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Digging a hole through earth is a common thought experiment, often used to explain effects of gravity.

But what would happen if someone finally dug the hole?
Sure, he took care to stabilize and isolate it inside with two and a half layer of unobtanium.

But he forgot to check what exactly is on the other side of the earth, the antipode.
And, just like surprising 85% of earth [travel.SE: antipodes], it turns out to be ocean.

The tunnel is under vacuum, and about the diameter of a two track railway tunnel. In the last step building it, obviously the ocean floor broke, and water is streaming in.
What happens then?

I'm aware that a body would oscillate through the core - the water can not, if the ocean does not run dry assume - at least if the water coming in can mostly fill the diameter.

First, I thought the water will just fill the hole, and directly end up in an equilibrium regarding gravity and pressure. But that assumes the water, filling the diameter at the ocean ground will do so in the whole tunnel; But doesn't the stream of water get thinner, as it accelerates up to the center of the earth, and thicker from there? So will it oscillate for a while?
Would it change if we assume the hole in the ocean floor is much smaller than the tunnel's diameter?

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    $\begingroup$ Another candidate for what-if.xkcd :-) . $\endgroup$ – Carl Witthoft Jul 14 '14 at 12:09
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    $\begingroup$ You are positing that they can deal with the heat, pressure and in some places fluid environment of the deep Earth but are going to have trouble with a minor inconvenience like an ocean? Seriously? Just wave the same unobtanium at it that you use on the liquid part of the core. Problem solved. $\endgroup$ – dmckee --- ex-moderator kitten Jul 14 '14 at 13:29
  • $\begingroup$ You'll get a hypersonic schock wave propagating through the wall. When the hole breaches the ocean floor at, say, 10 km depth, you have water at 1000 bar entering the vacuum. The water that rushes in will accelerate and the friction will heat it up, making it superheated steam. $\endgroup$ – Count Iblis Jul 14 '14 at 15:13
  • $\begingroup$ @CountIblis I can't resist suggesting you misspelled "schlock (wave) " there :-) $\endgroup$ – Carl Witthoft Jul 14 '14 at 17:08
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By my rough calculations, a fall to the center of the Earth will raise the temperature of water by some $7,200 K$. This is extremely hot, but I believe it falls short of a full plasma ionization. Additionally, I would put the pressure at the center somewhere around $30 GPa$. That's a "G", not an "M". I can't even look up properties because these conditions are so extraordinary. However, I believe that we will be looking at a supercritcal fluid either way.

I take this question to refer to the "fast" event where water is let in. If it was left to reach thermal equilibrium over a billion years, the answer might be slightly different.

Regarding the fall itself, I would go ahead and assume that the fluid quickly reaches its own form of terminal velocity. Even with a 12 meter pipe, and viscosity reduction due to phase transition, the Earth is huge relative to the pipe. Frankly, that means that there's little kinetic energy by the time it reaches the center as opposed to thermal energy. In order for it to make it back up to the other side of the hole, it has to be pushed through by the back-pressure, which is also enormous. This process will take much longer than the time required to complete half of an Earth orbit (about 40 minutes), which is what would be involved for a free-fall. Maybe a few hours, it's an interesting question.

Also keep in mind that energy is liberated in the process. You stored up a lot of energy by building a vacuum tube through the Earth. As the water falls, that gravitational potential energy is converted into thermal energy in the fluid.

Play this movie out, and you realize that the fluid won't stop moving at the center of the Earth... or the surface on the other side. "But wait!" you protest, this like like a pendulum, it has to climb the same gravity well that it fell through. Yes, but its density changed due to the heating. That means that the falling fluid is much more dense than the climbing fluid. The climbing fluid will undergo a relatively isentropic expansion process, but its enthalpy will always end at a higher value than the original inflow (because the only net heat flow is frictional heating).

Because of this, a geyser would sprout at the land-based entrance to the tube. It should probably release a volume similar to the total volume of the tube, and it will be at massively super-critical temperatures. This will be very dangerous, but I predict that it'll fall short of climatic scale impacts.

In fact, the geyser would eventually stop, because all the energy is coming from liberating a finite gravitational potential energy source. Then there would be some near-term equilibrium state where the tube's fluid is significantly heated, but the driving force has basically puttered out. As this process happens, the flow rate goes down, and so does the thermal energy of the geyser. Eventually it might just be regular steam that it belches out, and then water, although very hot water. Toward the very end of this process, it'll just be lukewarm water pushed out by a more mild driving force. However, depending on the elevation difference, this could continue to pump a large volume, larger than the volume of the pipe I think.

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  • $\begingroup$ How did you end up at $7,200K$? $\endgroup$ – Bernhard Jul 15 '14 at 11:29
  • $\begingroup$ @Bernhard The difference in gravitational potential is energy per unit mass, so divide by specific heat (which is only an approximation), which is energy per mass per Kelvin. You're left with Kelvin. $\endgroup$ – Alan Rominger Jul 15 '14 at 12:23
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There would probably be no nice oscillations, it would be rather chaotic and it would be insanely diffcult to calculate exactly. One big question is the incredible temperature in the core which would have significant effects on the whole process. Let's assume the temperature can be somehow isolated by the impossible material that you are using.

The water stream indeed does get "thinner", as it accelerates towards the core, but in fact it would not turn into a thin continuous stream. It would rather become a collection of drops, something like a rain. If you had a huge hole open into a massive vacuum tunnel, the water would be entering the hole quickly and it would boil and evaporate under the vacuum during the fall.

The volume of the tunnel would be around 1.44 cubic kilometers (assuming diameter of 12 meters), so this would first need to be filled with vapor to provide enough pressure to stop the boiling (part of the water would disappear), but after some time the water (in the shape of rain) would finally get to the opposite side of the Earth. The first drops would start falling back towards the core, but they would immediately start colliding with the drops following them, which would slow down their free fall.

Now the vapor becomes a problem, because there is more and more water entering the tunnel. I'm assuming the opposite side of the tunnel is sealed (otherwise you won't have vacuum in the tunnel at the beginning), so the vapor (that first provided pressure) would become compressed, as most of the tunnel becomes filled with liquid water and more is getting in. That would affect the behavior of the system. You would end up with a "bubble" on one side of the tunnel. But if you open the seal (once the pressure inside the tunnel equals the atmospheric pressure outside) to let the vapor get out, things become a little bit simplified. But not much.

After some time the water fills the whole tunnel and the pressure it the core is massive. I'm assuming the opposite end of the tunnel ends at the same distance from the Earth center as the ocean level on the entry end. The water from the ocean floor is entering the tunnel with some initial velocity, as it is under pressure already.

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