What would the kinematic of a black hole falling on an object look like?

Question

I wonder what would the kinematics of a black hole falling on an object be.

To make it realistic, I got told that the black hole needs to have a radius bigger than atoms, so that it can swallow them. I think atoms have a radius of $10^{-10}m$, so let's make a BH with a radius of $10^{-9}m$ which if I'm not mistaken gives it a mass of $10^{18}kg$ which is still far lighter than the earth ($10^{24}kg$).

Since the BH is lighter than the earth, from a stationary observer, it's the BH that we see falling towards the earth. At long range, my understanding is that the BH looks just like a normal object falling. But how does it behave as it gets closer ? Does it end up touching the ground, and piercing through it and going to the core, or something else?

Answer 1

1. Ignore the fact that the BH can swallow atoms. It probably won’t do so very quickly (from both the spacing between atoms and time dilation effects as it moves quickly through space after reaching the Earth). Think about it like this: a high-speed nanometer-sized black hole would create extreme time dilation for those atoms that it’s about to “hit”, such that atoms being pulled in from space adjacent to the BH’s path don’t end up actually within it, so the total volume “consumed” would just be a 1 nm-by-~6500 km column, which will not have significant mass.

2. Ignore the physical matter of the Earth completely. The BH’s mass is so much higher than that of an atom that hitting an atom (or a quadrillion atoms) elastically or inelastically is not really going to affect it’s momentum substantially. Its momentum will be proportional to its mass, which is a mind-boggling $10^{18}$-ish kilograms, compared to around $10^{-27}$ kilograms for hydrogen atoms. Hitting an atom of hydrogen (at rest) could thus reduce its momentum by a factor of at most $1-\frac{10^{-27}}{10^{18}}=1-10^{-45}\approx1$.

The BH will pass through the Earth, not slowing down substantially, and then fly out the other side. If it was launched with negative total mechanical energy (i.e. within Earth’s gravity well), it will remain in orbit. If it was launched with positive or zero total mechanical energy (i.e. outside Earth’s gravity well), it will fly off towards infinity.

What you have left is an idealized mass moving in an idealized gravity well, perhaps one best described by a twofold interior/exterior Schwarzschild/Lense-Thirring metric (or just Newtonian gravity if you don’t care about perfect accuracy). The kinematics will be described by basic orbital mechanics: the black hole’s total mechanical energy will remain constant (barring very minor changes to its mass as it absorbs, which will be very little on each pass). If you somehow keep it from evaporating (and thus destroying the planet in a supernova-like explosion), it would, eventually, consume the entire planet through this process. It would take longer than the Earth has to live due to a dying Sun is my bet, but without information on the tightness of its orbit I can’t fact-check that.