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In many Science Fiction books and games, there are worlds or planet-like objects that are rings that orbit (or float around in space) (some examples being the Halo (from Halo) or the Ringworld (from the book called Ringworld by Larry Niven).

After reading through a lot of the Ringworld book, it got me wondering. What would the size of a ring have to be to orbit a black hole, staying structurally stable. The size of the black hole would be stable (not growing) and being the size of an intermediate-mass black hole.

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  • $\begingroup$ Depends. Do you only have highly resilient machines on this ring, or also squishy flesh-humans? $\endgroup$ – Serban Tanasa Jan 14 '15 at 18:37
  • $\begingroup$ Both, but the humans would be the important part. $\endgroup$ – Shadow Z. Jan 14 '15 at 18:39
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    $\begingroup$ As a side-note: Halo doesn't orbit its host planet in the same way that Ring World does. $\endgroup$ – Thebluefish Jan 14 '15 at 19:34
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    $\begingroup$ What do you mean by structurally stable, and how is your ring world configured? Is it spinning? If you spin your ring fast enough, the force of gravity will equal the centripetal force and it won't experience any effects from gravity if it's sufficiently thin. $\endgroup$ – ckersch Jan 14 '15 at 20:13
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    $\begingroup$ Niven's Ringworld would already be structurally unstable if it weren't made out of unobtanium. $\endgroup$ – Philipp Jan 14 '15 at 21:15
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The answer is simple, although a bit of a cop-out: Exactly as big as if it were around a star.

It's a common misconception that black holes have stronger gravity than anything else. But (ignoring general relativity) gravity is always $$F_{grav} = G\frac{m_1m_2}{r^2}$$ where $G$ is a constant, $m_1$ and $m_2$ are the masses of the objects, and $r$ is the distance between their centers.

What this means is that it doesn't matter if the object is a black hole or a regular star with the same mass - the gravity you feel from it at $r$ away will be exactly the same. What makes black holes special is that their mass is compacted into an area small enough that there is a zone between the actual surface (assuming there is one) and the place where $F_{grav}$ is such that the $\text{escape velocity} = c$. (This is the space where light can't move fast enough to escape, which is why the hole is "black".) The edge is called the event horizon. Anything further away than that can (at least theoretically) escape, and behaves perfectly normally.


As for the question of how big it needs to be to circle a star, that depends on how much tidal stress it can support. Going back to the gravity equation, tidal forces can be described as the difference in $F_{grav}$ based on the difference in $r$ from the nearest edge to the furthest edge. So the closer edge is "pulled" harder than the further one. (Objects in orbit are moving fast enough to be at the balance point between being flung away and "pulled in")

There's a value called the Roche limit which determines how close something big enough to hold together under its own gravity can get before the tidal forces will rip it apart. Separately, there's tensile strength for how much force a given material can support before being pulled apart. You'd need some combination of these to figure out how close the ring itself can get.

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    $\begingroup$ For some reason I imagined that sentence ending differently: "What this means is that it doesn't matter if the object is a black hole or a regular star or <an inane object like a rabbit>". I was sort of startled when it didn't. $\endgroup$ – Mooing Duck Jan 15 '15 at 0:57
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    $\begingroup$ F_grav cannot equal c. F is in units of Newtons, c is in units of meters per second. $\endgroup$ – Almo Jan 15 '15 at 3:45
  • $\begingroup$ It should be: the escape velocity equals c. $\endgroup$ – Sanchises Jan 15 '15 at 11:48
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    $\begingroup$ @MooingDuck or a mole of moles $\endgroup$ – Jon Hanna Jan 15 '15 at 18:09
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    $\begingroup$ Your new explanation is not only correct, but I think it is easier to understand. :) $\endgroup$ – Almo Jan 15 '15 at 18:40
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It would have to be big enough to safely encircle the star that collapsed to create the black hole.

When matter collapses to create a black hole, it does not extend its gravity well. It has the same mass as before. It is just concentrated in a much smaller area. So a ring world encircling a star that collapses will continue to work and orbit just like it did before, unless of course the actual collapse destroys it. Of course getting closer to the black hole is still dangerous. And you don't want to cross the event horizon.

