Suppose we drop a small black hole, but large enough that it accumulates mass more quickly than it radiates it, into the Earth. What happens? This question has already been asked here and here, but the answers seem to be wrong or non-authoritative.

Many of the answers assume that the small black hole will eventually consume the Earth, resulting in a black hole with the mass of the Earth. A calculation shows that this is impossible due to conservation of angular momentum. The maximum angular momentum an Earth-mass black hole can have is $7.9 \times 10^{30}$ kg m$^2$/s. This is much less than the angular momentum of the Earth: $7.1×10^{33}$ kg m$^2$/s.

So what really would happen? The conservation of angular momentum shows it can't consume the entire Earth, but it will consume at least a small portion of it. How much would it consume? And what happens to the rest?

  • $\begingroup$ There seem to be two different cases. 1) a black hole at rest at the center of the Earth. 2) Off center. For 2), matter would be thrown at it from the direction of earth's rotation. The black hole would acquire a linear momentum. - Well, not two discrete cases. A continuum. $\endgroup$ – mmesser314 Aug 26 '18 at 15:34
  • $\begingroup$ Assume the black hole is big enough to destroy the Earth on a reasonable time scale (within a lifetime, say). Although knowing how small a black hole has to be to be "safe" is also interesting. $\endgroup$ – Peter Shor Dec 8 '18 at 2:12
  • $\begingroup$ I've deleted my earlier comments, which used a sloppy estimate of the accretion rate. There is a good treatment of this kind of thing in Giddings, arxiv.org/abs/0806.3381 . See eq. 4.31 and appendix A. It looks like you need a black hole with a mass of about $\gtrsim 10^{-4}$ that of the earth, and then the time scale is such that it can eat a significant fraction of the earth's mass within a human lifetime. $\endgroup$ – user4552 Dec 8 '18 at 19:15
  • $\begingroup$ @BenCrowell Thank you for the reference; I'll take a look at it, and incorporate anything relevant into the question. $\endgroup$ – Peter Shor Dec 8 '18 at 19:18
  • $\begingroup$ @Benjamin: I assume that if it accumulates mass faster than it radiates it, the "micro" black hole will keep on getting bigger, and will eventually destroy the earth. $\endgroup$ – Peter Shor Jul 15 '19 at 23:31

What exactly would happen depends on a lot of factors. For instance, you cannot quite assume perfect spherical symmetry of the Earth, and how exactly this symmetry is broken changes the outcome. Also, if the black hole is as massive as the Earth (and thus 9 mm in size), half of the planet's material is suddenly attracted more to the black hole than to Earth, so setting the mass precisely is also important. There are, however, a few general statements that we can give (I am discussing small, but macroscopic black holes):

1) There is nothing that can "hold" the black hole, the only force that can push and pull the black hole around is gravity and gaining momentum by absorbing matter (I assume the black hole is uncharged). If you approach close enough, any Earth material will shred to pieces smaller than the black hole due to tidal forces. The first direction of the black hole will thus be to start drilling downwards unless convinced otherwise by gravitational interaction.

2) The black hole will accrete, but not as much or as fast as you might think. Black holes are not universal vacuum machines, and from a few tens of Schwarzschild radii away (which is the case we are considering here) it can really be treated as a Newtonian point mass. Then the accretion rate will be roughly given by the Bondi-Hoyle-Lyttleton accretion formula $$\dot{M}_{\rm BHL} = \frac{\rho G^2 M^2}{c_{\rm s}^3 + v^3}$$ where $c_{\rm s}, \rho$ is the sound speed and density of the surrounding material respectively, $M$ the black hole mass, and $v$ the speed of the bulk of the material with respect to the black hole. The BHL formula takes into account both the fact that moving matter tends to "miss" the black hole (the $v$ part), but also that there will be a certain "choked up" region in the accretion flow known as the Bondi radius where too much pressure builds up and only a little bit ends up flowing through.

3) The accretion and the gravitational forces end up exerting an effective frictional force that tries to force the black hole to synchronize the motion with the surrounding material as $\dot{p} = \dot{M}_{\rm BHL}v$. Notice that the "frictional acceleration" $\dot{p}/M$ still depends on $M$, so the size/mass of the black hole determines its "frictional coefficient".

Now for the influence of the black hole mass as compared to the mass of the Earth.

