# Was the filling of the Three Gorges Dam's impact on the Earth's rotation rate detectable?

I'm a big fan of Vsauce and the video How Earth Moves is just one example of science related to the Earth available there (there's plenty more).

But the statement in this video starting at 01:39 strikes me as a little surprising.

You know how a figure skater, spinning in place, can slow down their (rotational) speed by extending their arms out; by moving some of their body mass away from the middle of their body? Well the same thing can happen to Earth.

The Three Gorges Dam did exactly what a figure skater does when they move their arms away from their center. It transferred thirty nine trillion kilograms of water one hundred and seventy five meters above sea level.

NASA calculated that that massive amount of water moved, caused Earth’s rotation to slow down, so that every day of your life since that dam was finished, has been longer by 0.06 microseconds. (emphasis added)

If I calculated correctly, that would be a "slip" of about 1 centimeter per year at the equator.

There are many causes of change to the Earth's distribution of mass, it's moment of inertia, and therefore the length of a day. Tidal forces from the Moon can contribute substantially to changes in Earth's rotation as well, and there could be short term geological effects as well.

Is this roughly 0.06 microsecond/day shift well below the "noise" level, where in this case I'm using "noise" to refer to effects that are smaller than the error that can be modeled? So for example exchanges between sea ice and sea water can be measured and modeled to some level of accuracy, and the Moon's tidal effects can be measured and modeled to great precision, so even if the Three Gorges Dam effect turns out to be small compared to those, it could still be detected as long as its effect is larger than the error that those effects can be accurately calculated.

below: Screen shots from the Vsauce YouTube video Water.

• To be fair to VSauce, the video description linked to the Wikipedia page on the dam, which makes the same claim. Wikipedia cites this page from JPL, so there may be additional calculations. Not that I don't trust the math in the accepted answer, but there may be more nuances to this calculation that aren't accounted for. I don't yet have a primary source for the quote in that page, however.
– Cody
Dec 20, 2018 at 18:40
• The change in inertia of water compared to the change in inertia of the whole earth is not the correct comparison. The Crust of the Earth spins at a different rate than the core. Only consider the inertia of the crust. Feb 3, 2022 at 18:46

One needs to calculate the change in the moment of inertia of the Earth and use conservation of angular momentum (the rotation period is proportional to the moment of inertia). Most of the water will ultimately come from the oceans, effectively removing a thin layer of water.

Jerry Mitrovica discusses this effect (in reverse) in a Nautilus interview:

Is water moving off glaciers, slowing the Earth’s rotation, this time analogous to a figure skater putting arms out?
Right. Glaciers are mostly near the axis. They’re near the North and South Poles and the bulk of the ocean is not. In other words, you’re taking glaciers from high latitudes like Alaska and Patagonia, you’re melting them, they distribute around the globe, but in general, that’s like a mass flux toward the equator because you’re taking material from the poles and you’re moving it into the oceans. That tends to move material closer to the equator than it once was.
So the melting mountain glaciers and polar caps are moving bulk toward the equator?
Yes. Of course, there is ocean everywhere, but if you’re moving the ice from a high latitude and you’re sticking it over oceans, in effect, you’re adding to mass in the equator and you’re taking mass away from the polar areas and that’s going to slow the earth down.

The contribution of the removed water to the moment of inertia depends on the distance from the axis and hence on the latitude. This is a simple calculation if we assume the world is all ocean.

The moment of inertia of the lake is $$m(R\cos L)^2$$ and the moment of inertia of a spherical shell is $$\frac{2}{3}mR^2$$, where m is the mass of water, $$R$$ is the Earth's radius and $$L$$ is the latitude of the lake ($$30.82305$$ degrees for Three Gorges). The relative change in the moment of inertia $$I$$ of the Earth is then $$\frac{mR^2}{I}(\cos^2 L - \frac{2}{3}) = \frac{39 \times 10^{12} \times(6.37 \times 10^{6})^2}{8.04×10^{37}}(\cos^2 L - \frac{2}{3})$$ $$=1.97×10^{-11}(\cos^2 L - \frac{2}{3})$$

Multiplying by the number of microseconds in a day ($$8.64 \times 10^{10}$$) gives $$1.7(\cos^2 L - \frac{2}{3}) = 1.7 × 0.071 = 0.12$$ microseconds.

