Tjyylli

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.

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 :

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

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 $frac v c approx 1 - 2times 10^{-19}$. 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 $2times 10^9$ 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 $3times 10^{13}$ 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.

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 70 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...

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.

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:

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

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 ($s = rtheta$), 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.