If the black hole is sufficiently light (much lighter than the Earth), it will tear a decent hole into the Earth, sink into its core (slowed down by the "accretion friction"), and may live there for a prolonged time accreting at the BHL rate before triggering some sort of instability in Earth's structure.

There should, however, be an intermediate-range of masses such that the tidal forces are large enough to make the Earth break off a significant piece and the orbit of the black hole and the two or more pieces of the Earth correspond to a gravitational three(or more)-body problem. In such a problem, one of the masses can receive sufficient angular momentum to be ejected, the rest of the bodies to be left in a tighter orbit. This is especially true for lighter bodies in the system. So it is in principle possible that the black hole would be pushed away while leaving behind a spun-down shattered Earth. On the other hand, the same type of gravitational interaction can lead to a piece of the Earth to be ejected while making the black hole and the leftover piece in a tighter grip (which almost certainly involves shattering it to even smaller pieces).

At some point, only pieces of comparable or smaller mass are left in the vicinity of the black hole. This may have also been the case from the very beginning since the black hole might be about as massive as the Earth from the very start. The leftover pieces will then gradually be shattered and some of them may be ejected with higher angular momentum by collisions and many-body gravitational interaction. Eventually, the collisions cause the debris to flatten into a disk orbiting the central black hole. Occasional collisions between the pieces of debris will then cause further exchange of angular momentum and a fraction of the matter is continuously travelling outwards, away from the black hole, and other inwards, towards the black hole. Macroscopically, this is seen as spreading of the disk at its outside edge to higher radii, and further accretion onto the black hole at its inner edge, which is located roughly at the innermost stable circular orbit of the BH. In certain scenarios, the collisions within the disk may drop to such low rates that the disk "freezes out" into Saturn-type rings.

I have not talked about the radiative processes a lot. It is quite likely that at many points mentioned above, a considerable amount of heating may occur. This may come about A) purely from collisions, or B) from a triggered nuclear fission/fusion. I believe that scenario B) is unlikely, since we are starting from a black hole above the surface of the Earth, so the situation will be very asymmetric from the very start and there will thus be plenty of channels for the matter to find a way to escape such high compressions.

Either way, if a lot of radiation is released early on in the processes mentioned above, it can either heat up and evaporate many of the mentioned players, or completely blow them away by radiation pressure. Any gas component that is still bound to the system after that will very quickly circle towards the black hole and join an accretion disk around it. If an accretion disk has a large fraction of gas, it is unlikely it will ever "freeze out". Instead, it will keep the accretion process running while transforming the binding energy into heat and radiating it out as photons as we know it from active galactic nuclei or X-ray binaries.

In none of these scnearios, gravitational waves will be of any significance - unlike what has been stated in other answers. There are really quite general arguments that take you to the conclusion that if we are looking at systems say for as much as thousands of their orbital periods, gravitational waves can matter only for interacting systems of compact objects, that is, objects with radii not laughably far from their Schwarzschild radii at given mass. If all of the actors are not compact, they will simply smash into each other or get tidally disrupted before they can conspire to radiate enough gravitational waves to influence their dynamics in any appreciable manner.

| cite | improve this answer | |
  • 4
    $\begingroup$ BHL accretion formula is fine for a star flying through a sparse gas cloud, but applying it to a tiny black hole flying through dense solid material would mean that the material is compressed past nuclear densities near the horizon. Assuming $\rho\simeq5\,\text{g/cm}^3$, $c_s\simeq 1\,\text{km/s}$, the density near the horizon is $\rho_h = A\, \rho\, (c/c_s)^3\sim 10^{16}\,\text{g/cm}^3$ with some dimensionless constant $A$. $\endgroup$ – A.V.S. Jul 17 '19 at 7:21

So, to be brief there are going to be two things resisting the entirety of the matter falling into the black hole. The first kind of leads to the second so I'll begin with the first. Assuming the black hole is at the center of the Earth (these arguments stand up in an orbiting situation as well), as matter near the center of the Earth falls in it will become compressed and begin to generate high temperature that will then lead to high energy photons being emitted. These photons will then emit an ever-growing force on the matter on the outside, and begin to rip the Earth apart preventing more matter from falling in.