EDIT: At the surface of the sun, gravity is about 28 times that of earth. If the sun collapsed into a black hole, that same distance from the center of the hole will still be 28 times earth. but the new 'surface' will be a long way away.

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    $\begingroup$ So, (Theory) if the Sun did not cause damage to the Ring World when it was destroyed, and went straight to a black hole, it would sill be able to orbit it perfectly fine like nothing happened at all? $\endgroup$ – Shadow Z. Jan 14 '15 at 18:40
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    $\begingroup$ @ShadowZ. pretty much, minus the solar power available. $\endgroup$ – bowlturner Jan 14 '15 at 18:42
  • $\begingroup$ Lastly, how large would the event horizon be ( to tie this up in size). Would it be small enough so there would not have to be a modifier to increase the size more to avoid it? $\endgroup$ – Shadow Z. Jan 14 '15 at 18:45
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    $\begingroup$ Don't stars expand as they die, collapsing only after they've consumed most of their inner planets? Sticking with Niven's model including the indestructible metal underneath the soil, if you found a ring-world around a black hole, I think it would just be a silvery wedding ring, scourged of atmosphere, buildings, soil, and everything else which had been contained inside. $\endgroup$ – Henry Taylor Jan 14 '15 at 19:07
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    $\begingroup$ Stellar collapse into a black hole is always accompanied by a supernova. Specifically, a core collapse supernova. A ring world would definitely not survive. Also, the resultant black hole will be significantly smaller than the star that produced it, since most of the stellar mass gets ejected during the supernova. $\endgroup$ – ckersch Jan 14 '15 at 22:34
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If you're in orbit, you'll be in microgravity (aka free-fall) anyway, so you won't feel the acceleration. With a black hole, once you get pretty close, you get insane tidal effects that would rip you apart, but you'd have to be well within the surface of the original star, probably within a few thousand kilometers or so of the center for a few-solar-masses black hole for that to happen.


So you can pick any distance from the black hole where tidal effects are not overwhelming. If I assume arbitrarily for a second that you care to find out where the gravitational acceleration is the same as that of the Earth surface, i.e survivable easily (no spaghettification), and relatively easy escape velocity.

For a black hole (or whatever spherical object) the mass of the sun, $g=9.8m/s^2=g_{earth}$ at $3.68×10^9$ m 3.7 billion meters from the central singularity, well inside the orbit of Mercury in our solar system (58 billion m from sun). At this distance, tidal forces should be essentially nil $<10^{-8}N$ for a human-sized object.

The general formula for a small mass object being acted upon by a large mass M is: So multiply as desired given your chosen black hole mass. $g=G(M/r²)$

That would make your structure is $2\pi r$ that length, so about 23 billion meters around. That's so large that it would take a human about 1500 years to walk the length of it, if they walked 8 hours a day.


More interestingly, what's the minimum distance you could build and remain un-spaghettified? At about a solar radius, the tidal forces for a 2m tall, 70kg human would be about $10^{-5}$N, which I think is quite tolerable. The minimum is likely even closer than that, but I can't find my old Misner & Wheeler Gravity book to look up lethal tidal forces.

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  • $\begingroup$ You only care about the black hole's gravity if the ring is not orbiting and the black hole provides the gravity for the inhabitants. A normal Ringworld design uses spin to provide the equivalent of gravity to the occupants. $\endgroup$ – Loren Pechtel Jan 14 '15 at 19:13
  • $\begingroup$ Yeah, that's what I said. I picked that distance arbitrarily, just to make sure there are no tidal effects, and a relatively easy escape velocity. $\endgroup$ – Serban Tanasa Jan 14 '15 at 19:15
  • $\begingroup$ @LorenPechtel That's not quite too. You get tidal effects due to the difference in gravity at your head and at your feet even while in free fall. The question becomes how close to the black hole you can get before those forces become a big deal. $\endgroup$ – Tim B Jan 15 '15 at 9:34
  • $\begingroup$ @TimB Ok, I worded it poorly. You don't care about it's gravity, just about it's tides. I very much doubt you want to be close enough in to worry about the tides, though--the accretion disk will be pretty nasty! $\endgroup$ – Loren Pechtel Jan 15 '15 at 20:30
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This question is more specific than other answers have given it credit for. Consider:

What would the size of a ring have to be to orbit a black hole, staying structuraly stable. The size of the black hole would be stable (not growing) and being the size of an intermediate-mass black hole.

