# Would a micro black hole keep oscillating around the center of Earth forever?

If a micro black hole was dropped from the surface of the Earth, and that it was too small to absorb particles and eat the Earth, it would fall until emerging from the other side, then fall again and so on...

One would think that it may however collide with particles on the way, which would slow it down and make it ultimately settle in the center of the Earth.

How large would be such an effect, if existing at all? How long would a black hole take before settling depending on its mass?

Edit: “Micro” here means only means that it is too small to feed on Earth's particles, so it could still be very massive at human scales. A $$10^{12}\,$$kg black hole would still be smaller than a proton, after all.

Also, we are ignoring how such a black hole could have appeared at rest at the surface of the Earth in the first place.

• I don't know the specifics, but if a black hole were too small to suck in matter around it, it should also be small enough to evaporate very quickly. Things might get very messy in that scenario. – JMac Jul 15 at 16:55
• More on BH+Earth. – Qmechanic Jul 15 at 16:56
• @JMac Depends on what you cal quickly. It could still take thousands of years or more before evaporating, which is long enough for many oscillations. Even a 10^12 kg black hole would probably be too small to absorb much, and it would evaporate in trillions of years. That could be a pretty funny practical joke for aliens to do to an unsuspecting Earth, though: drop a small one and watch the hilarity each time it emerges, emitting slightly more Hawking radiation... – Eth Jul 15 at 17:02
• @Eth I guess that depends on how you are choosing to define "micro black hole" then. I was thinking more along the lines of what some people panicked about the LHC potentially creating. – JMac Jul 15 at 17:11
• What do you mean "too small to … eat the Earth"? Even a small one would "eat" the Earth unless I'm missing your point, and barring the evaporation issue mentioned already. – Brick Jul 15 at 17:16

We will use the value of $$10^{12}\,\text{kg}$$, proposed in OP, for black hole initial mass for estimates.

This value is large enough that the evaporation time through Hawking radiation for such a black hole is longer than the current age on Universe, moreover, the accretion rates if such a black hole is placed inside the Earth would be larger than the loss of mass through Hawking radiation by at least an order of magnitude, so such a black hole would be gradually consuming Earth, with its mass growing exponentially.

Characteristic time for such growth through accretion could be estimated through the Eddington luminosity limit: $$\tau_\text{E} = \frac{\eta}{4\pi} \, \frac{\sigma_\text{T} c}{ G m_p} \simeq 2.6\times10^7 \,\text{yr}.$$ Here $$\sigma_\text{T}$$ is Thompson scattering cross-section, $$m_p$$ is the mass of a proton and $$\eta$$ is the efficiency of conversion of accreting mass into radiation, which we assume to be about $$5\,\%$$.

Of course, mass absorption would slow our black hole down, but the main mechanism which would be responsible for the dampening of black hole oscillations is the dynamical friction (a.k.a. gravitational drag), because our black hole could lose its velocity not only through direct absorption of mass/energy but also through long-range gravitational interaction with the surrounding medium.

Characteristic decay timescale of such process is shorter (for oscillations which would reach the surface of the Earth) and could be estimated as $$\tau_\text{d}=\frac{v^3}{9\pi G^2 M \rho \ln \Lambda} \simeq 2.5\times 10^6 \times\left(\frac{M}{10^{12}\,\text{kg}} \right)^{-1}\left(\frac{v}{8 \, \text{km/s}}\right)^3 \, \text{yr} ,$$ where $$M$$ is the black hole mass, $$v$$ is its average velocity (for which we assume value of $$8\,\text{km/s}$$), $$\rho$$ is an average density of the material through which the black hole is moving ($$\rho\simeq 5 \, \text{g/cm}^3$$), and $$\ln \Lambda$$ is the gravitational Coulomb logarithm. We see that the dampening effect from the dynamical friction is proportional to the black hole mass (decay timescales are shorter for larger black hole), and is inversely proportional to the cube of black hole velocity, so once the black hole lost most of its velocity and is oscillating around the core, the dampening is greatly increased and the decay timescales would be much shorter.

Within a several million years our black hole would be oscillating with much smaller amplitude deeply in the Earth's core where it will be growing doubling in mass every dozen million years. And by the time the black hole accretion rates become large enough for the accompanying radiation to heat up the Earth surface to make it uninhabitable, several dozen million years would pass.

• Are you sure the Eddington luminosity formula is relevant here? We are talking about a tiny black hole moving through a solid (the Earth). – Andrew Steane Jul 15 at 21:47
• It should be considered only as an order of magnitude estimate. While the processes are no doubt complicated, they are largely not nuclear (no hadron jets, etc.) and fundamentally Eddington luminosity is about balance between two charged species, with very different charge to mass ratios. – A.V.S. Jul 16 at 7:09
• The question is specifically about black holes too small to grow faster than they evaporate. I thought 1 Gt would be small enough due to its Hawking radiation repelling close particles and slowing its accretion down enough to make it lower than its evaporation, hence why it was given as an example. If it is not the case, I will have to edit the question with the actual lower limit. How low would the mass have to be in this case? – Eth Jul 16 at 8:45
• Also, the question was about the dampening effect depending on the mass. How much would this change with a lighter black hole? This answer is already very interesting as it is, that said. Last thing, in the last equation: is M the mass of the black hole or the Earth + black hole? – Eth Jul 16 at 8:47
• I've added the dependence on mass and velocity for the dynamical friction timescale equation. – A.V.S. Jul 16 at 13:53

A $$10^{12}$$ kg Black Hole would be putting out a constant 356 megawatts of Hawking radiation in all directions and it would have an Event Horizon of only 1.4 femtometers (smaller than an atomic nucleus) No particles would even manage to get close to this thing, let alone absorbed by it.

However, it's not going to evaporate any time soon either. By the equation: $$t = 5120 M^3 \pi G^2 / (\hbar c ^4)$$ we can see it's life time will be $$2.6 \times 10^{12}$$ years, longer than the history of the Earth and quite a bit longer than it's expected future.

Now this bubble of hawking radiation will produce interesting effects as it vaporizes a pocket of matter inside the earth as it passes back and forth. Some of the light emitted by the black hole will be reflected back into it and in an asymmetric way. The matter behind the black hole will have had more time to be heated by the megawatts of radiation than the matter in front of it. This means that it will emit more radiation back into the black hole and thus increase it's speed through the momentum of the photons (similar to a light sail.) It won't be a large effect considering the small target of 1 fm$$^2$$ squared, but it will add up over time giving the black hole more and more momentum with each pass until it finally reaches escape velocity and leaves the Earth.

• You need to roughly calculate this effect in order to justify the claim that it will be sufficient to propel the black hole to escape velocity. – Andrew Steane Jul 15 at 22:18
• I'm not an expert on Hawking radiation, but it's not clear to me that a black hole surrounded by matter, in this case the Earth, radiates at the same rate as one in vacuum. There's some argument that goes into it about the temperature at infinity, which makes sense in vacuum and maybe makes less sense for this situation? How legitimate is that estimate in matter? – Brick Jul 16 at 13:23