In the spirit of what if at xkcd, I saw the following question posted on a thread in reddit/r/space:

What would happen if a black hole the size of a grain of sand existed on Earth?

The thread was deleted by mods for trivial reasons (it wasn’t posted in the weekly ‘general questions’ thread) so there’s little point in linking to the thread itself. But before it was deleted, all the answers were really bad: either saying it would do absolutely nothing, it would only sink to the center of the earth growing as it went, or it would immediately disappear in a flash of radiation.

All of those answers are a possibility, depending on the size of the black hole, but the biggest problem is that no one did any math or calculations to actually determine what would happen, they were just guessing random crap.

So let’s do the math ourselves. Fortunately for us there is already a website that can do most of the math for us: a Hawking radiation calculator. Basically it lets you input a property of a simple black hole (no spin or charge) and it will calculate the rest of the properties for you. Parameters are the mass of the black hole, radius of the event horizon, surface area of event horizon, gravity at the surface of the event horizon, surface tidal force, entropy, temperature, peak photon energy (due to Hawking radiation), effective density, luminosity (again due to Hawking radiation), time to free-fall from the event horizon to the singularity, and total lifetime of the black hole (due to losing energy via Hawking radiation).

So I put in a radius of 0.1mm for a black hole, and here are some of the results:

Quantity | Value |

Mass | \(6.7\times 10^{22}\, \text{kg}\) |

Surface gravity | \(4.5\times 10^{20}\, \text{m}/\text{s}^2\) |

Temperature | \(1.8\, \text{K}\) |

Lifetime | \(2.6\times 10^{52}\, \text{s}\) |

The mass of \(6.7\times 10^{22} \text{kg}\) is approximately 1/100th that of the Earth, or about the same as the moon. This shows how dense a black hole really is: something as massive as the moon, when compressed to a singularity, would have an event horizon only 0.2mm across.

So what would happen to this very tiny black hole? You might think it would pass harmlessly through the Earth, since it only has a very small cross-section of a circle with a 0.2mm radius, so whatever it passes through would be absorbed into the black hole, but everything else would be fine, right? Unfortunately, *wrong*.

A big problem is that even though its mass is no larger than that of the moon, because it’s compressed to a singularity the gravitational acceleration near the event horizon becomes *very* large. This is because the strength of the gravitational acceleration is proportional to square inverse of the radius. So if we double our distance from the center of mass, the gravitational force is is 1/4th as strong. But conversely if we halve our distance, the gravity is 4 times as strong. So as you can imagine, being able to decrease the distance all the way down to 0.1mm will make the gravity *very very* strong. In fact the gravitational force at the event horizon of our black hole is about \(1 \times 10^{19}\) times stronger than Earth gravity!

We can calculate the distance from our black hole that its gravitational acceleration will be the same as Earth surface gravity:

\(\displaystyle g_s r_s^2 = g r_2^2\)

Where \(g_s = 4.5\times 10^{20}\, \text{m}/\text{s}^2\) is the gravitational acceleration at the event horizon of our black hole, \(r_s = 0.1\, \text{mm}\) is the radius of the black hole, \(g=9.8\, \text{m}/\text{s}^2\) is the gravitational force at Earth’s surface, and \(r_2\) is the distance we are solving for. This comes out to be about 680km, so everything within that radius is going to be attracted to our black hole more strongly than it is to the rest of the Earth!

So this is looking bad. We could surmise that anything in that radius is going to be pulled into the singularity. How long will it take for something to be pulled into the black hole from this distance? We can estimate this with a bit more math:

Using good old Newton’s law of motion

\(\displaystyle F=ma\)

where \(m\) is the mass of the object and \(a\) is the acceleration, along with Newton’s law of gravity:

\(\displaystyle F=\frac{G m_1 m_2}{r^2}\)

where \(G = 6.7\times 10^{-11} \text{m}^3 / (\text{kg}\, \text{s}^2)\) is the gravitational constant, \(m_1\) and \(m_2\) are the masses of the two bodies, and \(r\) is the distance separating them. We substitute in the gravitational force into Newton’s law of motion, and substitute acceleration using the definition \(a=\frac{dv}{dt} = \frac{d^2 r}{dt^2}\). So now we have a 2nd order ordinary differential equation (specifically an initial value problem):

\(\displaystyle \frac{G m_1 m_2}{r^2} = m_2 \frac{d^2 r}{dt^2}\)

The \(m_2\) term, the mass of the falling body drops out of both sides, so our answer is only in terms of the body we are falling towards. Our initial conditions are \(r = 6.8\times 10^{5} \text{m}\) and \(\frac{dr}{dt}=0\) at \(t=0\).

This can be integrated to solve exactly for the time it takes for something to fall from distance \(r\) into the black hole, but we don’t even have to do calculus to get a rough estimate. If we ignore the derivatives, we can use this to get a rough scaling of how the time is affected by distance and the other parameters.

\(\displaystyle \frac{G m_1 m_2}{r^2} \sim \frac{r}{t^2}\)

We solve this relation for \(t\) and we get the following:

\(\displaystyle t \sim \left( \frac{r^3}{G m_1} \right)^{1/2}\)

This tells us that the time for an object to fall into the black hole is proportional to the distance it falls \(r^{3/2}\) and the mass of the black hole \(m^{-1/2}\). In other words if we increase our distance by 4x then the time to fall in increases by \(4^{3/2}=8\) times, and if the mass of the black hole increases by 4x than the time to fall in decreases by \(4^{-1/2}=1/2\).

