The Chelyabinsk meteor is estimated to have had a mass of 10,000 - 13,000 metric tons. A black hole of mass $13\times10^6\ \mathrm{kg}$ has a radius of

$$ r = \frac{2GM}{c^2} = \frac{(2 \times 6.674\times10^{-11}\ \mathrm{m^3\ kg^{-1}\ s^{-2}}) \times (13\times10^6\ \mathrm{kg})}{9\times10^{16}\ \mathrm{m^2\ s^{-2}}} \approx 1.93 \times 10^{-20}\ \mathrm m \approx \frac{1}{22000} r_\text{proton}$$

According to this Hawking Radiation Calculator a black hole of this mass has a lifetime of about 185000 seconds, a little over 2 days.

Assume for a moment that somehow such a black hole could exist and collide with the earth, am I correct in assuming that, since its interaction cross-section is so small, it would sail through the earth without much happening?

Looking it at from another point of view, calculating the gravitational attraction to the black hole at short distances, I find that at $1\ \mathrm m$ the force is negligible ($.0009\ \mathrm N$). However the inverse-square law applies, so at $1\ \mathrm{mm}$ the force is $867\ \mathrm N$, meaning maybe the "cross-section" isn't so small after all.

The Hawking Radiation Calculator also gives a luminosity of $\rlap{\raise{0.5ex}{\rule{17ex}{1px}}}\approx 3.56 \times 10^8\ \mathrm W$. At a distance of $\rlap{\raise{0.5ex}{\rule{5ex}{1px}}}1\ \mathrm{km}$ the intensity would be only $\rlap{\raise{0.5ex}{\rule{10ex}{1px}}}28\ \mathrm{W/m^2}$. You probably wouldn't want to get too close to it. (see below)

So what happens:

  1. Not much, sails right through
  2. Lots of fireworks but no lasting damage
  3. Immediate global cataclysm
  4. (1) or (2) initially but the black hole settles at the center of the earth and eventually consumes the planet
  5. (4) but Hawking Radiation prevents net matter inflow and the black hole eventually evaporates in a burst of energy... but then, how much energy?

Correction: I did something wrong the first time in the Hawking Radiation Calculator... the actual luminosity would be $\approx 2.1 \times 10^{18}\ \mathrm W$, greater by 10 orders of magnitude. At $1000\ \mathrm{km}$ the flux would be about $167\ \mathrm{kW/m^2}$. So basically a significant fraction of the Earth's surface under the black hole's path would be sterilized, and as it approached the surface it would induce fusion. Not a pretty sight, and we're getting closer to Option 3 for a lot of people.

  • $\begingroup$ Gravity is a very weak force. The electrostatic force is much, much stronger and even a tiny static charge on your black hole asteroid would make that EM field much more significant than the gravitational one. $\endgroup$ Commented Jun 21, 2018 at 22:26
  • $\begingroup$ The gravitational force on what mass at 1m? I tried to calculate the other mass from 13k tons and got about 1.1kg, which is an unexpected number. $\endgroup$
    – BenRW
    Commented Jun 21, 2018 at 22:28
  • $\begingroup$ Related: physics.stackexchange.com/q/2743/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Jun 22, 2018 at 6:58
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    $\begingroup$ @StephenG an interesting point. Would there be any EM interaction around a black hole? No force carriers could leave the black hole to 'tell' other charged particles about its charge. Definitely an area of ignorance on my part, just thinking aloud. $\endgroup$ Commented Jun 22, 2018 at 12:00
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    $\begingroup$ Why is this on SE instead of XKCD What If? $\endgroup$
    – jpmc26
    Commented Jun 22, 2018 at 13:58

4 Answers 4


Update: The values of luminosity of the blackhole and the composition of the Hawking radiation emitted by a hot blackhole claimed in this answer seem to be inaccurate. Check @A.V.S's answer for a more accurate description.

For a blackhole of a mass of a few thousand metric tons, the Hawking blackbody radiation would correspond to an astronomically high temperature of about $10^{16}$ K ($ \because T = \frac{1.227 \times 10^{23}}{M}$ ). The radiation from such a hot blackhole would mostly be in high energy gamma rays with each photon carrying TeVs of energy. Note that the temperature claimed above is orders of magnitude higher than that is required to start nuclear fusion.

