# What would happen to a modern skyscraper if it rains micro blackholes?

Set in present day New York City, an unknown spacecraft of alien origin expelled millions of micro blackholes each with the mass of a grape in the earth atmosphere. I like to know what happens if these millions of micro blackholes were to fall on building structures such as skyscrapers, would it trigger an extinction level event?

• Given the aliens could easily send waves of asteroids to destroy Earth's surface completely with practically trivial effort (at their tech level), mucking around with micro black holes (or any black holes) seems quite daft. – StephenG Apr 5 at 13:33
• @user6760 what do you want to happen or expect to happen? I presume you chose black holes for a reason. – Snyder005 Apr 5 at 19:26
• @StephenG - an entire can of micro black holes fits in the storage cupboard in the corner of the spacecraft's kitchen (which has a stasis field to keep food fresh, and keep black holes from evaporating), going out and dragging waves of asteroids is a lot more work than just opening a can of micro-blackholes and sprinkling them out a hatch. – Johnny Apr 5 at 23:10
• @Johnny As explained in answers, in an instant of time after you open the "can" (remove the magic statsis field) so short you could not measure it, all the micro black holes evaporate (with a huge out-pouring of radiation like a nuke). Dragging asteroids is what we in engineering call "safer", at least for the aliens - but still kills the pesky humans. :-) – StephenG Apr 6 at 1:18
• just one point - whatever effect you're thinking about (and don't forget, there is no effect - small black holes just evaporate) ... it would happen to the air before happening to anything else it is moving towards. – Fattie Apr 6 at 16:09

would it triggers extinction level event?

Since they'd evaporate more or less instantaneously (known as Hawking radiation), releasing energy according to the famous equation beginning E=, the spaceship would last a few microseconds at best, Earth would be fine.

Yes, the aliens in the ship would become extinct.

• According to this calculator (eguruchela.com/physics/calculator/…), they would last 1.6581375e-29 seconds. There is also the fact that their radius would be so small, that they wouldn't even interact with atoms most of the time. – Tyler S. Loeper Apr 5 at 13:28
• @TylerS.Loeper Gosh, we don't have an SI multiplier to express that, atto is feeling left out and lonely. – Chickens are not cows Apr 5 at 13:31
• How about they just pick the right size black holes that they evaporate with a boom after they reach Earth? That shouldn't be too hard. They'd only be grape sized for a tiny while, but that's fine. – John Dvorak Apr 5 at 14:17
• @John Dvorak a 1,000 metric tons black hole will have a lifespan of 84 seconds. It may be enough to reach the surface, but it's going to release an energy amount equivalent to teratons of TNT. – Alexander Apr 5 at 20:30
• @reirab As long as you're outside the event horizon, a black hole's effects are more or less the same as a non-black-hole object of the same mass (Hawking radiation notwithstanding). So, if you can work out how to manipulate the black hole without touching the event horizon, it's just like having a thousand-tonne marble on board. – anaximander Apr 6 at 7:16

Black holes evaporate by emitting Hawking radiation

a 1-second-life black hole has a mass of $$2.28 \cdot 10^5 \ \mathrm{kg}$$

A grape has far less mass than that, thus the black hole would evaporate way faster than that.

An intelligent life form dropping micro black holes on Earth would thus quickly annihilate its own bombing squad in a shower of gamma ray, proving that they were not so intelligent as we thought.

• So, basically, aliens playing with tech they don't quite fully understand yet resulting in unintended consequences. Basically an alien version of the early Cold War period. – reirab Apr 5 at 21:51
• @reirab, I mean, we managed not to blow ourselves up during the Cold War, so... score one for humanity I guess? – Gryphon Apr 6 at 1:29
• @Gryphon let's be real though: we came damned close. – Sora Tamashii Apr 10 at 6:32
• @SoraTamashii yeah, but we managed not to commit self-genocide at the last second every time we almost did. So it's still a point for us. – Gryphon Apr 10 at 12:34
• @Gryphon A-hecking-men. – Sora Tamashii Apr 12 at 0:30

The electromagnetic force from one electron on another and the gravitational force of this micro-black hole both follow an inverse square law. A grape about 1.5 cm in radius would have a mass of about 0.015 kg.

