I was inspired by the APOD pictures and discussion here and here. The conditions for the ‘experiment’ are:

For some mysterious reason, all the water on Earth (minus that in living things—the biosphere) suddenly & immediately converges into 17 identical sphere and let to drop down to the land below right away.

The places of those water balls are the centers of 17 biggest tectonic plates as listed here.

The bottoms of the huge balls just touch the highest points on the ground there. Also from my calculation, which may be wrong, their radius is 258 km.

The ISS is currently at its highest altitude of 460 km with ship(s) in dock.

*So, what will happen to us, the Earth and things on it? *

My prediction is very grim for humanity, thus I turn to the ISS as our last hope. Supposed that at t=0, the astronauts can choose the station’s direction of flying. Can they avoid all the 17 falling balls? Is there any chance they can land back on Earth without any help from Houston when things settle down, in case the water *does *settle down and not all turn into vapor? (That’s a possibility I can’t calculate and will need your help).

I’m looking for answers in xkcd style, which IMO is a textbook way to respond to absurd “What if?” questions. It’s detailed, with nice pictures to help our imagination, displaying formulas when necessary but also very rookie-friendly, tackle the problem from many different angles & POVs, and takes into account some easy-to-overlook stuffs. Too bad he doesn’t answer anymore.

Anyway, thank you just for reading! :)

Sorry for the late answer, I intended to write this 2 weeks ago but couldn’t find the time.

OK, so let’s look at the amount of potential energy locked up in our configuration of water spheres: I calculated the radius to be r=√34π⋅117⋅1.35⋅1018m3=266.644 km. By symmetry, the potential energy would be the same if all the water were located at the center of these spheres, i.e. 133.322 km above ground. Add negative potential energy of Earth’s oceans in its equilibrium state (1.844 km, half the average ocean depth), and we get a total of g⋅h=9.81ms2⋅135166m=1.325MJkg

That’s a lot of energy. Let’s assume the water is at 0°C initially (most ocean water is in the cold deep layers). To bring all this water to the boiling point, we’d need 100K⋅4.186kJkg⋅K=0.4186MJkg. That leaves us 0.9064MJkg to boil the water. Given the latent heat needed, we can boil 906.4kJkg2257.92kJkg=40.14% of the water.

Since Earth’s oceans is 262 times as massive as the atmosphere and we boiled almost half of it, we now have an atmospheric pressure of 105 bars, exceeding that of Venus. Of course, the boiling point of water increases with pressure, but the latent heat of vaporization decreases and these two effects pretty much cancel each other out. It does mean however that we’ll have a surface temperature of 315°C, approaching that of Venus. In other words, all life on Earth’s surface, including the hardiest extremophile bacteria, is toast (or rather steam buns). Absolute overkill actually, since DNA itself disintegrates completely at around 200°C. The only safe place would be deep underground where the heat can’t penetrate until it is radiated off into space.

What about seeking refuge in the ISS? Let’s see what the atmospheric conditions are at its orbital height of 400 km. We’ll use the barometric formula without temperature lapse because this isn’t an equilibrium state anyway. ρ400km=ρ0⋅exp(−g⋅M⋅hR⋅T0)=58.94kgm3⋅exp(−9.81ms2⋅0.018kgmol⋅400000m8.314JK⋅mol⋅588K)=58.94kgm3⋅e−14.44=3.14⋅10−5kgm3This is equivalent to the air density of our current atmosphere at 60 km height.

Here’s a video of the Mir space station at 80 km height.