Marine biologist Andrew David Thaler recently made headlines when he used Google Earth and other freely available software to simulate the effect of catastrophic sea-level rise. He called this project #DrownYourTown, and while many of the examples are far-fetched (flooding Denver, anyone?), his basic point was simple: as the effects of climate change increase, we’ll see ocean levels rise. Many people living in coastal cities will need to relocate as polar ice melts, and we see water encroaching on the coastlines.

But of course, anyone who has watched disaster movies or played SimCity knows that there’s a bit of morbid fun to be had in imagining impossible catastrophes — so the drowning of (say) Des Moines can feel satisfying, because it won’t happen. (Thaler’s model is merely static sea-level rise, not including the effects of tides, much less rivers and the like, which of course produce a common type of transient flooding.) As someone who lives in Richmond, Virginia, a city on a tidal river, I found some of the images of flooded cities to be a little disconcerting.

However, I can’t resist this sort of thing:

Alright physics tweeps, a challenge! Assuming the 8800 m sea level rise for Waterworld, how would that much extra mass affect the Earth?

— Andrew David Thaler (@SFriedScientist) October 25, 2013

I was half-asleep when this call to arms came in, but I did a quick and dirty estimate, which turned out to be wrong, so consider this blog post to be my correction! The question: how much water would be required to flood the entire globe to a depth of 8,800 meters? That’s deep enough to cover Mount Everest, as in the terrible movie *Waterworld*. Even the most dire of climate change models doesn’t predict drowning all the continents: as we’ll see, this is a *lot* of water.

The most catastrophic-yet-realistic predictions Thaler considered involve 80-meter sea rise. Now 80 meters is 262 feet, which is also known as pretty friggin’ deep. Richmond, with an average elevation of about 45 meters, is completely gone at that level of flooding (with the usual caveats about topography and the like). 8,800 meters, or 29,000 feet, is more than 100 times higher, but correspondingly much larger when considered in terms of the volume of water.

Physicists are well known for our simplified models; we invented the spherical cow, after all. To estimate the mass of water required to flood Mount Everest, I assumed Earth is spherical and ignored the variations in surface elevation. Then I estimated the volume of water, using a spherical shell with a thickness of 8,800 meters.

The (average) radius of Earth is about 6.4 million meters, so even drowning the planet to 8,800 meters doesn’t really increase the size of the planet that much. Therefore, we shouldn’t expect adding even a relatively large amount of water to affect Earth’s mass substantially: rock is much denser than water. So, without further ado….

### Drowning the whole world ~~in tears~~

Melting all the polar and alpine ice wouldn’t raise the sea level this much, so we’d have to add the water from some imaginary extraterrestrial source. (I’ll discuss another possibility below.) To flood the world to a depth of 8,800 meters would require adding about 4.5×10^{18} cubic meters of water, with a mass of approximately 4.5×10^{21} kilograms. (The details of this calculation are in the appendix below.) That’s an incredible amount of water: all of Earth’s oceans as they actually exist only contain 1.3 ×10^{18} cubic meters, with a mass of about 1.4×10^{21} kg. Since freshwater constitutes less than 3% of all water on Earth, we’d have to more than double the amount of surface water to flood Mount Everest.

The original question asked how all this extra water would affect Earth. Our planet’s mass is about 6.0×10^{24} kg, so we’d be adding about 0.075% of Earth’s mass in the form of water. In other words, even though we’re adding a *lot* of water, Earth is so massive that it doesn’t make a huge difference. Similarly, the increase in Earth’s radius is about 0.14%, another tiny amount.

Nevertheless, with our precision clocks on the ground and on satellites, we’d notice! Adding to the mass and radius would both increase Earth’s rotational inertia (or *moment of inertia*, if you want to use the funky physics jargon): the resistance to spinning. All else being equal (see the Appendix for what I mean by that), spontaneously increasing the moment of inertia would slow Earth’s rotation down, lengthening the day by 242 seconds, or about 4 minutes. Compared to the length of the day, that’s not much, but 4 minutes is a significant amount of time in the absolute sense; certainly it’s measurable!

