# How much damage would collision with a similar mass primordial black hole do to earth?

All this talk of Planet 9 perhaps being a primordial black hole (PBH) made me wonder how dangerous a PBH collision with earth would really be.

Specifically, if earth collided with a PBH of 1 earth mass, traveling fast enough that they didn't end up orbiting each other, what would happen?

Edit: From a paper asking Can one detect passage of small black hole through the Earth? the analysis was that for an approximately proton-volume and mountain-mass sized PBH

"It creates a long tube of heavily radiative damaged material, which should stay recognizable for geological time."

but is otherwise pretty hard to detect.

What would change if the PBH were the mass of the earth and the volume of a large marble? How slow would it have to be going before it really messed things up?

• The tidal effects and the energy released by the accretion of part of earth's mass as it passed near and through the earth would be disastrous, although just how disastrous I can't say. I'm guessing no atmosphere left and the remaining earth largely molten. Oct 11, 2019 at 17:29
• an earth mass black hole would have a diameter of about 1cm, rather less than a baseball (though qualitatively not much difference) Oct 11, 2019 at 18:44
• @joseph.hainline at a distance of 640km, the force of gravity of the earth-mass black hole is 100 gees. At a distance of 200 km, the gravity from the earth-mass black hole is 1000 gees. And so on, by Newton's Law of Universal gravitation. The damage is going to be less than if an Earth-mass planet smacked into Earth, but it's still going to be ridiculously devastating. Oct 11, 2019 at 19:28
• On the other hand, if the black hole is coming from the outer solar system, the encounter would be over in a fraction of a second. It is not obvious (to me) how much damage the gravitational (tidal) forces could do in such a short time scale. Similarly, this timescale seems short for the formation of accretion dynamics. I think my biggest worry would be the exchange of momentum between the black hole and the earth and the impact on the Earth's orbit. Oct 15, 2019 at 11:24
• @Chappo Coming in from the outer solar system would result in a relative velocity to earth on the order of 20-90 km/s. 90 km/s spends more than two minutes passing through the earth. Coming from interstellar speeds I probably wouldn't expect more than a few hundred km/s relative, as presumably the object is generally orbiting with the rest of the galaxy. Not a long period of time, but not a fraction of a second. That said, we're really getting into worldbuilding.stackexchange.com 's territory at this point. Oct 16, 2019 at 12:03

Consider an initially stationary particle of matter and suppose a 1 Earth mass black hole flies past it at speed $$v$$ on a trajectory that passes the initial position of the particle at distance $$r$$. The particle will be mainly affected by the gravity during a time period of roughly $$r/v$$ (up to some "geometric" constants), during which time it will accelerate at roughly $$GM/r^2$$. This means that it will move a distance about $$GM/r^2 \times (r/v)^2 = GM/v^2$$. So if this is small compared to $$r$$ it will be left "more or less" where it was, rather than being ripped away. So we can expect a hole of diameter a few times $$GM/v^2$$. $$GM$$ is about $$4\times 10^{14}$$ so to keep the hole diameter to say 1m you could need $$v$$ about $$2\times 10^7 m/s$$ which is about $$0.07c$$.
We can also estimate the typical velocity change achieved by our test particle from this interaction, which is $$GM/r^2 \times r/v = GM/rv$$, giving a KE per unit mass of about $$G^2M^2/r^2v^2$$.
So, suppose the Earth has density $$\rho$$ and radius $$R$$ the cylindrical shell of thickness $$dr$$ will mass $$R\rho rdr$$ (still ignoring "small" constants like $$2\pi$$). That shell will acquire kinetic energy $$G^2M^2R\rho dr/rv^2$$ from the interaction. Integrating from $$GM/v^2$$ to $$R$$ we get a total energy deposited in the parts of the Earth which are not "ripped away". $$\frac{G^2M^2R\rho}{v^2}\log\frac{Rv^2}{GM}$$
Using $$R = 6\times 10^6$$, $$GM = 4\times 10^{14}$$, $$\rho = 5000$$, and $$v = 2\times 10^7$$ (all SI units) we get about $$10^{25}J$$, not enough to actual destroy the Earth (Earth's gravitational binding energy is about $$10^{32}J$$) but about $$10^{10}$$ megatons or an earthquake of about 13 on the Richter scale.