I thought I had it.

I thought I had finally found the solar system with enough light to produce hundreds if not thousands of Earths. I thought that a black hole 12 billion times more massive than our sun with an accretion disk 439 trillion times brighter would be the perfect place to create a million Earths without concern of tidal locking.

Indeed, the habitable zone from such a system would start at 316 light years from the black hole and end at 455 light-years. Do the math, and that would result in a habitable zone measuring in at 139 light-years. Compare that to our solar system, which spans only 30 astronomical units, or 0.00047302 light years.

I thought I had finally found a suitable habitable zone that's big enough and with no time limit (unlike those giant or hypergiant stars.) But orbiting a black hole with a hyper-bright accretion disk has its own monkey wrench--x-rays. It turns out that they can't be deflected by either an electric or a magnetic field, making life on any of those habitable-zone planets impossible.

So without barring life from forming, what would protect an Earth-sized habitable zone planet from a black hole's x-ray bombardment?

  • $\begingroup$ tidal locking would give you at least one usable side of the planet, although I am not sure tidal locking behaves the same wit black holes. $\endgroup$
    – John
    Commented Sep 6, 2021 at 0:09
  • $\begingroup$ One usable side is not good enough if you want to seed that planet with Earth species. $\endgroup$ Commented Sep 6, 2021 at 0:24

3 Answers 3


The disk itself can shade you

Consider a luminous disk with an additional point light source in the center. If you are above or below the plane of the disk, it is very bright from both disk and point source.

But in the plane of the disc you receive only radiation from the edge. The rest of the disc is interposed between you and light emitting materials farther in.

If the planet is in the shadow of the accretion disc (and perhaps associated dusty torus) and orbiting in the same plane as the disc, radiation coming from the innermost and hottest parts of the disc and the corona of the black hole will be absorbed and scattered by interposed disc materials. Some radiation will make it through to the edge of the disk and some will energize materials within the disk to themselves emit energy but total energy will be much less than an object orbiting (or observing) from above or below the plane of the accretion disc.

from comments - I found a fine youtube video where they look at a flat LED bulb - it is a circle rather than a sphere. The bulb produces a dark band corresponding with its edge, where relatively less light is emitted as compared to areas illuminated by the full disk of the flat bulb.

Philips Slim Style LED Light Bulb Review and Teardown: Race to the Bottom


The dark band is more evident in the video than in this screenshot., It is a predictable problem with a light source that is flat. The edge gives off relatively less light. So too the accretion disk.

  • $\begingroup$ "If you are above or below the plane of the disk, it is very bright from both disk and point source." You mean like this? i.imgur.com/hgSzMPU.jpg $\endgroup$ Commented Sep 7, 2021 at 21:36
  • $\begingroup$ @JohnWDailey - no, that is the disc seen edge on. The gravity warps the light so we can see some disc also not just a line (so the shade might not be as good as I thought). But check out this movie - it starts with the image you link and then the perspective travels up over the disc so you are above the plane of the disc looking down - it looks like a disc. svs.gsfc.nasa.gov/13326 $\endgroup$
    – Willk
    Commented Sep 7, 2021 at 22:07
  • $\begingroup$ Then I'm afraid I don't follow you. $\endgroup$ Commented Sep 7, 2021 at 22:28
  • $\begingroup$ OK - I added a screenshot and a link to a video demonstrating what I mean about the shadow of the disk. $\endgroup$
    – Willk
    Commented Sep 8, 2021 at 0:09
  • $\begingroup$ Won't work, the radiation that is abosrbed by the disk will simply be Re-emitted, and it will be at short wavelengths. $\endgroup$ Commented Dec 19, 2022 at 15:21

As a comment.

There are different https://en.wikipedia.org/wiki/Accretion_disk 's including those which are thermal, similar to a star (as I understood) soo maybe handwave if it is backdrop/just a premise for the stuff to be. The shape of the thing may be a bigger deal and a disk is not necessarily a great case.

