<h1>The .06 microseconds figure doesn't sound accurate</h1>

<p>The <a href="https://en.wikipedia.org/wiki/List_of_moments_of_inertia" rel="nofollow noreferrer">moment of inertia</a> for a solid sphere is <span class="math-container">$$I = \frac{2mr^2}{5},$$</span> where <span class="math-container">$m$</span> is the mass of the sphere and <span class="math-container">$r$</span> is its radius. Applied to the Earth, this gives us <span class="math-container">$$I = \frac{2\cdot5.972\times10^{24} \text{ kg} (6371000 \text{ m})^2}{5} = 9.70\times10^{37} \text{ km-m}^2.$$</span></p>

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

<p>Assuming that the angular momentum (<span class="math-container">$L$</span>) of the Earth stays constant, then since <span class="math-container">$L=I\omega$</span>, <span class="math-container">$$I_{before}\omega_{before} = I_{after}\omega_{after},$$</span> or <span class="math-container">$$\omega_{after}=\omega_{before}\frac{I_{before}}{I_{after}}.$$</span> The angular velocity, <span class="math-container">$\omega$</span>, of the Earth (before) is <span class="math-container">$7.292\times10^{-5} \text{ radians/s}$</span>. 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 <span class="math-container">$6.776\times10^{-20} \text{radians/s}$</span>. For a rotation of the Earth this means the rotation will take about 80 picoseconds longer, or 0.00008 milliseconds. </p>

<h1>Conclusion</h1>

<p>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. </p>

CC BY-SA 3.0
2017-03-31T01:21:07.730Z
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](https://en.wikipedia.org/wiki/Moment_of_inertia_factor) of about [0.33](https://en.wikipedia.org/wiki/Earth) rather than $2/5$) there's still seems to be something wrong as stated. Time for someone to send a Vsauce Alert?
uhoh
CC BY-SA 3.0
2017-03-31T13:06:24.343Z
@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 :)
kingledion
CC BY-SA 3.0
2017-03-31T13:27:28.067Z
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
CC BY-SA 4.0
2019-01-12T19:08:56.877Z
This has been [Asked again](https://earthscience.stackexchange.com/q/15907/6535). 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.
Keith McClary
CC BY-SA 4.0
2019-01-12T20:16:25.187Z
Using your method I get .00008 microseconds, in agreement with @uhoh . You can avoid the floating point error issue by $(r-175)^2 -r^2 \approx 2 \times 175 \times r$.
Keith McClary
CC BY-SA 4.0
2017-03-30T18:53:46.720Z
2019-01-12T20:17:10.600Z
2019-01-12T20:17:10.600Z
kingledion
<p>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.</p>
<p>Jerry Mitrovica discusses this effect (in reverse) in a <a href="https://nautil.us/issue/62/systems/why-our-intuition-about-sea_level-rise-is-wrong-rp" rel="nofollow noreferrer">Nautilus interview</a>:</p>
<blockquote>
<p><strong>Is water moving off glaciers, slowing the Earth’s rotation, this time analogous to a figure skater putting arms out?</strong><br />
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.<br />
<strong>So the melting mountain glaciers and polar caps are moving bulk toward the equator?</strong><br />
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.</p>
</blockquote>
<p>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.</p>
<p>The moment of inertia of the lake is <span class="math-container">$m(R\cos L)^2$</span> and the moment of inertia of a spherical shell is <span class="math-container">$\frac{2}{3}mR^2$</span>, where m is the mass of water, <span class="math-container">$R$</span> is the Earth's radius and <span class="math-container">$L$</span> is the latitude of the lake (<span class="math-container">$30.82305$</span> degrees for Three Gorges). The relative change in the <a href="https://en.wikipedia.org/wiki/Rotational_energy" rel="nofollow noreferrer">moment of inertia <span class="math-container">$I$</span> of the Earth</a> is then
<span class="math-container">$$\frac{mR^2}{I}(\cos^2 L - \frac{2}{3}) 
= \frac{39 \times 10^{12} \times(6.37 \times 10^{6})^2}{8.04×10^{37}}(\cos^2 L - \frac{2}{3})$$</span>
<span class="math-container">$$=1.97×10^{-11}(\cos^2 L - \frac{2}{3})$$</span></p>
<p>Multiplying by the number of microseconds in a day (<span class="math-container">$8.64 \times 10^{10}$</span>) gives
<span class="math-container">$$1.7(\cos^2 L - \frac{2}{3}) = 1.7 × 0.071 = 0.12$$</span>
microseconds.</p>
<p>Why the difference from NASA's <span class="math-container">$0.06$</span> ? Note that the expression changes sign at <span class="math-container">$\cos^2 L = \frac{2}{3}$</span> or <span class="math-container">$L ≈ 35$</span> 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 <span class="math-container">$\frac{2}{3}$</span> term comes from the <em>&quot;all ocean&quot;</em> assumption.</p>
<p>As I understand <a href="http://www.leapsecond.com/pages/lod/2010-Chapanov-Gambis-Long-Variations-Earth-Rotation.pdf" rel="nofollow noreferrer">this paper</a>, the <span class="math-container">$\frac{2}{3}$</span>  term should be multiplied by <span class="math-container">$\frac{1.414}{1.38}$</span> to account for the shape of the oceans (search in the PDF for those numbers), resulting in <span class="math-container">$0.09$</span> microseconds.</p>