When does the gravitational force of the grape exceed the electromagnetic force between electrons ? It's when :

$$\frac r R < \sqrt{\frac {4\pi \epsilon_0Gm_em_h}{e^2}} = 6.3\times 10^{-8}$$

Meaning the black hole would have to pass less than one ten millionth of the distance between electrons to have a significant influence on one. Away from than range the electron will happily go about it's business hardly disturbed at all.

Even if a black hole passes this close the effect is only temporary. You're still nowhere near the event horizon of that black hole and so the electron will, at worst, be pulled away from it's normal motion and after some brief period when the black hole moves away it will simply recombined in some way with the bulk of atoms around it. It might cause a minute amount of damage on a molecular level (even allowing for millions of these micro black holes), but the net effect would be tiny, probably less that someone hitting a wall with their hand.

How about they expell at fraction of c so we take length contraction into question?

You seem to mean that to avoid Hawking radiation evaporation destroying these black holes before they even reach the black hole, they could be ejected at a high fraction of the speed of light.

So how high a speed is needed to avoid them evaporating before they travel 100 meters, assuming your aliens like low level flying ?

The fraction of the speed of light needed is :

$$\frac v c > \frac 1 { \sqrt{ 1 + \left( \frac {Tc} L \right)^2 } }$$

Where $$L$$ is the distance they must travel and $$T$$ is the lifetime of the micro black hole before it evaporates.

This works out at $$\frac v c \approx 1 - 2\times 10^{-19}$$. That's insanely close to the speed of light.

A million grapes of mass 0.015 kg will have a mass of 15,000 kg. But the energy required to get them moving at this insane fraction of the speed of light would be enormous. It equates to a mass about $$2\times 10^9$$ times 15,000 kg. Or to put it another way, the ship firing these micro black holes would need to have a mass-energy of about $$3\times 10^{13}$$ kg. The asteroid Vesta is substantially larger than this.

So this is actually a small mass in terms of asteroids and you could probably destroy Earth a lot more easily simply by grabbing some handy largish asteroids and sending them on their merry way towards Earth at some modest speed that's easily imparted with your spaceship.

Conclusion :

No need at all to mess around with ultra-relativistic micro-black holes when the universe provides you with much simpler and easy to handle "ammunition" in the form of basic asteroids.

• I was wondering about that speed, thanks for working it out. – Kevin Apr 5 at 20:30
• Ultra-relativistic black holes would be considerably less effective than throwing Vesta at the Earth: as you note, they'll mostly just pass through Earth without doing anything. – Mark Apr 5 at 22:23
• Interestingly, an actual grape going at that speed would be far more devastating than a grape-massed black hole. I don't have the time to do the math at the moment, but I'd guess it'd be enough to overcome the gravitational binding energy of the Earth. – Gryphon Apr 6 at 1:34
• 👍for addressing my comment too – user6760 Apr 6 at 2:15

All the answers so far take as read the veracity of Hawking Radiation. If we assume, for a moment, that this is false and that some undiscovered process prevents black hole evaporation (perhaps there is a layer of physics underlying quantum mechanics in the same way that QM underlies classical physics...); then what happens?

The black holes would fall to Earth like grapes (I assume you've removed their orbital velocity so that they don't just stay in orbit). They would accelerate like any falling body but, because of their tiny size, would not experience any air resistance. So they would arrive at the surface going at a fair old clip. If dropped from orbit, say about 400km up, this would be about 3 km/s. At the surface, what would happen? Nothing much, I'd guess... They're still so small that "solid" matter is practically a vacuum to them so they go straight through, down past the crust, mantle, core and then up the other side, out through Western Australia, back up to about 400km where they stop - and then tumble back down again. Eventually, they'd settle into a highly elliptical orbit around the centre of the Earth. The Coriolis force would make it look like the stream was scanning round the Earth every 24 hours.