This puts a definite range on the mass values of the black hole. This range starts well above the mass of our own sun. Stars over 3 times our sun's mass can turn into a black hole, but smaller than that, we haven't conclusively discovered any natural process than can create such a black hole.

Furthermore, if you are being actually literal with the definition of "intermediate", that means we start the scale at 100 solar masses.

The question also clearly specifies that this is a Niven-style construction. As such, the walls need to be high enough to hold in the atmosphere. Earth sea-level parameters give a characteristic height of about 8 km, and pressure falls off exponentially, so the radial dimension should be on the order of 50 to 100 km.

The event horizon of a 100 solar mass black hole would be at about 295 km radius. The photon sphere is the closest that you could orbit. It is an unstable orbit, but we can allow for active stabilization and control systems and easily wave this off. After all, the same was claimed from the original Ringworld concept!

At the photon sphere of our smallest intermediate-mass black hole (100 solar masses) will have tremendous tidal forces, and a 50 km structure is unworkable. Because of this, we would have no choice but to locate it at a more distant radius. But what accelerations can the wall tolerate? I'll say 1 g as a Fermi estimation to set the magnitude. Applying Newtonian tidal estimation:

$$ \Delta h \frac{ G 100 M_s }{ r^3 } = 1 g \\ r = 407,000 km $$

This is fairly nearly the distance between the Earth and the moon. This is the minimum radius for a ring which can withstand the tidal forces of an intermediate mass black hole.


That answered the question, but as the saying goes, there is a fly in the ointment. In doing this calculation I actually undermined the construction principle of the Niven-type Ringworld. In that design, we have unobtanium to hold against 1g of acceleration over a large radius. In the scenario I analyzed, the tidal forces alone are enough to produce Earth gravity over the scale of the walls. And this is true if you're in freefall orbit in the first place (actually, that changes the calc by a factor of 2 I think).

The logical thing to do would be to simply trash the unobtanium in the first place. You could get gravity by the tidal forces, so if this was a bike tire tube, you could live on the innermost circle of the tube or the outermost circle of the tube. You could even transit between the two, passing through zero gravity. Or, if you didn't want a full enclosure, you could expand the dimension to >100 km and have the space between the two surfaces unpressurized. This is fairly workable with conventional materials. Although, for this design you might need smaller tidal forces and, thus, a larger radius.

You don't even need a ring at all. The principle would work just fine for two space habitats held together by a tether.

But then you have other problems, like the lack of a sun to provide energy. I don't see any easy answer to that. There are some ways to get energy out of a black hole, but I these tend to focus on electricity production, and light production would necessarily be artificial.

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  • $\begingroup$ The power was not really an option. Also, side question, could a Ring World of that size survive the creation of the black hole from the star? $\endgroup$ – Shadow Z. Jan 15 '15 at 3:49
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    $\begingroup$ @ShadowZ. The radius of our sun is larger than the radius of the ring I calculated, so most simple answer is "no". But by definition, intermediate-mass black holes are not just formed by stellar collapse. They are formed by mergers of stellar-mass black holes. Could the ring survive the mergers? I can't back this up, but I suspect the answer is "heck no". $\endgroup$ – AlanSE Jan 15 '15 at 3:56
  • $\begingroup$ Could it be made out of a material to survive the collapsing of the star? I want to turn this idea into a Ring World around a star that collapsed so life returned to it as it was around the black hole. I'm fine as long as the structure stays, and it can eventually be remade to working potential. Any fictional or non-fictional material would work. $\endgroup$ – Shadow Z. Jan 15 '15 at 4:01
  • $\begingroup$ Would the one gee calculation mean your head sees one g extra acceleration than your legs? If so, is that survivable? $\endgroup$ – Serban Tanasa Jan 15 '15 at 11:59
  • $\begingroup$ @SerbanTanasa It transitions from microgravity to 1g over 50 km, not the height of a human. It would feel like roughly constant gravity to the people there. $\endgroup$ – AlanSE Jan 15 '15 at 12:03

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