As stated in several sources, it's supposed that in every galaxy there is a black hole in the middle.

My question is, why do these black holes in the middle of galaxies not suck up all the surrounding matter in the galaxy?

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    $\begingroup$ Obligatory xkcd. $\endgroup$
    – Raidri
    Commented Dec 20, 2017 at 14:34
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    $\begingroup$ @Raidri There's also a good one about the relationship between mass and gravity in the What-If book. IIRC it discusses what effects a chunk of "neutron star material" would do locally on earth. Long story short: Don't touch it. If you haven't already, I highly suggest picking up a copy (of the book, not the neutron star). $\endgroup$
    – thanby
    Commented Dec 20, 2017 at 15:49
  • $\begingroup$ Please don't use the comment space for answers. $\endgroup$
    – called2voyage
    Commented Dec 21, 2017 at 13:20
  • $\begingroup$ Similar: astronomy.stackexchange.com/questions/21443/… $\endgroup$
    – K Split X
    Commented Dec 21, 2017 at 15:47
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    $\begingroup$ Last I knew, it was not supposed that "in every galaxy there is a black hole in the middle." Notice the emphasis on every. Should edit "every galaxy" to "many galaxies" $\endgroup$
    – Aaron
    Commented Dec 21, 2017 at 22:46

11 Answers 11


You shouldn't think of black holes as "sucking things in". Black holes interact with matter through gravity, just the same as any other object. Think of our Solar System. All the planets orbit around the sun because it has a lot of mass. Since the planets have some lateral motion (they're not moving directly towards or away from the sun), they circle around it. This is known as conservation of angular momentum.

When talking about gravity, all that matters is the mass of the objects involved. It doesn't really matter what kind of object it is*. If you were to replace the sun with a black hole that had the same mass as our sun, the planets would continue on the exact same orbits as before.

Now, the black holes at the centers of most spiral galaxies do accumulate mass. Some of these black holes have accretion disks around them. These are swirling disks of gas and dust that is slowly falling into the black hole. These gas and dust particles lose their angular momentum through interactions with gas and dust nearby and by radiating energy as heat. Some of these black holes have very large accretion disks, and can generate huge amounts of electromagnetic radiation. These are known as active galactic nuclei.

So, long story short, black holes don't "suck". They just interact with things gravitationally. Stars, gas, and other matter in the galaxy has angular momentum, so it stays in orbit around the center of the galaxy. It doesn't just fall straight in. This is the same reason the Earth orbits around the Sun.

*Disclaimer: When you talk about things like tidal forces, you do need to take into account the size of the objects. But for orbital mechanics, we don't need to worry about it because the distances between the objects are generally much larger than the objects themselves.

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    $\begingroup$ To add, if the Sun became a black hole, and if it managed to do so without losing mass, Earth and the rest of the solar system will continue to orbit around it as they always have. You only end up getting "sucked into" the black hole if you get really close to it, at which point the gravitational force is tremendous. But you have to get really close. For illustration, the Earth would need to be compressed to ~1 cm wide to become a black hole. In real life, even if you tunnel to the Earth's core, you won't feel that kind of gravity since the mass is above you (see the shell theorem). $\endgroup$
    – Allure
    Commented Dec 19, 2017 at 22:59
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    $\begingroup$ "Some of these galaxies have very large accretion disks" I think you meant "Some of these black holes..." $\endgroup$
    – jpmc26
    Commented Dec 20, 2017 at 0:00
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    $\begingroup$ Possible pedantic caveat - on a long enough timescale one might expect the various objects orbiting a galactic black hole to eventually fall into it as friction with the interstellar medium gradually depletes their angular momentum. Of course, the density of the interstellar medium is so low that "a long enough timescale" is probably absurdly, ridiculously long (almost certainly longer than the lifespan of any object in the galaxy). $\endgroup$
    – aroth
    Commented Dec 20, 2017 at 2:34
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    $\begingroup$ I have just realized that a galaxy is essentially just a giant accretion disk for the supermassive black hole in the middle... $\endgroup$ Commented Dec 20, 2017 at 13:27
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    $\begingroup$ @JohnDvorak Not the same thing at all. The black hole + accretion disk system is vastly dominated and is driven by the gravity of the black hole. Big as the supermassive black hole may be, it's just a speck of dust at the scale of the whole galaxy. The galaxy is a very different system from an accretion disk - it's held together by its own total gravity, as opposed to merely orbiting a black hole. $\endgroup$ Commented Dec 21, 2017 at 0:09