Occasionally, one of them would strike a nucleus head-on and capture some of its quarks, so it would grow slightly. This process would have some positive feedback (bigger the event horizon gets, more chance of interaction) but I'm not sure what the time constant would look like. Anyone fancy doing the calculation? I'm guessing it might be aeons before it eats the Earth...

• With 10^51 neutron-big particles in the 10^21 m3 of earth (10^30 per cubic meter), each about 10^-15 m big (volume thus about 10^-45 m3 for one, or 10^-15 m3 per cubic meter of earth), we should have a chance per meter of about 10^-15 of hitting one of those neutron-big particles with one black hole. Thats not a lot, but it's per meter and for only one black hole, so a million black holes orbiting inside earth (about 10^5 m/s)should accrue some hits – bukwyrm Apr 9 at 16:51
• 1/10 000 chance of a hit per second, with the above values, and a one in three chance every few hours. So not a lot after all ... – bukwyrm Apr 9 at 17:01

If we take "size of a grape" to mean a gram, then a million of them would be 1 metric ton. When they evaporate, they release energy several orders of magnitudes higher than a nuclear bomb. So New York would be devastated, but the earth as whole would not have large effects, although it might produce radioactive elements that would increase cancer rates.

• You forget that those micro black holes won't make it to New York. They'll evaporate the alien ship instead. – cmaster Apr 6 at 9:22
• @cmaster The question says "an unknown spacecraft of alien origin expelled millions of micro blackholes each with the mass of a grape in the earth atmosphere." So the black holes making it to the atmosphere is stated as being part of the hypothetical. The equivalent of nuclear bombs exploding in the atmosphere is going to cause problems. – Acccumulation Apr 6 at 16:55
• Yes, but that's already the part that simply doesn't work out. The aliens won't manage to get such tiny black holes into the atmosphere. It's not like you can create them directly where you need them out of nothing. If you want a black hole that can travel from the alien ship into the atmosphere before exploding, it needs to be many orders of magnitude bigger, and then it'll explode with such tremendous force that you won't need a second black hole to destroy the city. – cmaster Apr 6 at 18:03
• @cmaster - They could get them into the atmosphere if the containment system used to transport the microholes is designed to be ejected from the ship and wait until it enters the atmosphere (or descends to some lower altitude) before releasing them. Of course, at that point, you've basically turned it into a bomb anyhow and there are probably easier ways to make a non-black-hole-based bomb which would be just as effective. – Dave Sherohman Apr 8 at 8:15
• @DaveSherohman You cannot contain black holes. Black holes cannot have a meaningful electrical charge, you can only move them by gravity, i.e. by moving masses within their vicinity. And you most certainly cannot contain black holes of grape size, they'll simply blow any containment away. A BH with a mass of 1000 tons evaporates on the timescale of a second. Lifetime grows with the cube of it's weight, so the last 100 tons only take a millisecond to evaporate. 100 tons worth of energy within $1ms$! - Even if those number don't blow your mind away, they will blow away any containment. – cmaster Apr 8 at 19:10

While this may or may not be an extinction event, it has the potential to be very bad news for at least some of those on the surface of the Earth, even if not the whole surface. The trick is that it will depend crucially on the alien ship's location, since as mentioned in the answers, the ship will be destroyed on the instant the black holes are released.

A typical grape has a mass of about 5 grams (cite: https://www.reference.com/food/many-grams-grape-weigh-fcc1e34fdbbcf843). 5 grams times a million is 5 megagrams (about five typical-sized road vehicles). The explosion into Hawking radiation can be considered effectively as very much like the instant detonation of an antimatter or nuclear explosive which ends up with the conversion of that same total mass to energy. By