Of course, there’s another effect that will play a role: the Moon’s tidal force on Earth. The tidal force is the difference between the pull on the sides of Earth nearest and farthest from the Moon, due to the Moon’s gravity. The net effect of the tidal force is to slow Earth’s rotation; our planet rotated faster in the past. Increasing Earth’s size by adding water will increase the tidal force very slightly, slowing our rotation even more. I didn’t do that calculation for this post, but the increase in the length of the day will be less than the 4 minutes above.

Now, as a final note: while there’s certainly not enough surface water to flood the entire planet, is there enough water in Earth’s interior to do the job, assuming it could be liberated somehow? I’ve had trouble coming up with specific numbers, possibly because we only have rough estimates, but it appears there’s more water in Earth’s mantle than there is on the surface. It’s not in the form of subterranean lakes: it’s more small numbers of water molecules mixed in with the crystal structure of the mantle rocks, which are under such high temperature and pressure that they behave like plastic.

This water was likely part of the raw materials that built Earth about 4.5 billion years ago, so it was never on the surface, and never will be. However, since we’re considering fantastical scenarios anyway, let’s pretend it somehow can bubble up and flood the world. (Creationists, don’t even.) In that case, Earth’s mass doesn’t increase — the mass is just moved around — but the size still goes up a little. As it turns out, the length of the day is more sensitive to increases in size than mass on this scale, so we’ll still end up with about 4 extra minutes.

Yet again, this is not a real scenario: it’s a fun “what if?” exercise. Nevertheless, the dangers of climate change are real. A substantial fraction of Earth’s population lives near the coast: many large cities across the globe are vulnerable in the event of rising oceans. While an 80-meter sea-level rise is not going to happen overnight (and may not actually come to pass), a rise of just a few meters would still have devastating consequences in many places. Maybe not Richmond, maybe not Los Angeles, but Bangladesh and India, Florida and Pacific island nations will suffer.

### Appendix: Calculation details

It’s pretty easy to calculate the volume and mass of the water required to flood Earth to a depth of 8,800 meters. Play along! I’ll try to make each step clear, and point out where I’m making estimates for simplicity.

First, let’s find the volume. As I pointed out above, we can think of this extra water as a spherical “shell” swathing Earth, with thickness equal to 8,800 meters. The easiest way to calculate the volume of this shell is to pretend we have a big sphere with a radius equal to Earth’s radius plus 8,800 meters, then subtract Earth’s volume. (The diagram above shows how that works.) However, we can also be clever, since we know that Earth’s radius is a lot bigger than 8,800 meters.

The volume of a sphere is where *R* is the sphere’s radius. In the diagram above, we called the depth of the water *h*, so the volume of floodwater to drown the world isHowever, anyone who has done computer programming knows that it’s easy to introduce errors when you subtract two numbers that are very close to each other in value. So, let’s do a trick: using high school algebra, we can pull the common factors out, then multiply and divide the whole equation by *R* (which is the same as multiplying by 1):This may *look *more complicated, but it’s not! The stuff outside the curly brackets is just the volume of Earth, which you can look up online.

Everything inside the curly braces is easy to calculate: *h* is 8,800 meters, or 8.8×10^{3} m, and *R* is 6.4×10^{6} m, so *h/R* = 0.0014. And we’re (nearly) done: once we plug in Earth’s volume, we get a floodwater volume of 4.5×10^{18} m^{3}. The mass of this water can be found by multiplying the volume by the density of seawater, which is about 1,025 kilograms per cubic meter. The result: a mass of 4.5×10^{21} kilograms. (I’ve assumed the density is constant, even though it would be slightly higher at the bottom of the new global ocean and lower at the top.)

For calculating the length of the day, we need a bit of physics. Without an external force (actually torque) acting, the rotational momentum of Earth before and after the water is added will be the same. For reasons of tradition, we write rotational momentum as *L*, and rotational inertia as *I*:I skipped a few steps, but *P* is the length of Earth’s day, also known as the period of its rotation. We could calculate the inertia before and after, but if we rearrange, we see that what we really need is the ratio:Again, glossing over the details, the rotational inertia of a regular object is proportional to its mass multiplied by the square of its size, soWe already found *h/R*, and the ratio of the water mass to Earth’s mass is 7.5×10-7, which is so tiny that the first expression in parentheses is basically equal to 1. As a result, we find that That’s 242 seconds longer than the normal Earth day, or about 4 minutes.