I once heard of a Japanese cartoon/movie/show where space pirates threatened to compress the planet Jupiter into a black hole and thus destroy half the Milky Way galaxy.

It sounds like an interesting idea, but... even if you could compress Jupiter into a black hole, its mass would remain the same, meaning that Jupiter (now a black hole) would still continue to move around our sun in its same orbit, and Jupiter's moons would still continue to orbit Jupiter as they did before.

Many people think that once a star collapses into a black hole, its "sucking power" (its gravitational force) increases. This is simply not the case. Believe it or not, many stars are less massive after they turn into a black hole than before, when they were shining stars. This is because, at the end of their lives, some stars shed a significant portion of their outer layer into space right before they collapse into a black hole.

I've read that if you compressed the Earth to the size of a cherry, its density would be so great that it would turn into a black hole. Assuming that were true and it was actually done, the black hole of Earth would still continue orbiting the sun once every year, and Earth's moon would continue to orbit the Earth about once every 29.5 days. (Now, the spin of the new black-hole-Earth about its axis would probably be different, but the time it would take to orbit the sun would not change.)

Surprisingly, once the Earth got compressed into a cherry-sized black hole, less space debris would fall into it than before (when the Earth was the size of... well, Earth). This is because the newly-formed black-hole-Earth would take up much less space (volume) and asteroids and comets would be more likely to miss the cherry-sized (or slightly-larger-than-cherry-sized) volume that, if not missed, would cause the debris to be sucked into the black hole.

If the debris missed the black-hole-Earth by even a kilometer (which might seem like a large distance to us, but is very miniscule in astronomical terms), it would be flung off in a different direction, possibly never to return.

So, basically, a common misconception people have regarding black holes is that nothing has more gravity than a black hole, and that stars that form into black holes all of a sudden have increased gravity and therefore get more "sucking power." This is simply not true. Black holes still have the same mass as before (sometimes less, depending on how they're formed), and how much "sucking power" they have is still dependent on how much mass they're made up of.

While it may be true that the most massive stars in the universe are indeed black holes (if you'd even call them stars at that point), there exist many stars that are more massive (and therefore have more "sucking power") than many black holes.

So the fact that our galaxy's center probably contains a super-massive black hole doesn't mean that the black hole would suck up any more matter than if it were the same amount of mass which happened to not be in black-hole-form.

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    $\begingroup$ Another inaccuracy about space pirates destroying half the Milky Way with a black hole is the fact that any changes in Jupiter's gravitational field would not propagate faster than light and thus would take tens of thousands of years for all this destruction to be finished. $\endgroup$
    – Michael
    Commented Dec 22, 2017 at 22:41
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    $\begingroup$ Also, isn’t destroying the galaxy sort of counter to the usual aims of piracy? $\endgroup$
    – rickster
    Commented Dec 23, 2017 at 21:38
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    $\begingroup$ Nice and clear reminder of a common misconception. Thanks for pointing out that a black hole is no less no more than a gravitational object. $\endgroup$
    – Benj
    Commented Dec 24, 2017 at 0:18

Gravity follows the law of inverse square. To put it simply if you double the distance from a gravity source you quarter is effect. So if you double the distance you are from the earth you feel 1/4g. It's important to note that as the distance increases it will never be 0, it will always be some non-zero value no matter the distance.

So at galactic distances the force of gravity the central black hole has very little effect.

