How fast a (relatively) small black hole will consume the Earth?

Question

This question appeared quite a time ago and was inspired, of course, by all the fuss around "LHC will destroy the Earth".

Consider a small black hole, that is somehow got inside the Earth. Under "small" I mean small enough to not to destroy Earth instantaneously, but large enough to not to evaporate due to the Hawking radiation. I need this because I want the black hole to "consume" the Earth. I think reasonable values for the mass would be $10^{15} - 10^{20}$ kilograms.

Also let us suppose that the black hole is at rest relative to the Earth.

The question is:
How can one estimate the speed at which the matter would be consumed by the black hole in this circumstances?

Answer 1

In the LHC, we are talking about mini black holes of mass around $10^{-24}kg$, so when you talk about $10^{15}-10^{20}kg$ you talk about something in the range from the mass of Deimos (the smallest moon of Mars) up to $1/100$ the mass of the Moon. So we are talking about something really big.

The Schwarzschild radius of such a black hole (using the $10^{20}$ value) would be

$$R_s=\frac{2GM}{c^2}=1.46\times 10^{-7}m=0.146\mu m$$

We can consider that radius to be a measure of the cross section that we can use to calculate the rate that the BH accretes mass. So, the accretion would be a type of Bondi accretion (spherical accretion) that would give an accretion rate

$$\dot{M}=\sigma\rho u=(4\pi R_s^2)\rho_{earth} u,$$

where $u$ is a typical velocity, which in our case would be the speed of sound and $\rho_{earth}$ is the average density of the earth interior. The speed of sound in the interior of the earth can be evaluated to be on average something like

$$c_s^2=\frac{GM_e}{3R_e}.$$

So, the accretion rate is

$$\dot{M}=\frac{4\pi}{\sqrt{3}}\frac{G^2M_{BH}^2}{c^4}\sqrt{\frac{GM_e}{R_e}}.$$

That is an order of magnitude estimation that gives something like $\dot{M}=1.7\times10^{-6}kg/s$. If we take that at face value, it would take something like $10^{23}$ years for the BH to accrete $10^{24}kg$. If we factor in the change in radius of the BH, that time is probably much smaller, but even then it would be something much larger than the age of the universe.

But that is not the whole picture. One should take also in to account the possibility of having a smaller accretion rate due to the Eddington limit. As the matter accretes to the BH it gets hotter since the gravitational potential energy is transformed to thermal energy (virial theorem). The matter then radiates with some characteristic luminosity. The radiation excerpts some back-force on the matter that is accreting lowering the accretion rate. In this case I don't thing that this particular effect plays any part in the evolution of the BH.

Answer 2

This question is addressed in Giddings and Mangano, http://arxiv.org/abs/0806.3381 . See eq. 4.31 and appendix A. For a $10^{20}$ kg black hole, the rate of accretion comes out to be about $10^{13}$ kg/s.

This is greater than the estimate in Vagelford's answer by a factor of $10^{18}$. The reason for this factor is that Giddings essentially uses the Bernoulli equation to model the mass flow, and in this model the mass does not just flow in at the speed of sound $c_s$ all the way until it hits the event horizon. If I'm getting the gist of the calculation right, mass flows in at the speed of sound up until it reaches a certain radius, which is greater than the Schwarzschild radius by a factor of $(c/c_s)^2$. Even without digging in to the details of the calculation, this sort of makes sense. Infalling matter is going to move at relativistic velocities, $\sim c$, as it approaches the horizon, not at $c_s$. They refer to this effective radius as the Bondi radius, and the difference between their estimate and Vagelford's is basically that they use this radius where Vagelford uses the Schwarzschild radius. This causes their accretion rate to be greater than Vagelford's estimate by a factor of $(c/c_s)^4$, or about $10^{18}$.

Using the Giddings result, it takes on the order of years for the black hole to double its mass. I haven't integrated the relevant differential equation, but since the rate goes like the square of the black hole's mass, it looks like it would only be a matter of decades before the black hole consumed a significant fraction of the earth's mass. (It may not consume all of it because of conservation of angular momentum, the ejection of some mass, and other astrophysical processes that occur near the end where the earth's structure is severely disrupted.)

Answer 3

Since I have much better answer from Vagelford -- I'll write my own version.

When matter falls on the black hole it gets fractioned and radiates. As far as I know (correct me if I'm wrong) one can estimate the radiated energy as $\simeq 0.05mc^2$. Where $m$ is the mass of the falling matter.

The Earth's matter is pulled by the black hole gravitation and pushed away by the radiation. Moreover, for the matter flow $J$ we have "negative feedback" system:

• bigger $J$ -> more radiation -> more matter is "pushed away"
• smaller $J$ -> less radiation -> more matter is "pulled in"

The equilibrium between those forces corresponds to already mentioned Eddington luminosity:
$L (J/s) = 1.3\cdot 10^{21} \frac{M}{M_{sun}}$

Equating $L=0.05Jc^2$ and going to $r_{sh}(m) = 3000 \frac{M}{M_{sun}}$, I obtain:

$J (kg/s) = 100 r_{sh}(m)$

It is remarkable, that the "consumption speed" for the $10^{20} kg$ black hole ($r_{sh} = 148.5\mu m$, look here) will give you $1.48\cdot10^{-5}$ kg/s. Which is just order of magnitude larger than the estimate by Vagelford.

