Set a hundred years into the future, an alien mothership antimatter engine has malfunctioned and to prevent a disaster they decided to eject the byproduct, a micro blackhole as massive as an empire state building towards Earth. Let's not speculate what the alien are thinking or what exactly they are scheming, I like to know how on earth can we notice our impending demise since the micro blackhole trajectory allows it to avoid collision until impact unless we can help it?

  • $\begingroup$ For fun and giggles, consider my answer to a related question. Some of the data applies here. $\endgroup$ – JBH Jan 4 at 8:12
  • 11
    $\begingroup$ It's falling from interplanetary space towards Earth? It will "impact" the Earth at a dead minimum of 11 km/s and almost certainly much faster than this. 17-70km/s is more likely. It will fly through earth, leave a microscopic hole surrounded by what appears to be a continuous nuclear bomb's heat, and continue on its merry way around on orbit around the Sun. As it will only last 130 years, it is ludicrously unlikely to have a second encounter with anything. $\endgroup$ – PcMan Jan 4 at 18:25
  • 3
    $\begingroup$ Is the black hole as big as the Empire State Building or does it just weigh as much as it? $\endgroup$ – MonkeyZeus Jan 4 at 20:35
  • 10
    $\begingroup$ @MonkeyZeus OP uses the phrase "as massive" which, if used correctly, means of equal mass, not volume. $\endgroup$ – DKNguyen Jan 4 at 21:18
  • 3
    $\begingroup$ I guess you should have a look at the corresponding What if: what-if.xkcd.com/129 $\endgroup$ – Vincent Fourmond Jan 5 at 17:47

Dont do anything.

1: a BH of that size would evaporate in 129 years (which surprised me I thought it would be shorter). This makes my point of it evaporating in microseconds moot so ignore point 1.

2: a BH is tiny. At the mass of the empire state building it is smaller than atoms, and its questionable it will actually hit anything as it passes through the earth. If the BH didnt evaporate you could shoot someone through the head and they wouldn't even be damaged. Facepalming would be more dangerous.

Edit: as Jason Goemaat points out in the comments, the gravity at a few mm distance from the BH would still rip a small hole through your body. The point of the example was to illustrate the small size of the BH: At the mass of the Empire state building (365000 tons) a BH is 5.4×10-19 meters in size. An atom is on the order of 1×10-10 meters in size. A difference of 9 magnitudes!

3: the small size and evaporation create a seeming paradox: the BH is almost impossible to feed. Because its so small only a few atoms could be pushed inside at a time, assuming the atoms dont push off against each other enough not to fall in. On top of that the BH's evaporation pushes out harder than the BH pulls things in, you need neutron star pressure or a very specific particle accelerator to keep it fed.

Edit: for some information on BH's check something like this: https://www.vttoth.com/CMS/physics-notes/311-hawking-radiation-calculator

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – L.Dutch - Reinstate Monica Jan 5 at 18:07
  • 1
    $\begingroup$ The luminosity is (according to the linked calculator) 0.64 megatons-of-TNT per second. At 64 km/s that still means 10 tons of TNT on every metre. That's not apocalyptic, but it is a lot more dangerous than a facepalm. $\endgroup$ – Jan Hudec Jan 6 at 6:39
  • $\begingroup$ @L.Dutch-ReinstateMonica, comments may not be for extended discussion, but the fact an answer is disputed is totally lost by moving the comments to chat. $\endgroup$ – Jan Hudec Jan 6 at 6:40
  • $\begingroup$ @JanHudec as the answer states, "If the BH didnt evaporate you could shoot someone through the head". That means its luminosity is non-existant when it doesnt evaporate, its to illustrate the size of the BH. Someone has used the comments to point out that its gravitational pull at a few mm distance would still cause damage, so I'm going to add that to the answer. $\endgroup$ – Demigan Jan 6 at 7:16
  • 1
    $\begingroup$ The answer is still failing to account for the biggest effect it would have. $\endgroup$ – Jan Hudec Jan 6 at 8:39

We'd definitely notice.

A 365000 tonne black hole has a luminosity of $2.6*10^{15}$ watts. On the K-scale of civilizations this is about 0.95.

1.0 is roughly consuming all of the energy the Sun deposits on the Earth.

Even an advanced Earth-based civilization cannot pass, and practically cannot reach, 1.0 -- before 1.0, you literally cook the biosphere, as you are using more energy than we can radiate off into space. A Dyson-swarm scale civilization can pass 1.0 without cooking itself, but in such a civilization Earth is an energy backwater.

So this thing is putting out more power than all of human civilization on Earth is, and doing so with ridiculously high frequency photons. That ain't gonna be hard to spot; it will be hard to ignore really.

