{\text{Earth}}^2$$ $$r=346r_{\text{Earth}}$$ So Earth would have to be 346 AU away from Rigel to receive the same flux as it does from the Sun. Put it in a binary system with two Rigel-equivalents and that figure is multiplied by $\sqrt{2}$, becoming roughly 490 AU. That's well into the Kuiper Belt - right in the area of Planet Nine.

A binary system of blue supergiants is UW Canis Majoris, with spectral types O7.5 and O9.7. Their combined luminosity is approximately 260,000 solar luminosities, and they orbit close together, likely at about 0.16 AU. Therefore, their circumstellar habitable zone should be comparable to that of Rigel's; you won't be able to orbit closer than about 350 AU and still have a habitable planet.

---

Put Earth around the central blue supergiants in the system and things get interesting. It's going to receive 120,000 times the flux that it receives from the Sun, so it will be hot. To put that in perspective, the Earth would have to be about 0.00288 AU from the Sun to receive that kind of heat. I would expect temperatures to be - well, probably many hundreds of degrees, no matter what scale you use. Life on the surface is out of the question.

Any life will have to be belowground. Even the extremophiles will be feeling pretty warm. There won't be any water, so subterranean life is our only option. I doubt that anything bigger than small bacteria would have a shot. Even something like a worm would have a whelk's chance in a supernova. Oh, and this assumes that such a small planet could form around these stars (I wouldn't bet on it) and that life could develop quickly enough.

Answer 2

You have only two stable orbital options here :

• Earth orbits around one of the blue stars, and the other one is far away. The day/nigh cycle with the main star is classic, but the second will be visible during some nights and maybe even during the day. Depending the time of the year, you could have a season with bright nights, a season with dark nights, and two seasons in-between.
• Earth orbits around the barycenter of both stars. You always see both stars very close in the sky, except when they go down or up the horizon. You have a balanced day/nigh cycle, depending of the orbital speed and planet rotation.

In both cases, stars are very bright and emit strong radiations, putting high pressure on lifeforms. Species will avoid exposure to direct sunlight and develop special protection mechanisms, like ADN-multiple replicas. In the unlikely case of a tidally-locked planet, the dawn area will be an interesting location, with reduced exposure but enough energy.

Now, we have an issue regarding the lifespan of blue stars. Our earth is 4.6 billion years old. Simple cells appeared 1 billion year thereafter, closely followed (well, only 200 million years) by cyanobacterias which ruled the day. 1.5 billion years after this, multicellular life appeared. A blue star is unlikely to stay for that long, so your lifeforms are likely to be very primitive.

Answer 3

Well the absolute magnitude of Rigel from Wikipedia is -7.92+/-0.28, compared to the Sun's absolute magnitude of 4.83. In the sky it would have an apparent magnitude of -26.74 - 4.83 - 7.92(+/- 0.28) this is an average visual magnitude of -41.29.

Each 5 points of magnitude lower is 100 fold increase in intensity, that means that Rigel would give you between 97,000 and 163,000 times the power output of the sun with an average at 126,000 x.

Using the fourth power law for radiation and surface temperature of earth at 300K on a warm day, you'd get a range of temperatures from 5000 - 5800 degrees C. In practice, Rigel has more gravity than the Sun, which would pull it into a closer orbit.

How close would the earth orbit Rigel of 21 solar masses, if there was no push given to the earth and the sun was suddenly replaced by Rigel?

We define M as Sun's mass, r as 1AU, m as Earth's mass, and G as the gravitational constant.

From http://www.schoolphysics.co.uk/age16-19/Mechanics/Gravitation/text/Kinetic_energy_in_orbit/index.html

The total energy in a stable orbit GPE, is -GMm/r assuming an energy 0 to be a theoretical infinite distance away, which would require no rotational velocity not to be drawn in. KE is given as GMm/2r, giving a total energy of -GMm/2r. This would mean 2KE + GPE = 0 for a stable orbit.

As we are only interested in ratios, we will not convert into base units. With Rigel of 21M, GPE is initially -21GMm/r and KE is initially GMm/2r at its original orbit around the Sun, before it begins to accelerate inwards towards the bigger and heavier Rigel.

As the earth falls inwards, it will gain KE and lose GPE in equal amounts.

$\text{GPE loss} = -21\,\frac{\text{GMm}}{r} + 21\,\frac{\,\text{GMm}}{R}$,

so $\text{New KE} = \frac{\text{GMm}}{2r} - 21\frac{\text{GMm}}{r} + 21\frac{\text{GMm}}{R}$.

$\text{New GPE} = -21\frac{\text{GMm}}{R}$

Now we need to find R such that $2 \cdot \text{New KE} + \text{New GPE} = 0$

$2 \cdot (\frac{\text{GMm}}{2r} - 21\frac{\text{GMm}}{r} + 21\frac{\text{GMm}}{R}) - 21\frac{\text{GMm}}{R}) = 0$

We can multiply out the G, M and m in order to be able to follow more easily.

$2 \cdot (\frac{1}{2r} - \frac{21}{r} + \frac{21}{R}) - \frac{21}{R} = 0$

$2 \cdot (0.5/r - 21/r + 21/R) - 21/R = 0$

$\frac{1}{r} - \frac{42}{r} + \frac{42}{R} - \frac{21}{R} = 0$

$\frac{-41}{r} + \frac{21}{R} = 0$

which makes $R = \frac{21}{41r}$ which is 0.512AU.

This increases the flux density hitting earth by nearly 4 x and the surface temperatures by 1.4 x assuming a constant radiation heat loss for earth. This would cause surface temperatures to rise to 7100 - 8200 Celsius.

In practice, it would probably form some sort of elliptical orbit, getting closer and further away than this, varying between 1AU where it started, and a value where total energy was the same, but kinetic energy enough to fling it back out to 1AU again. If someone else wants to calculate this, go ahead.

Either scenario would more than cook the earth, causing it to glow white hot and maybe even boil it off into space!

Answer 4

Re blue stars not having a long enough life: how about blue stragglers? That fits in with mult-star systems, and opens up even more exotic features.