Answer 4

If the black hole simply swalled matter, and didn't lose any energy, it probably isn't too hard a calculation, just assume the earth is unsupported mass that falls into the BH, which grows in mass as it adds more stuff. The problem, is we know this isn't how it would happen, and some significant fraction of swalled mass will be released as energy, maybe one to a few percent of mC**2. So the energy liberated from swallowing mass, is orders of magnitude greater per unit mass than an H bomb. Clearly most of the planets mass would be blown away, and only a small amount would end up incorporated into the BH. I'd bet this would happen extremely rapidly, and the shock wave that rips the planet apart would probably only take a few seconds. Note freefall time to the center of the earth is probably more like a half hour (order of magnitude), so most of the planet wouldn't even begin to fall before the released energy blasted it apart.

Answer 5

It would take a long time if we do a back of the envelope calculation.

• the black hole would exert a force of 1g at around 20 km (assuming 10^20 kg of mass).
• if we can reasonably assume that the mass inside this sphere is going to be absorbed quickly, that would mean the black hole mass increases correspondingly.
• on the other hand this extra mass can be calculated to be around 10^20 kg too. So we can expect the 1g radius not to increase significantly.
• I believe that mass with less than 1g of pull will take a long tine to spiral inside the black hole, as its size (Shwartzchild radius) would be in the micrometer scale and the sizes involved in the kilometre scale.

Hope this helps!

Answer 6

I just want to add that oscillations about the center of the Earth are dampened due to the momentum of the entering mass.

Figures for volume of the mass continually eaten by the black hole differ by orders of magnitude going by previous posters. But the consumed material as it falls will depend on the cross section of this volume times the radius of the Earth, or cross section times density for linear mass density of path of destruction.

That mater has zero kinetic energy and a potential energy as a function of height. Gravitational field is directly proportional to radius (due to continuous spherical distribution) so potential energy is $r^2$ function. I write gravitational potential of BH as $C R^2 m$ where R is radius of Earth and m is mass (kg) of BH and C is some constant I'm not going to address. Denote linear density of path of destruction as $l$ (kg/m), and integrate $C r^2$ potential to find $1/3 C R^3 l$ to center of Earth, or $2/3 C R^3 l$ to other side of Earth.

Assume it eats material perfectly and there are no other interactions. It begins with $C R^2 m$ energy (Newtonian!) and m mass. It acquires $2 l R$ mass in one trip (assuming acquired mass is small relative to total and thus nearly touches surface again). We find the deficit in specific potential energy at the end of its trip: (P.E._end/end_mass) / (P.E._start/start_mass)-1.

$$\frac{ \frac{ C R^2 m+\frac{2}{3} C R^3 l}{m+2 R l} }{ \frac{C R^2 m}{m} } - 1$$

$$=\frac{m}{m+2 R l} \frac{R^2 m+\frac{2}{3} R^3 l}{R^2 m} - 1$$

$$=\frac{1+\frac{2}{3} \frac{R l}{m} }{1+2 \frac{R l}{m}} - 1 = \frac{1+2 \alpha }{ 1+\frac{2}{3} \alpha} - 1$$

where $\alpha = R l/m$ dimensionless parameter representing fraction of initial mass added by trip.

We assume $\alpha \ll 1$ and Taylor expand at $\alpha=0$ to find

Specific energy deficit after one trip $= -4/3 \alpha$

Looking more closely at alpha, write $\alpha = R A \rho/m$, where A is the cross sectional area I referred to and rho is the density of Earth.

$$R = 6.4 \times 10^6 m$$ $$A = 1 cm^2$$ $$\rho = 4.0 g/cm^3$$ $$m = 10^{20} kg$$

$\alpha = 2 R l/m = 2.56 \times 10^{-13}$ (fraction of BHs mass accumulated in half-trip, sounds good)

For change in height due to trip, use mgh approx and find

$$( - \frac{4}{3} \alpha ) 6,400,000 m = 2.18 x 10^{-6} m$$ Lower

It falls 2.18 micro meters lower at the end of the trip. Now, this scales directly with the area eaten, and thus with the square of the radius at which material is captured. To get a factor of 1e6, that radius would need 10 meters versus 1 cm.

Thus, dampening really IS SMALL, and the fate of the Earth would be dictated by how it eats matter while traveling at high speeds through the core. I'm going to go tell people now that the reason the LHC is underground is so that a BH won't poke out the surface if an accident happens. I love spreading disinformation.

Edit: This was my first answer given on physics SE, so I've gone back and put the equations in the right format, although the organization of the answer probably reflects its bizarre history.

Answer 7

The black hole of mass $10^{20}$ kilograms is not as dangerous as it may look. It is $10^{28}$ Planck masses, so the radius is $10^{28}$ Planck lengths or $10^{-7}$ meters. The gravitational acceleration near its event horizon is $10^{-10}\times 10^{20} / 10^{-14}$ which is $10^{24}$ meters per second per second. Even meters from the black hole horizon, the acceleration reaches the Earth's acceleration on the surface. Centimeters from the black hole horizon, the acceleration would be enough to break the matter.

The black hole would obviously try to find the minimum of the gravitational potential produced by the Earth, so it would sit and eventually stabilized at (and oscillated around) the Earth's center. If the Earth's center were imagined to be solid, I find it plausible that the black hole would break and eat some matter that is meters away from the micrometer-sized black hole. And the rest of the solid Earth could just sit around.

However, this is clearly not what would happen because the center of the Earth is liquid - because of the immense pressure. The liquid metals from meters around the black hole would simply flow to the black hole by reasonable speeds. The black hole would drink a lot of this fluid and its density would decrease in much of the Earth's core. It's questionable whether the solid layers of the planet could survive this reduced pressure. I can imagine that a black hole could actually be sitting inside the Earth and slowly drinking the liquid iron.

On the other hand, we know for sure that this is not what generically happens inside the celestial bodies because it would also take place inside gas planets, and those could be fully swallowed in a relatively shorter time, producing lots of rotation along the way.