Its gravitational effects are quite limited; within a mm we are talking 2400 earth-Gs of gravity, and a cm away there is 24 Gs, but the raw outpouring of energy from its evaporation outpaces it: Nothing is getting within a mm of this black hole, except the Hawking Radiation it emits (and things interacting with that radiation).

Despite how bright it is, it lasts over 100 years. Its final evaporation -- from blue-whale sized to nothing -- happens at the end of the 100-odd years, and puts out as much energy as a magnitude 11 Earthquake in a second. Power output is non-linear, so much before that point it isn't putting out geological event power levels.

A magnitude 11 Earthquake is nuts. Luckily, if that Earthquake happened, it would happen near the core (if the Earth captures it, it means it slows it down in a single pass enough that by the time 100 years have passed, the black hole is captured by the Earth's core).

As it falls through the atmosphere, it lets off enough energy to match that of a decent sized nuclear bomb. Nothing world-shattering, but definitely noticeable. (take $10^{15}$ W times 100 seconds; we get $10^{17}$ J. A megaton nuke is $5*10^{15}$ J; this thing is like a medium-small sized nuclear bomb going off every second. But at 11 km/s, it will pass from relatively far away in outer space to inside the planet in mere seconds; and it will burrow through the Earth so fast, that the surface effects will be modest (on a nuclear bomb scale definition of modest))

If it exits the Earth (see below), it does the same thing at the exit point. So a nuclear bomb crater, fried electronics in a hemisphere, etc.

The energy it deposits as it goes through the Earth is not enough to trigger anything geological in scale, like cataclysmic Earthquakes or a Volcano. Geological event power levels are simply insane.

(If it hung around for longer, it would be enough to make above-hurricane-scale climate disruptions; but it is either going to end up deep in the Earth, or flying out in space, so that won't happen.)

Now, will the Earth capture it?

So as the black hole enters the Earth, it is going to cook everything nearby. You'll end up with a high pressure plasma wave surrounding it. Now basically none of this matter will reach the black hole -- the Hawking radiation pressure is much too high for that -- but good old newton's law means that all force the black hole applies is applied back on the black hole. And there is going to be some asymmetry caused by burrowing through solid rock.

The black hole will act like a larger object in how it interacts with the Earth. How much larger is an important question.

Based on the Newtonian Impactor math -- basically, that objects stop when they push aside roughly as much mass as their own mass when penetrating another object -- if the black hole "acts like" a 10 cm diameter or larger object when flying through the Earth, it is going to slow down and be captured by the Earth. If it acts smaller, it passes right through.

So I don't know if the Earth with capture it. That is above my pay grade in a number of areas of physics. I'd argue it is plausible it passes through the Earth, and it is plausible that there is enough 'drag' to stop it.

If it is captured, are we doomed?

First, will it turn the Earth into a black hole? No, the surface pressure of a black hole this size or smaller is so large that nothing the Earth can do can prevent it from evaporating. You'd have to shoot it into a Neutron Star or something equally exotic for it to grow; infalling matter has to outpace the Hawking radiation pressure.

The other possibility is that it will destroy the Earth through its evaporation. But the final explosion is going to be strong, but even a modestly strong explosion at the center of the Earth isn't going to cause apocalyptic damage on the surface.

Lastly, will it cook us? I mean, it is 0.95 K-scale; if it hung around on the surface of the planet, it would risk cooking the biosphere. But the planet core has a lot more heat capacity: the total energy is releases is enough to heat up the Earth by a tiny faction of a degree (doesn't matter which units). Even the core won't change much. The final explosion will be enough to be heard around the world, and maybe the effect might be big enough to mess with Earth's Magnetic field (the power output of Earth's magnetic field is under 1000 kW; on the other hand, the dynamo driving it is probably ridiculously bigger).

The energy density of the black hole might do exotic things to nuclear chemistry, like creating strange isotopes; but I'd suspect the amount would be small, and the isotopes light. So there could be some danger in creating toxic particles; the transit of the atmosphere is short however, and most of the radioactive element danger from nuclear bombs is from fission by-products, not produced by the explosion. So I doubt this will be a large concern.

What we can do

We could just ignore it. Unless Earth has a massive population increase, much of the world isn't all that occupied. It would take a crazy lucky strike to hit an area where we'd care all that much (most likely, some fish would die; if not, it would make a couple of craters in a place like Siberia or the Sahara desert; failing that, a rural area with a few 1000 fatalities; it would suck, but not that much).

But if we want it so miss, gravity-based nudging of trajectory is plausible. Making your craft capable of surviving the energy output at a close enough distance is hard, but interplanetary trajectories are really easy to make them miss a target. It requires ridiculous pin-point accuracy in order to hit a planet at interplanetary ranges.

A ridiculously tiny nudge would move it.

