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I have just found perhaps the biggest thing in the universe: A black hole named SDSS J010013.02+280225.8. It itself is 12 billion times more massive than our sun, and its accretion disk is 439 trillion times brighter. This could mean the universe's biggest habitable zone, right?

Here is the scenario: A black hole identical in dimension to SDSS J010013.02+280225.8 is orbited by co-orbital of trinary star systems within its habitable zone. All the stars are white dwarves, each one only 20% more massive than our sun. Therefore, the "sunlight" comes only from the black hole's accretion disk.

In regards to the black hole's mass, its accretion disk's luminosity and the amount of X-rays being emitted, how far and how wide would the habitable zone be?

MolbOrg
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JohnWDailey
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    The title is misleading. I think you mean "around a black hole", not in one, because you can't really survive getting inside. Though some people have argued that the universe is actually the inside of a black hole. However, for folks outside, there's no way of knowing what's going on inside, or indeed, if there even IS an inside :-) – jamesqf Sep 05 '21 at 03:27
  • Interestingly TON 618 is about 5x more massive but 0.25x as brigjt by those numbers. – The Square-Cube Law Sep 05 '21 at 03:56
  • @TheSquare-CubeLaw Could you expand on that comment? – JohnWDailey Sep 05 '21 at 03:57
  • You said you found the biggest thing, but the size of a black hole is proportional to its mass and TOM 618 is 66 billion solar masses massive (2 billion more solar masses than all Milky Way stars combined). https://www.youtube.com/watch?v=0FH9cgRhQ-k – The Square-Cube Law Sep 05 '21 at 04:14
  • A remark.. so this black hole's accretion disk, which is many light years away, provides for the heat to let life exist on the planet. As opposed to a near sun, it would show as a single spot, a permanent, extremely bright star. Wouldn't that have impact on the viability of evolution on this planet ? Will all life need to be blind ? I imagine looking into the black hole once would result in loss of eyesight.. – Goodies Sep 05 '21 at 11:41
  • @jamesqf u have edit priveleges, so as anyone since xx rep - it such a clear cut - just use it – MolbOrg Sep 05 '21 at 19:59
  • @MolbOrg: I have better manners than that. Even if I didn't, I'm not sure what the OP wanted to say. – jamesqf Sep 06 '21 at 16:04
  • @jamesqf helping others now is equivalent to have no manners - what a lovely time are we living, lol. – MolbOrg Sep 06 '21 at 16:34
  • @MolbOrg: Changing what people have written, without their consent, is helping? I don't think so. Making a polite suggestion in a comment seems much nicer. – jamesqf Sep 07 '21 at 16:45
  • @jamesqf idk, it is one of the things I like about .se and except one case where it was 50/50 disagreement (understand the reason, but it was intencional - reverted and incorporated that edit it was interesting still) suprisingly only positive expririence of users editing my stuff. I like both aspects - technical ability and that people can participate. You identified things correctly from very beginning, and indeed I was misled by title(q still is interesting, no regrets) same as some of other 30-40 people before me. U "respect" one and neglected others. Sure do what u do best. – MolbOrg Sep 07 '21 at 17:43

2 Answers2

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Taking "439 trillion times brighter" to mean the "stellar" luminosity is 439 trillion times the sun, we can make some rough calculations.

The inner boundary of a stellar habitable zone is approximated by:

$$inner = \sqrt{L_s \over 1.1}$$

Where Ls is luminosity is solar units and the result is in AUs.

Plugging in the given value, we get:

$$inner = \sqrt{4.39 \times10^{14} \over 1.1} = 2.00 \times10^{7} AU$$

There are approximately 63,241 AUs in a light year so we can simplify that to 316 ly.

The outer boundary is approximated by:

$$outer = \sqrt{L_s \over 0.53}$$

Plugging in again:

$$outer = \sqrt{4.39 \times10^{14} \over 0.53} = 2.88 \times10^{7} AU$$

Which simplifies to 455 ly.

Making the width: 455-316 = 139 ly.

CAVEAT: these formulas were never meant to apply to this situation so I have no idea if they're right.

legio1
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  • Have you also taken the excessive amounts of x-rays from the black hole into consideration? – JohnWDailey Sep 05 '21 at 03:48
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    I did not. I meant to include a note about that but forgot. Basically, I couldn't find any good info on the x-ray component of the emissions of this little beastie. Given the nasty properties of quasars, I'd suspect your habitable zone would get routinely fried with high-energy radiation making the whole thing moot. – legio1 Sep 05 '21 at 04:13
  • @JohnWDailey You're going to have to stop them with atmosphere. – Loren Pechtel Sep 05 '21 at 04:13
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    At this distance you shouldn't be bound by the black hole's gravity. – The Square-Cube Law Sep 05 '21 at 04:16
  • @TheSquare-CubeLaw 439T/12B = 36,583 times as bright at our gravity level. You need to be 191x as far out. Since escape from Earth's orbit is 16,600m/s you will still have an escape velocity from the black hole of 87 m/s. In the billions of years it would take to go around once there certainly will be disruptions but it's still weakly bound. – Loren Pechtel Sep 05 '21 at 04:28
  • @LorenPechtel "Weakly bound"? Could you clarify on that? – JohnWDailey Sep 05 '21 at 04:31
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    @LorenPechtel What sort of atmosphere? I suspect that "identical in composition and thickness to Earth's" wouldn't be enough. – JohnWDailey Sep 05 '21 at 04:32
  • I'm thinking just about mass. Take the sun, increase mass by 10 orders of magnitude. But also increase distance from light minutes to 300 light years, that is 9 orders of magnitude. Gravity pull svales linearly with mass but is inversely proportional to the square of distance. Back of napkin, that hole's pull on Earth would be 10/81, or 12% that of the Sun. Weak bound, the slightest perturbation will have the planet escape. – The Square-Cube Law Sep 05 '21 at 04:42
  • @The-Square-Cube Law no idea if this makes any difference mathematically, but the planet is orbiting a star 20% heavier than the sun. Perturbations would have to catch the entire system orbiting the black hole, not only the planet.. – Goodies Sep 05 '21 at 08:53
  • @goodies I had skimmed through the question so I didn't see that detail. But even then: perturb just the star, and the whole star system will escape. However I did review my numbers, and back of napkin calculation we'd need a prpgrade push in the order of 5 km/s for that; That's quite a lot, requiring a stellar flyby at least to achieve. – The Square-Cube Law Sep 05 '21 at 09:57
  • @Loren Pechtel Orbital velcoity is proportional to (M/r)^.5, so it's over 1000km/s in this case. – blademan9999 Dec 19 '22 at 14:55
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legio1's answer is jaw-dropping but I see nothing in the math to disagree with. However, I am going to disagree anyway and say there is no habitable zone--the problem is the luminosity of the black hole won't hold steady enough for life. Planets might at times be of a suitable temperature but that state won't persist over long periods.

Loren Pechtel
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    "won't persist over long periods." Why? Also have seen claims that change in lyminosity of the sun, over time, was important for our life(and or evolution, do not remember exact statement) – MolbOrg Sep 05 '21 at 20:05
  • The luminosity of a star slowly grows over billions of years. A black hole has no innate luminosity, it's entirely dependent on what falls in. That's not going to be remotely consistent, the habitable zone will fluctuate too much for life. – Loren Pechtel Sep 06 '21 at 01:07