This would then lead to the matter being flung away from the black hole, some may escape but a detailed sim. would have to be performed to get a good fractional estimate. The rest of the matter that didn't fall in and wasn't ejected would then form an accretion disc which would begin to lose angular momentum via frictional losses. The matter wouldn't fall in until the matter had a small enough angular momentum to be theoretically allowed to pass the horizon.

A more detailed explanation can be found here https://slate.com/human-interest/2014/02/if-a-coin-sized-black-hole-were-placed-at-earth-s-center-what-would-happen.html, which seems like a credible source as it was written by a Stanford professor.

| cite | improve this answer | |

Consuming the Earth is possible if some mass is ejected, carrying away some of the angular momentum. That's exactly what would happen. Here is a simulation of a black hole consuming the Earth.

| cite | improve this answer | |
  • 3
    $\begingroup$ That video has several pieces of misinformation in it. For example, it says that the lifetime of a black hole the mass of the Earth is only a few days (at 10:05). In fact, it's around $10^{49}$ years. Why should I trust anything else it says? $\endgroup$ – Peter Shor Aug 26 '18 at 16:41
  • $\begingroup$ Good point. The video is eye candy and the simulation is probably far outside its intended range of parameters. However, my point was that angular momentum can be conserved if the interaction between black hole and Earth tosses some of the Earth's mass into orbit, and the video illustrates that. $\endgroup$ – S. McGrew Aug 26 '18 at 16:45
  • 1
    $\begingroup$ Yes, that's a possibility. It can't consume the entire Earth, but it will consume at least a small portion of it. So I'm revising my question to ask explicitly: how much will it consume? $\endgroup$ – Peter Shor Aug 27 '18 at 4:05
  • $\begingroup$ A guess: We can consider how much mass needs to be launched from the Earth's radius, tangent to the surface at an available speed, in order to carry away 99.9% of the Earth's angular momentum. Maybe the maximum available launch velocity is <2x the orbital velocity of the black hole. Classically, as the system shrinks the orbital velocity goes as the inverse of the square root of orbital radius, so angular momentum shed per unit launched mass is proportional to the square root of radius - which means the angular momentum needs to be shed early to maximize mass consumed by the black hole. $\endgroup$ – S. McGrew Aug 27 '18 at 13:08
  • 6
    $\begingroup$ Universe Sandbox is just a game with a crude Newtonian physics engine, that's not a legit simulation $\endgroup$ – Gendergaga Dec 8 '18 at 0:13

A merger process will generate gravitational waves, which can carry angular momentum away from the system. Whether or not this means that the whole Earth is consumed may in part be a question of perspective. The gravitational waves will also carry away mass-energy, so that will cause the final black hole to be a bit less than if the waves weren't created. On the other hand, the matter that constitutes Earth will end up in the black hole.

I see the other answer that claims some matter will be ejected by the process. I find that unlikely, but I admit that I'm not motivated to compute it rigorously. Even if some matter escapes by this mechanism, however, the premise of your question seemed to have missed the waves, which should generally be relevant.

| cite | improve this answer | |
  • 2
    $\begingroup$ "I find that unlikely" -- why? Plenty of matter gets thrown out of quasars. Why wouldn't the same thing happen on a smaller scale in this case? $\endgroup$ – Nathaniel Jul 15 '19 at 21:45
  • $\begingroup$ @Nathaniel I think the geometry of this with the black hole dropping directly at the Earth and the black hole not specified to be spinning and everything electrically neutral makes mass ejection unlikely in this case. I agree that in other geometries and in more realistic astrophysical settings mass can be and is ejected in mergers. $\endgroup$ – Brick Jul 16 '19 at 12:49
  • $\begingroup$ hmm, but the black hole would start spinning even if it wasn't initially, because the matter falling into it has angular momentum. I imagine the falling matter would turn into a superheated plasma pretty quickly too. But I admit I don't really understand why mass gets ejected in quasars in the first place, so I don't really know if it would in this case. $\endgroup$ – Nathaniel Jul 16 '19 at 13:12
  • 1
    $\begingroup$ Yeah, that's why I hedged that part of the answer. Since the bounty got put on after I answered, maybe this will attract attention of someone willing to compute or estimate those details. I was trying to give a short answer that I thought was an improvement over the answer available at the time that I first wrote not write the definitive study now sought by the bounty text. @Nathaniel $\endgroup$ – Brick Jul 16 '19 at 13:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.