Radiation by itself is not necessarily a life stopper, water is good enough to block the stuff, and it is most likely the place for one to form, and then there are plenty of subsurface bacterias and from there enough roads to multicellular options. How far it can go, in terms of complexity - it depends. May it reach intelligent enough creatures(at least ants or molds) - it depends.

It took about 500 million years to evolve photosynthesis, so before that, they used resources in a not renewable way and for that life can start deep enough, if conditions are right, and x-rays are a nonissue. The spectrum in for some accretion disk was like 3-15 keV x-rays, not sure but would not call it that harsh, it's definitely not MeV's which makes secondary stuff and all is quite complex. Those low keV x-rays will be absorbed in straightforward ways. And if it is enough of those may be your future life will use it.

But another moment, let's assume 100% x-rays, intensity like the sun 1.3kW/m2, of those keV ones. For water attenuation coefficient is order of 0.1/m. To decrease this flow by a factor of 130'000'000×3600 it requires the depth of that water to be about 2.7m(surprise surprise - water is sooo goood). And if my voodoo isn't that much wrong, it corresponds to 0.1 mkSv/h, more or less what we have on earth.

Soo, not necessarily such a big problem, but not sure about the numbers, needs to check it out. With more energetic x-rays, like cosmic ones we have here, you add 10-15m on top of that. So sea life can exist.

X-rays can be used for photosynthetic activity, or if intensities like next to fuel rods of reactors can be part/driver of abiogenetic restoring of consumed chemicals, by splitting end products of that life, which does not exclude x-photosynthetic process deeper down (few meters deeper).

So, not conclusive, but it looks like x-rays are overestimated.


This answer is not as much an answer as it is several questions you should consider and, I hope, answer.

Part One: The Radiation Problem.

As I remember, Earth's atmosphere stops most X-rays and gamma rays from reaching the surface. That is why X-ray and gamma ray telescopes are in satellites and space probes above the atmosphere.

MolbOrg's answer claims that water blocks X-rays. He calculated about 2.7 meters of water would be necessary, or about 10 to 15 meters more for hard X-rays.

Any habitable planet will have water vapor in the atmosphere. Water vapor is much less dense than liquid water. But the water vapor in the atmosphere extends much higher than the deeps of water needed to block X-rays. If water is only 1,000 times as dense as atmosphereic water vapor, 2.7 kilometers of atmosphere would equal the protective effect of 2.7 meters of water, and 12.7 to 17.7 kilometers of atmosphere would equal the protective effect of 12.7 to 17.7 meters of water.

Of course atmosphereic density, and thus the density of atmospheric water vapor, falls off rapidly with increased height.

Possibly high flying birds and high floating bacteria would be killed on your worlds, and possibly regions with dry air, like deserts, would be more dangerous than the rest of the planetary surface.

And possibly there are other common gases, ices, and forms of dust which could exist in the atmosphere in concentrations high enough to block x rays and yet small enough not to poison life.

What about ultraviolet rays? Earth's atmosphere greatly reduces the amount of ultraviolet radiation which reaches the ground. It is claimed that if a relatively nearby gamma ray burst hit Earth the gamma rays would break up the ozone in the ozone layer, and so much ultraviolet light would reach the surface that life above the surface of the ground and the surface of water would die. So I wonder what the intensity of gamma ray radiation would be in the habitable zone of your black hole.

So I don't know whether a planet with an Earth-like nitrogen, oxygen, etc. atmosphere would be totally safe for life or totally dead and lifeless due to the intense ultraviolet, X-ray, and gamma ray radiation. You may need to do a lot more research.

Part two: The Absolute and Relative Dimensions of a Habitable Zone.

You may need to do some research to calculate how many planetary orbits can fit within the circum black hole habitable zone of your super giant black hole.

I notice that the outer limit of your habitable zone is about is about 1.439 times as far as the inner limit. It is simple to calculate the relative size of the habitable zone of an object from the ratio between its luminosity compared to the Sun's.

As this list of estimates of the inner and outer edges of the habitable zone of the Sun shows, there is considerable uncertainty about its dimensions, and thus considerable uncertainty in calculating the habitable zones of other luminous astronomical bodies.