$$E = mc^2$$

that is about $$4.5 \times 10^{20}\ \mathrm{J}$$ released, or 450 EJ. For comparison, the TSAR (largest nuclear device ever exploded) was about 0.21 EJ, so this is roughly 2000 times more explosive. While not as large as the Chicxulub asteroid strike (about 400 000 EJ), this is still a considerable bang, more than enough that were it to occur on or near Earth's surface (i.e. the ship is "hovering" just above the skyscraper in question), it would lead to the complete annihilation of not only all of New York City (so way worse than just "a skyscraper"), but probably also the whole surrounding states, if not the entire Northwestern US due to the blast and thermal waves. So insofar as the "skyscraper" is concerned, the answer for it is: complete, instantaneous vaporization into a high-energy ball of plasma, similar to that from a very, very, very, very big nuclear explosive, along with a large amount of the surrounding city and probably also the ground it is sitting on. The expansion of this huge ball of plasma generates a very large blast wave that they lays waste to the surrounding countryside, out probably to a radius larger than New York State itself, plus formation of a large crater similar to that from an asteroid impact and resultant release of ejecta.

The other plausible alternative is if we imagine the craft is in orbit. For Low Earth orbit, we are talking about a height of 400 km above the surface. From geometry, we can then figure the amount of energy deposited at a point by using the inverse-square law, since the source will be approximately pointlike at this distance:

$$I(r) = \frac{I_0}{4\pi r^2}$$

where $$I_0$$ is the initial intensity, $$I(r)$$ that at distance $$r$$. Taking $$r = 400\ \mathrm{km} = 400 000\ \mathrm{m}$$, we get that the point on Earth's surface directly below the craft gets struck with about 223 MJ of energy per square meter, delivered effectively instantaneously. The farthest point that will experience irradiation of energy by the explosion is that for which it is just on the horizon, something we can calculate by considering when the line from the exploding craft to the point on the ground is at a right angle (so the tangent) to the line from the Earth's center to the same ground point. Geometrically, this forms a right triangle with the right angle that at the observation point, the hypotenuse is the line from the Earth's center to the craft (thus equal to $$R_E + 400\ \mathrm{km}$$) and the adjacent side is the line from the Earth's center to the observation point itself (thus equal to $$R_E$$). The length of the opposite side is then $$\sqrt{(R_E + 400\ \mathrm{km})^2 - R_E^2}$$ which with $$R_E = 6371\ \mathrm{km}$$ gives the straight-line distance as ~2300 km. To get the precise ground distance we need to take into account the curvature of the Earth's surface, and we can do that by taking the angle at the Earth's centre: since we have the hypotenuse and adjacent, we get $$\cos(\theta) = \frac{\mathrm{adj}}{\mathrm{hyp}} = \frac{R_E}{R_E + 400\ \mathrm{km}}$$ which gives $$\theta \approx 345\ \mathrm{mrad}$$ and multiplying it by $$R_E$$ to get the circular arc length ($$s = r\theta$$), which gives the ground distance as still being pretty close: ~2200 km.

Thanks to the cosine law of the angle of incidence, of course, radiation at this point will be effectively zero, so we can estimate that a radius of 1000 km will be subjected to radiation levels exceeding 100 MJ/m^2, delivered virtually instantly, chiefly as hard X/gamma rays. This will, for the most part, be absorbed in the atmosphere, but may cause interesting shock heating and chemical effects that I imagine cannot be good for anyone who happens to be underneath them. At the very least, you get a huge cloud of oxides of nitrogen ($$\mathrm{N_2 O}$$, $$\mathrm{NO_2}$$) produced immediately, like smog - poison. I'm less sure of how to calculate how much and moreover what the effects of that will be once dispersed globally, but I can't imagine they'd be too good, either. This effect can be considered similar to that of a nearby Gamma-Ray Burst (GRB) impinging on the planet; cited as a possible cause of the Ordovician mass extinction (though I might have also heard something more recently that this has been tracked to a different cause), though here affecting only a considerably smaller area - that would have affected an entire hemisphere. Nonetheless, it might give you some clue that this is probably not going to be too good a day (week, month, year) for anyone. It may not kill everybody, but it's also not going to be anywhere close to as innocent and harmless as so many other answers and comments here seem to be painting it to be.

And this also, I should tell you, varies with just how much "millions" actually is. This was for one million. If it's 100 million, then we are getting to around 10% of Chicxulub, and talking a lot worse.