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?
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 &quot;geometric&quot; 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 &quot;more or less&quot; 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 &quot;small&quot; 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 &quot;ripped away&quot;. $$\frac{G^2M^2R\rho}{v^2}\log\frac{Rv^2}{GM}$$
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.
A black hole moving 1000 times slower (typical solar system velocity) would essentially destroy the Earth.