Part Three: The Possible Number of Planetary Orbits.

As I remember, each planet has a forbidden region around its orbit in which no other planet can have a long term stable orbit.

The width of the forbidden region is calculated from the masses of the primary and the planet, and the distance between the primary and the planetary orbit.

I believe that the more massive the primary is and the stronger its gavity is at the distance of the planet, the smaller will be the planet's forbidden region.

Your black hole is 12 billion times as massive as the Sun, so that would work to make the forbidden zones around planetary orbits much smaller.

Your habitable zone extends from 316 light years to 455 light years from the black hole. There are over sixty three thousand astronomical units (AU) in a single light year - about 63,239.2493 in fact. So 316 light years is about 19,984,180 AU, and 455 lightyears is about 28,774,690 AU, according to this converter.


I believe that the effect of gravity falls off with the square of the distance. Since your distances are tens of millions times one AU, their squares should be equal to hundreds of trillions. Therefore, the gravity of an object at 316 light years or 19,984,180 AU is only about 2.5039596 times ten to the minus 15 power as intense as at a distance of 1 AU, and the gravity of an object at 455 light years or 28,774,690 AU will be only about 1.20077545 times 10 to the minus 15 power as intense as at 1 AU.

If the black hole is 12 billion times as massive as the sun it would be 12,000,000,000, or 1.2 times ten to the 9th power times as massive as the sun. So its gravitational influence on planets in its habitable zone should be about 1 millionth as strong as the gravity of the Sun at 1 AU, and the forbidden regions of the planets should be much larger.

So I think that you probably need to to make the black hole many times more massive, or the accretion disc many times less luminous, or both, to produce a system where the forbidden regions of planets will be smuch smaller and many more habitable planets can fit within the habitable zone.

If the black hole is X times as massive as the Sun, and the accretion disc is X times as bright as the Sun, the gravity and the radiation at a distance of the square root of X AU should equal the solar gravity and solar radiation at a distance of 1 AU.

So I think that you need to make the ratio of the black hole's mass to the Sun's mass higher than the ratio of the accretion discs's luminosity to the Sun's luminosity to fit more planetary orbits into the accretion discs habitable zone.

The requirements for a planet to be habitable for humans (and thus also for multi celled land dwelling oxygen breathing animals in general) are discussed in Habitable Planets for Man, Stephen H. Dole, 1964.


On pages 49 to 52 he discusses the spacing of planets in the solar system and their forbidden regions. Dole cites as a source a paper (Dole, 1961). According to the bibliography, that is Dole, S. H. "Limits for Stable Near-circular Planetary or Satellite orbits in the Restricted Three-body Problem", ARS J., 31, No. 2 (February, 1961), pp. 214-219. Apparently is the American Rocket Society Journal, which may help you find the formula Dole used.

Part Four: Exoplanet Examples of Planetary spacing.

Discovered systems of two or more exoplanets orbiitng a star have vast differences in the relative and absolute spacing of the discovered planets. Of course those systems could have as yet undiscovered planets which would change what we know about the planetary spacing.

And i don't know how well the known examples of planetary spacing agree with Dole's formula for calculating planetary forbidden regions.

Some very small stars happen to have a number of planets in orbits very close to the star and thus to each other.

Kepler-42 has three planets whose orbits have semi-major axes of 0.006, 0.0116, and 0.0154 Au. The differences are 0.0056 AU and 0.004 AU. if a habitable zone was 1 AU wide, it could have 178.5 planetary orbits separated by 0.0056 AU each, or 250 planetary orbits separated by 0.004 AU each. The ratios between orbits are 1.9333 and 1.327 respectively.


Kepler-70 was reported to have two planets with orbits of 0.0060 and 0.0076 AU. The difference between orbits would be 0.0016 AU, and the ratio would be 1.2666. A habitable zone one AU wide would have room for 625 planetary orbits spaced at 0.0016 AU.

And a third planet was suspected to orbit between them.