For a more exact answer, we need to do the actual calculus. We integrate this twice and use the boundary conditions, then we can solve for the time to fall into the black hole from the distance \(r\):

\(\displaystyle t=\left( \frac{2 r^3}{3 G m_1} \right)^{1/2}\)

For our black hole at the distance of 680km, this comes out to be about 200 seconds, or a little over 3 minutes. You’ll notice that the equation has the exact same form as the estimate earlier, it only differs by a constant (in this case \(\left( \frac{2}{3} \right)^{1/2}\)). This is pretty typical of this kind of analysis. The quick estimate gives you an idea of the relationship of the different parameters to each other, and the full analysis refines it to give you the actual answer you can use in your calculations.

As a bit of an aside, why did we have to use calculus to solve for this in the first place? The reason is that the force due to gravity is continually changing as the object falls closer to the black hole. If the force were constant, then the acceleration would also be constant and we could just look up the formula from an introductory physics textbook or website if we couldn’t remember it. But because the force and hence acceleration is continually changing, the only way to solve it with calculus, which incidentally was developed by Isaac Newton to solve this exact kind of problem, i.e. celestial mechanics.

Back to our black hole. As our black hole is pulling in Earth’s mass into its gaping maw at a radius of about 680km, the black hole itself is going to be falling towards the center of the Earth, absorbing things as it goes. This classic calculation tells us it will take about 42 minutes for it to reach the center of the Earth. Of course when it gets there it won’t stop moving, so it will keep on moving and orbit the center of the earth with a total orbital period of 168 minutes, absorbing matter all along the way. Technically the Earth, black hole, and moon will all orbit the center of mass of the three-body system, but from the point of view of someone on the Earth it will appear as I described it.

Gravity will pull in the remaining Earth’s mass to form a continually smaller sphere as the black hole absorbs its mass, but between the rotation of the earth and the orbital period of the black hole it probably won’t be too long before most of it is absorbed into the black hole. Verdict: earth is sucked into our black hole over the course of a few hours/days.

But wait, there’s more! It isn’t *just* the extreme gravitational force of the black hole pulling Earth’s mass into it that we need to worry about, we also have tidal forces.

The basic concept of tidal forces is pretty straightforward. From our Newton’s law of gravitation above, we know that something closer is attracted more strongly than something far away. This also means that the portion of an object (i.e. comet, moon, planet, human) that’s closest to the massive object is attracted more strongly than the portion of it that’s farthest away. This force can actually pull the object apart if it’s stronger than the force that’s holding it together. For really big objects (planets, moons, etc.) it’s the objects own gravity that keeps it together, for smaller objects it’s simply its own tensile strength.

To calculate when that actually happens, I found a nice formula here:

\(\displaystyle \sigma < \frac{G m\rho R^2}{\Delta^3}\)

Here \(\sigma\) is the tensile strength of the material, \(\rho\) is the density of the material, \(R\) is the radius of our object that’s being pulled apart (assuming a sphere), and \(\Delta\) is the distance between the two bodies. So if our tensile strength is less than this quantity on the right-hand side, then it will be pulled apart due to the tidal forces.

Let’s look at our black hole and Earth. Assume that Earth is made of granite: tensile strength is \(4.8 \text{MPa}\), and its density is \(2750 \text{kg}/\text{m}^3\). We can use this to calculate what the maximum size a granite boulder can be before it’s ripped apart due to tidal forces from our black hole.

At a distance of just 1m from the black hole, any rock larger than 0.03mm will be ripped apart due to tidal forces. At 100m any rock larger than about 3cm will be ripped apart. At 1km the maximum size is about 1m. If we go all the way to the other side of the earth at about \(1 \times 10^7 \text{m}\), the maximum size of a boulder is \(1 \times 10^6 \text{m}\), or about 1/3 the diameter of the moon. So if you’re on the other side of the Earth when this thing hits you yourself won’t be pulled apart into spaghetti, but Earth itself will rapidly break up into planetoid-sized and smaller chunks. Verdict: the Earth gets rapidly torn apart into planetoid-sized chunks in a matter of minutes, turning the whole mess into a hot ball of molten rock. Which then gets sucked into the black hole over the course of the next several hours/days.

There are a couple of other things that I’m missing. The most obvious is that all this mass can’t instantly fall into the black hole, it will all get in its own way. Also it won’t necessarily fall straight in, it has to conserve its own momentum so it will circle the black hole as it falls in. This will cause it to rapidly heat up, turning everything into super-hot magma, gas, and plasma that will form an accretion disk around the black hole. This will spew out all kinds of high energy x-rays and gamma rays, and we may even get coronal ejections from the poles of the black hole as well.

What about the rest of the solar system? The moon probably won’t change that much, it will still orbit the center or gravity of the system. The day side will get bathed in a whole lot of high energy radiation, but the night side won’t significantly change. It should still be tidally locked so that the same side will continue to face the Earth (or the black hole and accretion disk) with the night side facing away. Incidentally, the night side of the moon would probably be the safest place for humanity to survive. Anywhere else in the solar system is going to get massive x-rays and gamma rays from our accretion disk whenever its visible in the sky, but anyone on the dark side of the moon will always have the mass of the moon between themselves and the black hole. Another option would be deep underground habitats on other celestial bodies in the solar system: Mars, Ceres and other large asteroids, various gas giant moons, etc. It’s hard to say how much shielding you would need, since I don’t know how to estimate how much energy would be put out by the accretion disk that used to be the Earth.

Overall verdict: Earth is totally destroyed in a matter of hours, all that’s left is a super-hot super-radiation-spewing accretion disk falling into the black hole. Anyone not on the dark side of the moon or in a deep underground habitat somewhere else in the solar system is fried to a crisp in the radiation.

**Edit:** of course the *real* question is, where did a black hole with the mass of the moon come from? That much mass just doesn’t appear by itself. Possibly Ceres or a trans-Neptunian dwarf planet became a singularity due to a super-science experiment gone wrong, then collided with the Earth or something.