How Do Large Water Reservoirs Effect The Length of the Day

[alex3]

Homework Statement

Verbatim from the problem:

"About 10000km^3 of water is held behind dams in reservoirs around the world.Most reservoirs are at mid-latitudes, whilst the bulk of the world’s oceans are concentrated near the equator. By using conservation of angular momentum, estimate by how much the overall movement of water into reservoirs has changed the length of the day."

Given \rho_{E} (density of the Earth), R_{E} (radius of the Earth) and radius of gyration 0.58 R_{E}. The density of water is given as 1 g cm^{-3}, and we assuming the same for seawater.

Homework Equations

r^{2}_{g} = \frac{I}{M}
\rho_E = \frac{M}{V}
V = \frac{4}{3} \pi R^{3}_{E}
L = I \omega = r^{2}_{g} M \omega

The Attempt at a Solution

I have a feeling it's far more complicated, but I thought we have two relations for the angular momentum L. The angular momentum must remain constant, but the moment of inertia will change depending on the distribution of mass of the Earth, so the angular frequency must change. So:

I_{s} \omega_{i} = r^{2}_{g} M \omega_{f}

Where \omega_{i} and \omega_{f} represent the initial and final angular frequencies, respectively. I assume that the MoI before the water is dammed is I_{s}, and the given radius of gyration is calculated after the water is distributed. I_{s} is the moment of inertia of a sphere,

I_{s} = \frac{2}{5} M R^2

Then, simplifying:

\omega_{f} = \frac{2}{5 . (0.58)^{2}} \omega_{i}

But this gives quite a large difference, the coefficient is ~1.19. I'm assuming there's a better method, but I can't fathom it.

 

[D H]

Your 2/5MR² expression is the moment of inertia for a uniform density sphere. The water behind those impoundments is not going to form a uniform density sphere because of the solid Earth.

You are forgetting about the solid Earth.

You also are not using the given fact that "the bulk of the world’s oceans are concentrated near the equator."

 

[alex3]

What I was postulating was the the radius of gyration I was given, calculated using

I = M r^{2}_{g} was for the Earth with the distributed water (behind dams). The I_{s} was before the distribution.

So, this obviously isn't the right method. Could I use the MoI of a sphere, and exploit its additive property to add on two rings; one for the oceans and one for the water behind dams? Then compare a before/after, an MoI before the water was taken from the oceans, and one after?

 

[D H]

Close. You aren't adding the water. You are moving it. Hint: "The bulk of the world’s oceans are concentrated near the equator."

 

[D H]

One more thing: You cannot forget about the Earth itself.

 

[alex3]

So I have a large mass of water M_{W} in a ring around the center of the earth. Then, I take some of that water, M_{D}, and move it behind dams 45 degrees above the equator. Each situation has an MoI, so the solution comes from that.

Does that sound doable? I'm assuming that the mass of the water in the equator will cancel out in the end, as it isn't given.

 

[D H]

You are still forgetting about the Earth itself!

 

[alex3]

Woops :) Surely then, I just need to add to MoI of the Earth to both sides though, which would cancel? (I guess the MoI of the Earth would be equal to that of a solid sphere).

 

[D H]

It won't cancel. Hint: What is the conserved quantity here, and how is it calculated?

 

[alex3]

Angular momentum is conserved, and it's calculated

L = I \omega = r^{2}_{g} M \omega

So I presume I'll need an old L, before the water was moved, and a new L for after it. But.. Working out either would seem tricky. I know the solid sphere MoI.. Should I be adding a ring around the circumference to 'act' as the oceans (as my initial L)?

 

[D H]

Sounds like a plan -- except you do not need a new L for after moving the water. You already have the angular momentum. It is the conserved quantity here^1[/color].---------------------

^1[/color]Note that over the long haul the Earth's angular momentum is not conserved. Angular momentum is being transferred from the Earth's rotation about its axis to the orbit of the Earth and Moon about their common center of mass. This is a very slow process, however. You can ignore that here and assume that the Earth's angular momentum vector is constant.

 

[alex3]

So, by the sounds of it, I'll want to calculate an L after the water has been moved, by adding the MoI of a solid sphere with the ring of the ocean and the ring of the dams, but I'm a little stumped on calculating the MoI before the water has moved, should I calculate it by using the known radius of gyration? Apologies if I sound ignorant, I'm trying best!

 

[D H]

NO! Calculate the angular momentum of the Earth + tiny bit of water system in one location and equate this to the angular momentum of the Earth + tiny bit of water system in the other location. I said "tiny bit" because compared to the mass of the Earth, that 10,000 cubic kilometers of water is rather small.