This only explains part of it. The other part is conservation of angular momentum.

The force of gravity and the angular momentum is what is responsible for orbits. In orbital mechanics you raise your orbit by adding speed, not altitude. Your adding angular momentum which raises your orbit. To lower your orbit you reduce you speed which reduces your angular momentum and your altitude.

So for things to "fall" into a black hole they must be travailing at a speed where their orbit intersects the event horizon. This is rarely the case or those "things" wouldn't truly be in orbit to begin with. So the very fact that all the "stuff" that makes up the galaxy orbits the central black hole means that it cant just fall into it.

These 3 things are always in balance in a stable orbit, force of gravity, speed and altitude ( or distance from the gravity source ). If you change one of them the other 2 must also changes. If you decrease speed your altitude goes down, and the gravity increases. If you increase the gravity the speed must also increase or the altitude will decrease.

So you see things can't just fall into the black hole. That said it is my view that eventually everything in the galaxy will fall into the central black hole, however this will take many billions of years.

Of course this is overly simplifying things, and I am by no means an expert on this stuff. But it's something I can picture in my mind, the balance between momentum and gravity.



You must also factor in the dark matter which is interacting gravitationally with all of the "hot matter" that can be seen in the galactic disk. Dark matter was discovered by carefully mapping the orbits of objects in galaxies and finding that the matter that could be seen could not account for the observed orbital motion. One of the mysteries of dark matter is that it is not pulled into the black hole the way hot matter is. The dark matter has the practical effect of balancing out some of the gravitational pull of the supermassive black hole at the galaxy's center.

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    $\begingroup$ It depends on how the matter is distributed. Dark matter is generally thought to occur in a sphere around the galaxy. For a spherical distribution of matter, you only need to concern yourself with the portion interior to your orbit; what's outside won't affect you. So dark matter will still have the same result of pulling you inwards. $\endgroup$
    – Phiteros
    Commented Dec 21, 2017 at 7:31

Well, I'm not a physics student, but I think people usually nurture the misconception of "sucking power" of a black hole for a reason.

Let us consider Newton's equation for gravity:

$F = {Gm_im_j\over r_{ij}^2}$ for two bodies i and j, and $r_{ij}$ is the distance between the centre of mass of two bodies.

Now, if the Sun today suddenly decided to become a black hole without shedding weight, it won't affect Earth's orbit, because even if Sun's volume has changed, $r_{ij}$ remains constant.

The reason why Black Holes "suck" is because, since they occupy extremely less volume compared to planets and stars, you can get the $r_{ij}$ component to be really really small.

Correct me if I'm wrong.


For galaxies with large black holes, the surrounding matter is in orbit around the black hole(s), the same way that the moon orbits the earth.

The question is a direct analogy to "Why doesn't the Moon fall to the ground?" or "Why don't the planets fall into the sun?". The black hole is more massive than the Sun, but its effects are of the same type.

  • $\begingroup$ That is not a typical situation at all. In the vast majority of cases the black hole is a tiny fraction of the total mass of the galaxy. There are some outliers, sure, but they are the exceptions. $\endgroup$ Commented Dec 21, 2017 at 19:30
  • $\begingroup$ Perhaps it's fairer to say that the black holes and other matter are all in orbits with each other. Anyway, the important point is that there's a good analogy with the question most of us asked as children. $\endgroup$ Commented Dec 22, 2017 at 8:56
  • $\begingroup$ I'm sorry, but it's not a good analogy. What you're describing is literally a handful of cases out of countless millions. $\endgroup$ Commented Dec 22, 2017 at 19:10

One quick answer for your question would be event horizon or Schwarzschild radius. Anything which is fairly close to this radius/horizon will eventually be sucked up by the black hole.


The simple answer is that everything else in the galaxy is going sideways fast enough to escape being sucked in. Instead, the force of the suck (if you like) causes the stars' paths to be pulled into a circle around the black hole.