As most meteors have velocities well above the escape velocity of earth, the blackhole might just pass through the earth in a hyperbolic orbit around the core of the earth ( I assume that there is no other mechanism, electromagnetical or otherwise that would cause the blackhole to lose energy but I could be wrong). $2 \times 10^{18} W$ of energy is enormous, infact an order of magnitude more than the amount of solar radiation received by the earth and all that energy is in high energy gamma rays. This could be the end of all life forms on earth and would permanently deface earth.

Update: Not just a fraction of the earth and or a 'lot of people' as you say in your question's update, I believe that the power output would be enough to sterilize the entire planet or atleast the macroscopic life forms. Consider the fact that all the nuclear bombs ever tested and deployed on Earth till now together total about $10^{18} J$. So that would mean the power output of the blackhole would be equivalent to blowing up all those bombs every second. Moreover the temperature of the explosion and the energy of photons produced would be orders of magnitude higher than that produced by any normal fission or a fusion nuclear bomb. The runaway reactions that would follow such an event would have direct and indirect consequences on life on Earth. Also consider that fact that the power output and the temperature of the black hole increase more and more as the black hole evaporates and reduces it mass.

  • $\begingroup$ I recalculated your numbers and the lumosity for the mass of the blackhole you are taking about seems to be close to $2 \times 10^{18}$ W. Such a high amount of gamma radiation would definitely be classified as global cataclysm. It could be might as well be the end of life on earth. $\endgroup$ Commented Jun 22, 2018 at 3:58
  • $\begingroup$ I went back and redid the calculation... you are correct and somewhere I did something wrong. I updated my question. $\endgroup$ Commented Jun 22, 2018 at 4:38
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    $\begingroup$ It is funny that the main threat does not come from the black hole sucking the earth in, as one might expect naively, but from losing mass rapidly via Hawking radiation. From this point of view, a black hole that is much larger is far less dangerous. $\endgroup$
    – Thern
    Commented Jun 22, 2018 at 14:09
  • $\begingroup$ Would the black hole take on additional mass as it passes through the Earth and would that be enough to extend its lifetime in any significant way? $\endgroup$ Commented Jun 22, 2018 at 14:50
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    $\begingroup$ The radiation from such a hot blackhole would mostly be in high energy gamma rays with each photon carrying TeVs of energy Wrong. Precisely because the energy is that high, radiation would be in the form of a variety of elementary particles. And since quarks & gluons have a lot of degrees of freedom, it would mostly be hadron jets. $\endgroup$
    – A.V.S.
    Commented Jun 22, 2018 at 20:14

The temperature of Hawking radiation in units of energy (for a $1.3\cdot10^{7}\,\text{kg}$ black hole) is $$T=\frac{\hbar c^3}{8\pi G M} \approx 813\,\text{GeV} .$$

As explained in my answer to the question

What is the relative composition of Hawking radiation?

for such high temperatures Hawking radiation consists mainly from quarks and gluons that quickly hadronize producing jets. And since quarks and gluons have a lot of degrees of freedom (due to color and flavor) as a result, total power radiated by a black hole with such a high temperature would be much larger than the number from Xaonon's Hawking radiation calculator. The paper cited in my above-mentioned answer:

  • MacGibbon, J. H., & Webber, B. R. (1990). Quark-and gluon-jet emission from primordial black holes: The instantaneous spectra. Physical Review D, 41(10), 3052, doi.

provides the following figure for total power of Hawking radiation (I will omit uncertainties): $$ P_\text{tot}\approx 3.2\times10^{24}\left(\frac{T}{\text{GeV}}\right)^{2.1}\,\text{GeV sec}^{-1}. $$ For a $13\,000$ metric tons black hole, this gives $P_\text{tot}\approx 6.2\cdot10^{20}\,\text{W}$, two orders of magnitude larger that calculator's figure. (Actually, the power should be even more if emission of Higgs bosons is included). Such power corresponds to at least $7 \,\text{tons}\,\text{sec}^{-1}$ mass loss for the black hole, and so the total lifetime (from $13\,000\,\text{tons}$) would be less than 10 minutes. This power also greatly exceed Eddington limit, so there would be no matter absorption.

About half of this power would be released in the form of neutrinos/antineutrinos (and small fraction in form of gravitons) and so would not be (noticeably) interacting with any matter. The rest of the energy would be in the form of energetic gamma rays and relativistic hadrons and leptons and would definitely produce a planetary cataclysm.

The exact structure of such cataclysm depends on the geometry of a collision. Even if we assume a glancing flythrough of the black hole when the final stage of its explostion happened far enough from the Earth surface, the energy absorbed by Earth would still be enormous.

