Hypothetical water pressure question

As for the temperature point: If you take a fluid and compress it, it will heat up… But there’s nothing that says it’ll stay heated up. A low-viscosity fluid like water would be great for convective heat transfer, so any heat you initially had in the core would be quickly transported to the surface, whence it would be radiated into space. So absent some source of new heat (say, a nearby star), your liquid planet is going to end up getting awfully cold.

As for no “down”, this is correct, but irrelevant. Pressure does not have a direction, or equivalently, it pushes equally in all directions at a point. If I put a ping-pong ball in deep water (not necessarily at the center of a planet; the bottom of the Marianas Trench would do just fine), it’ll be pressed on in all directions: The top surface of the ball will be pushed down, the bottom surface will be pushed up, the left surface will be pushed to the right, the right surface will be pushed to the left, and so on. In short, all parts of the ping-pong ball will be pushed towards the center of the ball (not the center of the Earth), with the likely effect of crushing the ball. In much less than 20 minutes.

I should think that radioactive decay of naturally-occurring tritium would provide some heat, but I have no clue how to calculate it. From reasearching a prior Staff Report, I know that about 1/6700[sup]th[/sup] of the hydrogen found in water will be tritium–at least in our solar system. I don’t know enough about the tritium decay process to work out how much energy is released to contribute to heating.

I believe that tritium beta-decays to helium-3, so the energy released per decay would be the mass of a tritium nucleus minus the mass of a He-3 nucleus and an electron (we can neglect the mass of the antineutrino). But I can’t seem to find those masses to sufficient precision. And we can’t neglect the energy of the antineutrino, essentially all of which would be instantly lost to space, and I don’t know how much energy, on average, goes into the antineutrino.

It’s rather moot, anyway, since tritium has a half-life of only around 20 years, so it’d be almost entirely gone before the planet could shed its primordial heat. The only reason we still have any on Earth is it’s being continually replenished via energy from the Sun and cosmic rays.

Tritium decay releases 18.591 keV. On average, a little less than two-thirds of that will be wasted on the antineutrino. That leaves about 6 keV, or 10[sup]-15[/sup] joules, per decay available for heating. Taking QED’s quoted isotopic ratio and a half-life of 12.33 years, I find that a gram of water yields about 10 microwatts. But, yes, it will all be gone in a matter of decades without a source.

Well, the equation you’ve stated is the Ideal Gas Law (of which Gay-Lussac’s law is a special case dealing with temperature and pressure variations in a fixed volume). The Ideal Gas Law assumes “ideal” behavior, i.e. only momentum transfer via Van der Waals forces with no electrostatic interaction or nonequilibrium molecular states, which works pretty well for the noble gases or nonionized gas-phase fluids at low pressures and high temperatures, but not so well for gases that are relatively closer to the liquid state like air, and not at all for liquid water. (For the nonce, the Peng-Robinson Equation of State is a better approximation of the behavior of real-world gases under most conditions, including near condensation.)

With liquids, which are essentially incompressible, doing work (in the technical sense of the word) on a static fluid is difficult–nearly as much so as trying to compress a solid. The variation of temperature with pressure of a liquid is very small under normal pressures, and so the notion of a fluid becoming hotter as it is compressed, as gases do, is outside of our normal experience. Although this phase diagram for water doesn’t go much above 1 atm, you can see how both the solid and liquid states begin to turn upward at D and C respectively, offering very little increase in temperature with respect to pressure. Mineral oil in a hydraulic loop pressurized to 3000psi or 3500psi gets hot of course, but that’s mostly due to viscosity within the moving fluid; if it were pressurized to that temperature statically it wouldn’t get very hot or suffer much in the way of energy losses (which is a persuasive argument for the use of closed-loop hydraulics).

At the incredible pressures inside of a planetary-sized body of water, however, the temperatures from formation would be enormous, and the fluid would behave somewhat compressibly–in fact, it would be essentially indistinguishable from a highly compressed (and very nonideal) gas. We can SWAG the core temperature at formation by calculating the amount of energy required to create the static pressure that we would find at the center of the body (and since the pressure/temperature relation is so slight we can ignore the co-influence for our back-of-the-Publisher’s-Clearinghouse-envelope calculations) and from that, assuming an adiabatic (insulated) and quasistatic volume, estimate the temperature change between the initial temperature and the resulting core.

The implication, of course, is that our water-planet hasn’t radiated all of it’s heat away, but I’m assuming that this thing has just formed and hasn’t had time to come to an equilibrium; in any case, such a fluid body with a heavy molecular weight wouldn’t be stable, even over a short term. If it is in orbit of a star, tidal forces will draw it into an elongated barbell shape, and then shear the globules clean away in short order, ad infinium. And there are no doubt, as the OP notes “about eight million reasons why this couldn’t occur,” but since we’re playing with a thought experiment we’ll let it go at that. Besides, I need to see about the cat; Erwin dropped it by this morning in a box and since it stopped meowing I’m a little worried that it’s waveform might have collapsed.


Well, Kevin Costner would be the main character. Sorry. :wink:

Only if the planet is inside the star’s Roche limit. And for that matter, a rocky planet inside the Roche limit would be torn apart, too. If we put our water planet at the same size and orbit as Earth, the tides would only be a little higher than on Earth, and that only because the planet’s density (and therefore mass) would be less.

I do agree that the core of this planet would probably not be a liquid per se, but I also don’t think it would properly be called a gas. I don’t know what its equation of state would be, though.

I say your internal pressure would be equal to the pressure of all that water, since you’re mostly water yourself. Water loses virtually no volume when compressed, though I’m not sure if there is a limit. I decline to consider the temperature issue.

If you’re breathing air at the same pressure, everything’s OK until you start to go back to the surface. Then you get the bends because the non-oxygen gasses in your air probably got dissolved into your blood under the high pressure, and now they’re bubbling out. Oh, and you’d better open your pipes and breathe out the air as it keeps expanding in your lungs. Don’t worry, you’d be surprised at how much is in there.

But if you took a huge breath of air at the surface and went down to where the pressure is 1000 times as high, well the air in your lungs would get compressed to 1/1000 the volume, and your ribs would break, so don’t do it that way, especially since it’s hard to hold your breath long enough to go a thousand miles or more under water.


Quercus, CJJ, QED* and others. Thanks. Now I get it. It only took a moment to imagine the planet sectioned into hemispheres and consider the forces required to hold everything that way - and then imagine standing between them when they smacked back together! squish.

I love this place. :slight_smile:

You wouldn’t get crushed if your lungs were full of something dense, like water or oil. Either of which can carry oxygen, just as your blood can.
That’s how deep sea creatures survive, after all.

That is, they don’t have lungs, but their bodies are basically full of water with sacs of oil, and are not compressable.

Urg…yeah, that’s right. Mea culpa.