This phenomenon is "orbit". As other answers pointed out, it's the same reason the Earth does not fall into the Sun, or the Moon fall onto the Earth, and why the International Space Station is hurtling along at around 17,150 miles per hour. They're all going sideways, the force of some large object is turning that sideways motion into circular motion, and if they weren't going fast enough then they'd curve ("fall") towards that large object and crash into it.

It's like if you whirl a bucket on the end of a string. The bucket is going sideways, but the string is pulling it towards you. The bucket doesn't fly off away from you because of the force from the string, and so it curves in a circle. The force from the string happens to not be enough to collapse the bucket inwards and hit you.


This is a common misconception about black holes: that they somehow 'suck up' everything around them or pull things into them. In reality you could replace the Sun now with a black hole of the same mass and not notice any immediate difference. It's not like it would suddenly start hoovering in the planets around it, that's just not how it works.

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    $\begingroup$ Please do not repeat (only) what is already in other answers $\endgroup$
    – user1569
    Commented Dec 22, 2017 at 14:41

Be patient, eventually it will unless the galaxy's rate of expansion exceeds the black hole's gravitational growth as it consumes the matter around it.

In that scenario the galaxy will eventually diffuse, with its matter continuing to travel away from the black hole until it encounters another galaxy, at which point it stands a good chance of eventually getting sucked into that's galaxy's black hole. Nothing survives forever.. :-)

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    $\begingroup$ What makes you think that the galaxy is expanding? Sure, the universe is expanding, but that's only happening on the very large scale. Individual galaxies and even galaxy clusters aren't expanding because their gravity is stronger than the expansion at that scale. $\endgroup$
    – PM 2Ring
    Commented Jan 9, 2018 at 0:18

its all about ENTROPY which is proportional to the surface area of the event horizon of a black hole(see below for a heuristic quantum argument due to Moffat/Wang as to why this is so).

Assuming a Schwarzschild solution gives a radius of 2Gm for the event horizon with m the black hole mass and G Newton's constant. Adding mass to a black hole thus increases its entropy. Given an isolated system of finite total energy, it has a finite maximum entropy which acts as an attractor for the dynamics of the system,placing a limit on the horizon.

J von Neumann defines a quantum version of entropy as follows: Let f be a normal state of a local algebra of observables O(D) acting on the Hilbert space H. Then we can write this f as a convex sum of pure states. For a system of finite energy this sum is finite since H is then finite dimensional Von Neumann’s non-commutative equivalent of a partition is the density operator ie the weighted sum of the projections onto the minimal vector spaces corresponding to these pure states Then we have the well-known equivalence;
For such a normal state f, the von Neumann entropy is defined as the entropy of the weights . We interpret it as an (inverse) measure of the amount of information that the quantum system in a given state will yield through measurement. The larger the entropy of the quantum system, the less information can be extracted. The von Neumann Entropy of a Black Hole The measurement process cannot be performed by an external observer to elements within the interior, beyond the event horizon. We thus partition the event horizon of the black hole with elements each of area k squared, where k is the Planck length and assume the Planck area corresponds classically to the minimal projection of the pure vector state . Let N be the total finite number of partitions. By the ‘no hair’ hypothesis there is no preferred location on the event horizon, so that each partition element must have the same weighting. The von Neumann entropy of this partition is thus proportional to S the surface area of the black hole.

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    $\begingroup$ While black hole entropy and the Beckenstein bound is an important and fascinating topic, I fail to see how it's particularly relevant to the OP's question. $\endgroup$
    – PM 2Ring
    Commented Dec 21, 2017 at 19:17
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    $\begingroup$ This is a copypasta of researchgate.net/profile/Charles_Wang28/publication/… with no relevance to the OP's question. $\endgroup$
    – MBR
    Commented Dec 22, 2017 at 9:35
  • $\begingroup$ I improved my answer to take account of the kind comments $\endgroup$ Commented Dec 22, 2017 at 14:08

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