CC BY-SA 4.0
2019-05-12T14:14:26.027Z
I'd totally forgotten about this answer, until I noticed this question just posted: [How to caclulate the chinese Three Gorges Dam impact on Earth rotation?](https://astronomy.stackexchange.com/q/31841/7982)
uhoh
CC BY-SA 4.0
2019-01-12T22:25:53.637Z
2022-08-09T12:00:36.693Z
2022-08-09T12:00:36.693Z
Keith McClary
<p>I'm a big fan of <a href="https://en.wikipedia.org/wiki/Vsauce" rel="noreferrer">Vsauce</a> and the video <a href="https://www.youtube.com/watch?v=IJhgZBn-LHg" rel="noreferrer">How Earth Moves</a> is just one example of science related to the Earth available there (there's plenty more).</p>
<p>But the statement in <a href="https://youtu.be/YMTLkDrcr28?t=99" rel="noreferrer">this video starting at <code>01:39</code></a> strikes me as a little surprising.</p>
<blockquote>
<p>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.</p>
<p>The Three Gorges Dam did exactly what a figure skater does when they move their arms away from their center. It transferred <strong>thirty nine trillion kilograms of water one hundred and seventy five meters above sea level</strong>.</p>
<p>NASA calculated that that massive amount of water moved, caused Earth’s rotation to slow down, so that <strong>every day</strong> of your life since that dam was finished, <strong>has been longer by 0.06 microseconds</strong>. (emphasis added)</p>
</blockquote>
<p>If I calculated correctly, that would be a &quot;slip&quot; of about 1 centimeter per year at the equator.</p>
<p>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.</p>
<p>Is this <em>roughly 0.06 microsecond/day shift</em> well below the &quot;noise&quot; level, where in this case I'm using &quot;noise&quot; 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 <a href="https://physics.stackexchange.com/q/251700/83380">can be measured and modeled</a> 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.</p>
<hr />
<p><strong>below:</strong> Screen shots from the Vsauce YouTube video <a href="https://youtu.be/YMTLkDrcr28?t=99" rel="noreferrer">Water.</a></p>
<p><a href="https://i.sstatic.net/uJPj0.png" rel="noreferrer"><img src="https://i.sstatic.net/uJPj0.png" alt="enter image description here" /></a></p>
<p><a href="https://i.sstatic.net/DuWpS.png" rel="noreferrer"><img src="https://i.sstatic.net/DuWpS.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2018-12-20T18:40:32.717Z
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](https://www.jpl.nasa.gov/news/news.php?feature=716), 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
CC BY-SA 4.0
2022-02-03T18:46:42.113Z
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.
Popes Climate Theory
<water><earth-rotation><dams>
Was the filling of the Three Gorges Dam's impact on the Earth's rotation rate detectable?
uhoh
# Was the filling of the Three Gorges Dam's impact on the Earth's rotation rate detectable?

(Asked by: username_0 on 2017-03-28)

I'm a big fan of [Vsauce](https://en.wikipedia.org/wiki/Vsauce) and the video [How Earth Moves](https://www.youtube.com/watch?v=IJhgZBn-LHg) 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`](https://youtu.be/YMTLkDrcr28?t=99) 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](https://physics.stackexchange.com/q/251700/83380) 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.](https://youtu.be/YMTLkDrcr28?t=99)

[![enter image description here](https://i.sstatic.net/uJPj0.png)](https://i.sstatic.net/uJPj0.png)

[![enter image description here](https://i.sstatic.net/DuWpS.png)](https://i.sstatic.net/DuWpS.png)

**Tags:** water, earth-rotation, dams

> **Comments:**
> **username_1**: 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](https://www.jpl.nasa.gov/news/news.php?feature=716), 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.  
> **username_2**: 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.  

---

## 2 answers

### Answer by username_4 on 2019-01-12. Score: 6 | Accepted answer

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](https://nautil.us/issue/62/systems/why-our-intuition-about-sea_level-rise-is-wrong-rp):

> **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](https://en.wikipedia.org/wiki/Rotational_energy) 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](http://www.leapsecond.com/pages/lod/2010-Chapanov-Gambis-Long-Variations-Earth-Rotation.pdf), 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.

> **Comments:**
> **username_0**: I'd totally forgotten about this answer, until I noticed this question just posted: [How to caclulate the chinese Three Gorges Dam impact on Earth rotation?](https://astronomy.stackexchange.com/q/31841/7982)  

---

### Answer by username_3 on 2017-03-30. Score: 6

# The .06 microseconds figure doesn't sound accurate

The [moment of inertia](https://en.wikipedia.org/wiki/List_of_moments_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.

> **Comments:**
> **username_0**: 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](https://en.wikipedia.org/wiki/Moment_of_inertia_factor) of about [0.33](https://en.wikipedia.org/wiki/Earth) rather than $2/5$) there's still seems to be something wrong as stated. Time for someone to send a Vsauce Alert?  
> **username_3**: @username_0 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 :)  
> **username_0**: 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!  
> **username_4**: This has been [Asked again](https://earthscience.stackexchange.com/q/15907/6535). 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.  
> **username_4**: Using your method I get .00008 microseconds, in agreement with @username_0 . You can avoid the floating point error issue by $(r-175)^2 -r^2 \approx 2 \times 175 \times r$.  

---