Remember that Tsar Bomba releases only about $2\,\text{kg}\cdot c^2$ of energy, while our black hole releases about 1000 times that energy (in the form of charged particles) every second. Even if we completely disregard the irradiation of Earth during black hole approach (the considerable share of energy would be absorbed by upper atmosphere over large surface area), a 5 second flight inside the atmosphere would produce a fireball equivalent to several thousands Tsar Bombas. The shock wave created would be destructive across the whole globe. If considerable energy is deposited inside the earth core, this would produce an earthquake of an enormous magnitude.

Nevertheless, while such cataclysm has the potential to wipe human civilization, in terms of heat capacities of world oceans (and the Earth itself) the total energy is relatively minor. So life inside the ocean would likely survive.


That meteor was observed quite clearly. And it was moving at speeds that didn't surprise for a meteor, so call it 20 km/s just to have a number.

3E8 watts is going to be visible for a quite good distance. I think we'd have seen it coming. If it didn't stay in the Earth, then in the 2 days of life it has, it gets only about 3E6 km. If that much mass turned to radiation inside that distance, I think we notice. It gets a lot brighter towards the end IIRC.

I think with that radius for the horizon, and that much radiation coming out, it does not much notice a little thing like rock to fly through. I think it winds up burning a hole ahead of it, with radiation pressure shoving the debris out of the way.

So I'm estimating 2-ish. Seriously intensely bright light, pin-point sized, goes pretty much straight along, comes out the other side. If it hits you, it will be pretty nasty. Probably a lot of serious hard radiation coming from it. It then travels a few times the distance to the moon, then goes with a final super big flash. How bad that final flash would be is more than I can calculate.

  • $\begingroup$ The question has been updated with a much higher luminosity. $\endgroup$
    – PM 2Ring
    Commented Jun 22, 2018 at 7:30
  • $\begingroup$ A share of that energy could be in the form of Neutrinos and a percentage could be in the form of gamma rays and even gravitations(?). How much would be in visible light is unclear at least to me. physics.stackexchange.com/questions/89983/… Now when it hits the atmosphere (for the split second/few seconds it passes through) Then you'd probably see a very bright meteor like something. $\endgroup$
    – userLTK
    Commented Jun 22, 2018 at 12:44
  • $\begingroup$ @userLTK The various gamma-ray telescopes would detect it, for sure, probably along with a lot of military spy satellites looking for nuclear tests. $\endgroup$ Commented Jun 22, 2018 at 21:30
  • $\begingroup$ Looking at A.V.S.'s answer, I'm kind of embarrassed by my answer. $\endgroup$
    – user93146
    Commented Jun 23, 2018 at 15:37

Before matter can be consumed by a black hole it should dissipate its momentum and here comes the accretion disk. However, a tiny black hole can't feed effectively, because its event horizon is too small, secondly, such a black hole would be radiating a lot of energy, which would prevent any matter coming any close. Also such a black hole would not be able to dissipate its momentum and will penetrate the Earth through and fly away.

Let's say I am wrong and it would sink all the way to the center of the planet. In that case all its energy would get deposited at the earth core. Evaporation of a black hole with mass ten to thirteen thousand metric tons would release about $10^{24}\,{\rm J}$, and if we spread this energy onto $6\times10^{24}\,{\rm kg}$ of earth core, its effect would be negligible. However, the black hole would be radiating over $2\times10^{18}\,{\rm W}$ which might be enough to produce and support a tunnel spewing hot plasma in the atmosphere and $10^{24}\,{\rm J}$ deposited in $6\times10^{21}\,{\rm g}$ of Earth's atmosphere would burn everything on the surface of the planet.

  • $\begingroup$ 1E4 metric tonnes mass is 9E20 Joules, equiv of about 2E11 tonnes TNT. That is 200,000 mega tonnes. I think we notice even if it's deep in the Earth. $\endgroup$
    – user93146
    Commented Jun 21, 2018 at 22:18
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    $\begingroup$ Putting the numbers into Wolfram Alpha and converting into Moment Magnitude scale for earthquakes, I get, M0 ~= (9e20 * 2e4), Mw ~= (log(M0)-9.1)/1.5 ~= 10.77. I think we'd notice, that big has probably never happened since humans diverged from other still-present primates, but I don't think we'd care if that was in the core and the damage wasn't focused on any meaningful epicentre. $\endgroup$
    – BenRW
    Commented Jun 21, 2018 at 22:52

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