# Are tidal power plants slowing down Earth's rotation?

Are tidal power plants slowing down Earth's rotation to the speed of the orbiting moon? (1 rotation per 28 cca days)

Are they vice versa increasing the speed of moon orbiting by generating some waves in gravitation field?

If yes, can you calculate how much energy must be produced by how many tidal power plants (compare it to average nuclear plant please) to slow down the Earth's rotation to 25 hours / day?

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xkcd.com/162 –  Ami Mar 6 '11 at 5:53

In principle, yes, the ultimate source of energy for a tidal power plant is Earth's rotational energy, so these plants are slowing down the Earth's rotation. By conservation of angular momentum, that means they are pushing the Moon further away as well, although I wouldn't phrase it as being due to "waves in the gravitational field," as that expression suggests a different phenomenon.

The Earth's rotational kinetic energy is about $10^{29}$ J, and the world uses something like $10^{22}$ J/year, so you could power the entire world for millions of years before you'd run out of rotational energy.

To answer your numerical question, you should work out the rotational kinetic energy of the Earth now, and also when the day is 25 hours long. The difference between those is the total energy required. The way to figure out the rotational kinetic energy is ${1\over 2}I\omega^2$. Here $I$ is the Earth's moment of inertia, which is about $0.4MR^2$ where $M$ and $R$ are Earth's mass and radius. $\omega$ is the Earth's rotation rate in radians per second -- that is, $2\pi$ over the time for one rotation.

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But are tidal plants slowing down the rotation more than normal tides would without the plants there? –  Mark Eichenlaub Mar 5 '11 at 20:12
@Ted Thanks for the energy consumption calculation. Are you saying the earth will stop rotating one day completely? That will happen even without the power plants, right? –  daniel.sedlacek Mar 5 '11 at 22:10
@Mark -- I think they must be in principle. I haven't done the calculation to see which of the two (very small) effects is smaller. –  Ted Bunn Mar 5 '11 at 22:36
@daniel.sedlacek -- It'll stop rotating relative to the Moon. That is, eventually the rotation rate will be such that the Earth always keeps the same face toward the Moon. The Earth will still rotate with respect to the Sun and the stars, though. The Moon has already done this, by the way -- that's why it always keeps the same face towards the Earth. This sort of "tidal locking" turns out to be pretty common in the solar system: quite a few moons have done it. (Mercury has done something similar, but in kind of a weird way.) –  Ted Bunn Mar 5 '11 at 22:39
Also, note that the full system (earth and moon ignoring the sun for now) conserve angular momentum. The loss of angular momentum by the earth is counteracted by the increase in angular momentum represented by the lunar orbit. The oribital energy of the moon is increased and that term must be deducted from the loss of rotational kinetic energy of the earth before determining how much energy must be dissipated by a unit transference of angular momentum between the two bodies. –  Omega Centauri Mar 6 '11 at 2:01

In order to slow down the rotation of a body angular momentum must transferred off that body. In the case of Earth and the moon this occurs from the difference in gravity across the Earth, or tidal force. A tidal power generating system simply converts a tiny fraction of energy in the tidal bulge of the Earth, mostly in the oceans, as it moves around the globe into mechanical or electrical power. The question is whether that induces a torque on the Earth.

These systems might serve to reduce the tidal bulge of the oceans a very tiny amount. So from the systems perspective the flow of water is impeded, the effective viscosity increased, friction increased and the tidal bulge reduced. This represents a tiny amount of energy reduced on the Earth in this form. However, this is a near infinitesimal amount of the Earth’s rotational kinetic energy.

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I am a bit puzzled by the answers.

The earth moon system has a specific distance at a given moment and the gravitational force on the earth a specific magnitude that gives a "spin" to the earth. Does the composition of the earth play a role to the amount of the spin? i.e. water versus land?

I may be completely wrong but the bulge of the moon goes at the same rate over land and sea, irrespective of friction. I would expect the angular momentum transfer to exist even if the earth were rock solid and no bulge possible, in the same magnitude, the same way the moon's angular momentum changes by the interaction with the earth's gravitation.

Thus my answer is that the problem with the power plants would be a matter of earth angular momentum conservation that will not be transferred to the moon, and thus the power plants would make no difference because angular momentum of the earth will be conserved whether they are there or not.

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For the Earth to transfer angular momentum to the Moon, work has to be done (otherwise the conservation of energy won't be satisfied). So if the Earth was rock solid with no tides, there would be no work done, and the moon could not gain angular momentum. I don't know whether tidal power plants increase the amount of work being done, or just divert it into energy usable by people. I don't understand the microscopic details of how this works, either. –  Peter Shor Mar 6 '11 at 11:22
@Peter Shor Could not the energy go to an increase of the spin of the earth? Like a giant top? I suppose that the asymmetry comes because the earth is not a point gravitational mass but is a sphere, which will be interacting gravitationally incrementally to the distance from the moon. In any case the only interaction between the moon and the earth is gravitational and I do not see how , lets say continental drifts as an example, will affect the moon as long as the gravitational mass does not change. –  anna v Mar 6 '11 at 11:59
If you look at the Earth-moon system, there are two parameters, the speed of the Earth's rotation, and the distance to the moon. There are also two conserved quantities: energy and angular momentum. If you want to keep both energy and angular momentum fixed, you can't change the parameters. Currently, the energy of the Earth-moon system is decreasing because it's going into powering the tides, which is only possible because the Earth is not quite a rigid body. –  Peter Shor Mar 6 '11 at 12:23
@Anna, of course the "rock solid" earth dissipates angular momentum too! An example is the moon, whose momentum was consumed eons ago. Any "earth" which is not an ideally elastic body will "crunch" angular momentum into heat. Maybe this is far more than the energy used up in the oceans? I dont know. –  Georg Mar 6 '11 at 12:32
Some misconceptions above. Angular momentum cannot be consumed, only transfered in this case between rotational and orbital. If both earth and the moon were perfectly rigid, there would be no angular momentum transfer. Likewise if both bodies were perfectly elastic (they don't dissapate mechanical energy) there also couldn't be angular momentum transfer. So one (or both) bodies must be both defomable, and inelastic (at least some mechnical energy is transformed to heat) for the angular momentum to be transferred. –  Omega Centauri Mar 6 '11 at 15:47