If you can move asteroids with any speed, you can just shoot an asteroid to get as close as you can to the target. Even if it is destroyed or just does a flyby, it will move the trajectory of the micro black hole. And a tiny nudge is pretty good. The direction of the nudge will be somewhat randomized; as the asteroid approaches the mini black hole, it will start being destroyed by the radiation coming off the black hole, and that destruction will change its trajectory.

But space is really really empty, so a randomized trajectory will almost certainly miss. Hitting a planet is a lot like making a hole in one by hitting a golf ball, when you are in New York and your target is in California. A random push on that trajectory and it won't be making a hole in one anymore with a near total certainty.

  • $\begingroup$ Using the calculator I posted above (assuming it's correct), the energy output doesn't seem nearly as large unless you mean in total? And across more than 100 years it doesn't seem that powerful until the last few years. $\endgroup$ – Demigan Jan 4 at 16:23
  • 1
    $\begingroup$ @Demigan I get 2.90771E15 Watts. That is a lot: about 10^15.5 W. On the K-scale, that is 0.95, more than human civilization. Or, another way of thinking about it, it is a megaton nuke every few seconds. $\endgroup$ – Yakk Jan 4 at 16:27
  • 8
    $\begingroup$ @yakk The total energy given off by the evaporating black hole, over its whole lifetime, including the last frenetic burst, is enough to.... heat up the Earth's core by 1/200th of a K. or the whole earth by 1/600th of a K. not in the least significant. $\endgroup$ – PcMan Jan 4 at 18:15
  • 3
    $\begingroup$ @Demigan A million tonne spaceship using a 350,000 tonne black hole drive (which is 1/3 of its mass!) with near perfect efficiency as a photon-drive gets 0.01 m/s^2 acceleration. That is enough to propel that 1 million tonne spaceship fast enough to reach the nearest star in about 100 years, round about when the black hole runs out of gas. Overkill? Or roughly what you need to have a hard-SF interstellar spaceship? $\endgroup$ – Yakk Jan 4 at 20:05
  • 4
    $\begingroup$ @user6760 it is your question, you have decided the parameters on what you want as a good answer (which could require more detail). My answer misses the "would we notice" part and his answer misses a "what can we do" part. So its entirely up to you to accept an answer even if it has fewer votes as long as it follows your own guidelines for a good answer. You can even wait and see if someone combines my and his answer and accept that instead. Remember: the majority is often wrong! $\endgroup$ – Demigan Jan 5 at 7:20

If the BH was massive enough to be a danger to Earth.... not a lot you can do. You can't really move the Earth out of the way. So the best you could do is send as many people as you could to colonies on Mars or the Moon.

As for when people would notice, that depends on the size of the black hole. As it gets closer, we'll start to notice distortions in light. Of course, it isn't clear how fast the BH is approaching Earth. If it is moving faster than light via alien tech, you wouldn't be able to see it coming at all.

If it's moving slower, and has been travelling millennia or whichever... note that the odds of it hitting Earth, or the sun, would be basically none, if it was shot out by chance, there's so much space for things to go, the aliens would basically have to try to hit Earth's solar system in order to hit it.

Of course, if it's too close or fast, there won't be time to react.

  • $\begingroup$ What about if it's moving fairly slowly but was calculated to specifically hit the Earth? $\endgroup$ – LukeN Jan 5 at 21:14
  • $\begingroup$ If you had a lot of time, then you would start putting all of Earth's resources towards bases on Mars. Some people might try Venus, but that would be a terrible idea. A temporary moon base could be set up (temporary could mean decades, in this case) for launching people to Mars. $\endgroup$ – Johnny Jan 5 at 21:16

Answers regarding what it will do have already covered the effects as it sheds mass as energy. That also covers how quickly we'll notice it since it will be blatantly obvious. However, you also asked how to stop it.

From the tone of your question, it sounds like impact is very soon. In that case, it's too late and all you can do is clean up the mess. However, if you detect it far enough away or it's approaching slowly, you could try to redirect it. How much force would you need to deliver? The more time until impact, that's less force you need to apply to change the trajectory... so it depends.

Even if you have time to spare, the problem will be how can you apply that force? Hitting it with a large mass isn't going work, because it would zip right through any material that you throw at it. However, if you could hit that moving point with absolute precision, then high-speed particles or even lasers could apply force.

As a nifty story, another concept is not to poke at it, but instead shepherd it away. In other words, put an even larger mass near the same position and velocity and then use the gravity of that new mass to pull it onto another course. For the rule of cool, you could redirect it to and "give it back" to those pesky aliens.

  • 1
    $\begingroup$ I would say that it would be exceedingly easy to redirect it. It's "only" 365000 tons of mass, and throwing 1 ton of mass on its left side will mean that 1 ton will vaporize as the BH passes, which creates a difference in pressure pushing the BH off-course (although I'm not sure what it will push. I guess differences in electromagnetics and weak forces?) $\endgroup$ – Demigan Jan 5 at 7:26

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.