Why the difference from NASA's $$0.06$$ ? Note that the expression changes sign at $$\cos^2 L = \frac{2}{3}$$ or $$L ≈ 35$$ degrees (pretty close to the latitude of Three Gorges). The Earth will actually speed up if the lake is at high latitudes and slow down if it is at the equator. The $$\frac{2}{3}$$ term comes from the "all ocean" assumption.

As I understand this paper, the $$\frac{2}{3}$$ term should be multiplied by $$\frac{1.414}{1.38}$$ to account for the shape of the oceans (search in the PDF for those numbers), resulting in $$0.09$$ microseconds.

# The .06 microseconds figure doesn't sound accurate

The moment of inertia for a solid sphere is $$I = \frac{2mr^2}{5},$$ where $$m$$ is the mass of the sphere and $$r$$ is its radius. Applied to the Earth, this gives us $$I = \frac{2\cdot5.972\times10^{24} \text{ kg} (6371000 \text{ m})^2}{5} = 9.70\times10^{37} \text{ km-m}^2.$$

A simplified estimate of moment of inertia for the water in the three gorges dam is $$mr^2$$. The change in moment of inertia from raising that water by 175m can be calculated by \begin{align}\Delta I &= m_{water}\left[(r_{earth}+175)^2-r_{earth}^2\right] \\ &= 39\times10^{12}\left(6371175^2-6371000^2\right) \\&= 8.69\times10^{22} \text{ km-m}^2.\end{align}

Assuming that the angular momentum ($$L$$) of the Earth stays constant, then since $$L=I\omega$$, $$I_{before}\omega_{before} = I_{after}\omega_{after},$$ or $$\omega_{after}=\omega_{before}\frac{I_{before}}{I_{after}}.$$ The angular velocity, $$\omega$$, of the Earth (before) is $$7.292\times10^{-5} \text{ radians/s}$$. The discrepancy between the before and after moment of inertia is so insignificant that I would be wary of floating point error on this calculation, but using Python2 I calculate the delta in rotational speed to be $$6.776\times10^{-20} \text{radians/s}$$. For a rotation of the Earth this means the rotation will take about 80 picoseconds longer, or 0.00008 milliseconds.

# Conclusion

The statements made by VSauce don't match up with Physics 101. I am not inclined to believe this claim unless there is some other explanation or compelling evidence. Given that the magnitude of the change in rotational speed is three orders of magnitude less than he claims, and that 80 picoseconds is about 1/4 of the time it would take your computer to add 2 integers, I think it is safe to say that this is not measurable.

• Yep, I tried it following your outline and got the same thing; three orders of magnitude off, and still a factor of 3 off if it was supposed to be per year instead of per day. While Earth's non-uniform density results in a less than 20% difference (Moment of Inertia factor of about 0.33 rather than $2/5$) there's still seems to be something wrong as stated. Time for someone to send a Vsauce Alert?
– uhoh
Mar 31, 2017 at 1:21
• @uhoh Thats a good point, I should have accounted for the Earth's variable density. Good thing it still worked out so I didn't look foolish :) Mar 31, 2017 at 13:06
• Not at all! This is just the analysis one should do first. The only reason I dug deeper was to look into the "per year" vs "per day" possibility for a last factor of three, but the uniform sphere is already so darn close!
– uhoh
Mar 31, 2017 at 13:27
• This has been Asked again. Your calculation would be correct if we dammed off an ocean bay of volume 39 cubic km and then pumped it up 175 m leaving the bay dry. I don't think that is what is happening, as I explain in my Answer there. Jan 12, 2019 at 19:08
• Using your method I get .00008 microseconds, in agreement with @uhoh . You can avoid the floating point error issue by $(r-175)^2 -r^2 \approx 2 \times 175 \times r$. Jan 12, 2019 at 20:16