<p>You need to calculate the change in the moment of inertia of the Earth and use conservation of angular momentum. In the case of <em>eg.</em> a large dam, (most of) the water will ultimately come from the sea, effectively removing a thin layer of water. The contribution of the removed water to the moment of inertia depends on the distance from the rotation axis and hence on the latitude. This is a fairly simple calculation if we assume the world is all ocean. A more precise calculation would have to take into account the shapes of the oceans. You also need the moment of inertia of the earth, which depends on the density as a function of depth.</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 (30.82305 degrees for Three Gorges). The relative change in the moment of inertia of the earth is then
<span class="math-container">$$\frac{mR^2(\cos^2 L - \frac{2}{3})}{I}$$</span>
(where <span class="math-container">$I$</span> is the Earth's moment of inertia, <a href="https://en.wikipedia.org/wiki/Rotational_energy" rel="nofollow noreferrer"><span class="math-container">$8.04×10^{37}$</span> kg·m<span class="math-container">$^2$</span></a>) or
<span class="math-container">$$1.97×10^{-11}(\cos^2 L - \frac{2}{3}) .$$</span>
Multiplying by the number of microseconds in a day (<span class="math-container">$8.64 \times 10^{10}$</span>) gives: 
<span class="math-container">$$1.7(\cos^2 L - \frac{2}{3}) = 1.7 × 0.071 = 0.12\,\mu\text{s}$$</span></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 (fairly 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 "all ocean" assumption. I presume NASA took into account the shapes of the oceans and got a more accurate value for this term.</p>

CC BY-SA 4.0
2019-01-12T12:29:39.427Z
The differences between a sphere and geoid probably affect calculations as well.
Spencer
CC BY-SA 4.0
2019-01-10T23:23:28.623Z
2019-01-14T09:23:44.513Z
2019-01-14T09:23:44.513Z
Keith McClary
<p>According to this article <a href="https://futurism.com/drops-of-jupiter" rel="nofollow noreferrer">Drops of Jupiter</a></p>

<blockquote>
  <p>Raising 39 trillion kilograms of water 175 meters above sea level will
  increase the Earth’s moment of inertia, and thus slow its rotation.
  However, the impact will be extremely small. NASA scientists
  calculated the shift of such a mass will increase the length of day by
  only 0.06 microseconds</p>
</blockquote>

<p>is it possible to know which equations NASA used or is it too complicated to calculate that for regular people ? (thus it would be NASA required to do the math)</p>

CC BY-SA 4.0
2019-01-08T20:58:24.873Z
I can't find the cited text in the source you point to. The numbers 39 and/or 175 just aren't in there. Are you sure you're quoting the correct article? Do you mean perhaps [this atricle](https://www.google.com/amp/s/amp.businessinsider.com/chinas-three-gorges-dam-really-will-slow-the-earths-rotation-2010-6)?
Spencer
<earth-rotation>
Which are the equations needed to calculate how much moving Earth's water with dams would change Earth's rotation speed?
Pablo
# Which are the equations needed to calculate how much moving Earth's water with dams would change Earth's rotation speed?

(Asked by: username_0 on 2019-01-08)

According to this article [Drops of Jupiter](https://futurism.com/drops-of-jupiter)

> Raising 39 trillion kilograms of water 175 meters above sea level will
> increase the Earth’s moment of inertia, and thus slow its rotation.
> However, the impact will be extremely small. NASA scientists
> calculated the shift of such a mass will increase the length of day by
> only 0.06 microseconds

is it possible to know which equations NASA used or is it too complicated to calculate that for regular people ? (thus it would be NASA required to do the math)

**Tags:** earth-rotation

> **Comments:**
> **username_1**: I can't find the cited text in the source you point to. The numbers 39 and/or 175 just aren't in there. Are you sure you're quoting the correct article? Do you mean perhaps [this atricle](https://www.google.com/amp/s/amp.businessinsider.com/chinas-three-gorges-dam-really-will-slow-the-earths-rotation-2010-6)?  

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## 1 answers

### Answer by username_2 on 2019-01-10. Score: 8 | Accepted answer

You need to calculate the change in the moment of inertia of the Earth and use conservation of angular momentum. In the case of *eg.* a large dam, (most of) the water will ultimately come from the sea, effectively removing a thin layer of water. The contribution of the removed water to the moment of inertia depends on the distance from the rotation axis and hence on the latitude. This is a fairly simple calculation if we assume the world is all ocean. A more precise calculation would have to take into account the shapes of the oceans. You also need the moment of inertia of the earth, which depends on the density as a function of depth.

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 of the earth is then
$$\frac{mR^2(\cos^2 L - \frac{2}{3})}{I}$$
(where $I$ is the Earth's moment of inertia, [$8.04×10^{37}$ kg·m$^2$](https://en.wikipedia.org/wiki/Rotational_energy)) or
$$1.97×10^{-11}(\cos^2 L - \frac{2}{3}) .$$
Multiplying by the number of microseconds in a day ($8.64 \times 10^{10}$) gives:
$$1.7(\cos^2 L - \frac{2}{3}) = 1.7 × 0.071 = 0.12\,\mu\text{s}$$

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 (fairly 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. I presume NASA took into account the shapes of the oceans and got a more accurate value for this term.

> **Comments:**
> **username_1**: The differences between a sphere and geoid probably affect calculations as well.  

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