If these planets exist, then the orbits of Kepler-70b and Kepler-70c have 7:10 orbital resonance and have the closest approach between planets of any known planetary system. However, later research5 suggested that what had been detected was not in fact the reflection of light from exoplanets, but star pulsation "visible beyond the cut-off frequency of the star." Further research6 indicated that star pulsation modes were indeed the more likely explanation for the signals found in 2011, and that the two exoplanets probably did not exist.


TRAPPIST-1 is a very dim star with 7 planets orbiting very close to it, and thus to each other. Three, e, f, and g, are considered to be in the habitable zone, and up to 6 could be in the optimistic habitable zone.

The orbits of the TRAPPIST-1 planetary system are very flat and compact. All seven of TRAPPIST-1's planets orbit much closer than Mercury orbits the Sun. Except for b, they orbit farther than the Galilean satellites do around Jupiter,[44] but closer than most of the other moons of Jupiter. The distance between the orbits of b and c is only 1.6 times the distance between the Earth and the Moon. The planets should appear prominently in each other's skies, in some cases appearing several times larger than the Moon appears from Earth.[43] A year on the closest planet passes in only 1.5 Earth days, while the seventh planet's year passes in only 18.8 days.[41][37]


The semi-major axes of the orbits of e, f, and g are 0.02925, 0.03849, and 0.04683 Au respectively. The differences are 0.00924 and 0.00834 AU, and the ratios are 1.315 and 1.216. A habitable zone one AU wide could have 108.2 planetary orbits 0.00924 AU apart and 119.9 planetary orbits 0.00834 AU apart.

So the habitable zone around a normal sized star could have room for hundreds of planetary orbits spaced as closely as the smallest known absolute spacing of planetary orbits.

And if there was a lower limit of 0.05 AU in separation of planetary orbits, which is actualy larger than some of the examples I just gave. there would be room for 1,000,000 planetary orbits in a habitable zone 50,000 AU wide, which would be less than one light year.

With a lower limit of 0.005 AU for orbital separation, there would be room for 10,000,000 planetary orbits in a habitable zone 50,000 AU wide, which would be less than one light year.

But if the ratio between orbits is the determining factore instead of the absolute distance, then planets would have to be spaced more widely, going by the known examples.

Part Five: A suggested Change.

I suspect that you consider making your black hole only one billion (1,000,000,000) times as massive as the Sun, and the accretion disc only one million (1,000,000) times as luminous as the Sun. A planet orbiting at a distance of 1,000 AU (1,000 is the square root of 1,000,000) willreceive as much radiation from the accretin disc as Earth gets at a distance of 1 AU from the Sun. The gravity of the black hole at a distance of 1,000 AU will be 1,000,000,000 times that of the Sun divided by 1,000,000 (the square of 1,000), or 1,000 times as strong as the Sun's gravity at a distance of 1 AU. If my calculations are correct.

And I think that will make the forbidden regions of the planets in the habitable zone much smaller in absolute distance and in relative ratios than in our solar system, so many more planetary orbits should be able to fit in the habitable zone.

And of course you could use different masses of the black hole and different luminosities of the accretion disc.

Part Six: Another Suggestion.

You are not the first person to ever try to design a solar system with a great number of habitable planets.

For example, there is a blog by Sean Rayomond about astronomical topics called PlanetPlanet. And it has a section called Ultimate Solar System with posts designing star systems with as many habitable planets as possible.

You may find all of the posts there interesting. I think that posts like






Are especially relevant.

I think that if you can solve various problems, you might be able to design a fairly scientifically plausible system with even more habitable planets than anything Raymond designed.

  • $\begingroup$ 10-15 m in my answer comes exactly from the case of interaction of those x-rays with our atmosphere, assumed 10t/m2 atmosphere equivalent for 10-15 tons of water over shorter distance. So it probably better to take original numbers for earth instead of reverse calculations. I forgot/ignored about atmosphere by being carried away by surprisingly good shielding properties of water. $\endgroup$
    – MolbOrg
    Commented Sep 8, 2021 at 8:46

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