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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?

Qmechanic
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daniel.sedlacek
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3 Answers3

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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.

Ted Bunn
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    But are tidal plants slowing down the rotation more than normal tides would without the plants there? – Mark Eichenlaub Mar 05 '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 05 '11 at 22:10
  • @Mark I think they are. It's the same principle as a child on a swing, I think. – daniel.sedlacek Mar 05 '11 at 22:18
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    @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 05 '11 at 22:36
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    @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 05 '11 at 22:39
  • @Ted: I can see where you are going with this, I just don't see a mechanism. The energy associated with the slowing has presumably been going into heat if the system is left to itself. What is the reason to think that more energy is being dissipated in the plants? – dmckee --- ex-moderator kitten Mar 05 '11 at 22:59
  • It has to be looked at one a case by case basis. Tidal power plants will be located in places where the tidal range is high, i.e. the local tidal basin has some sort of resonant frequency that is not too far from the tidal frequency. The tidal plant will perturb the amplitude of the local tide, but whether the net dissipation is increased or decreased depends upon the details. In any case there are probably only a few tens to maybe a couple of hundred gigawatts available, and this amount of dissipation is very small wrt the rotation kinetic energy of the planet. – Omega Centauri Mar 06 '11 at 01:54
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    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 06 '11 at 02:01
  • @dmckee-- I thought I was sure, but I can't seem to construct a coherent argument, so I'm not sure. Omega Centauri may be right on this. – Ted Bunn Mar 06 '11 at 14:40
  • Also, on @Omega Centauri's last comment: about 95% of the energy change is in Earth's rotational KE, not the Moon's orbital energy, but I agree that's not obvious until you run the numbers. I get $I_{\rm moon}/I_{\rm Earth}\approx 100$ for moments of inertia ("moon") referring to the Moon's orbital motion, not to the Moon itself) and $L_{\rm moon}/L_{\rm Earth}\approx 4$. If $\Delta L$ is transferred from rotational to orbital, then each component undergoes an energy change $L,\Delta L/I$, so the ratio of energy changes is the ratio of $L/I$'s, or about 5%. – Ted Bunn Mar 06 '11 at 14:54
  • @Omega: If you make that "understand each installation individually" comment an answer, I'll vote for it. Though, the longer I think about it the more I suspect that the average installation will add slightly to the dissipation--just by intuition without any math to back it up. – dmckee --- ex-moderator kitten Mar 06 '11 at 17:23
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    dmckee: I think a random installation will more probably add to dissipation, but there could be some circumstances where an installation moves a basin further away from a resonance, and thus lowers overall dissipation. I think we should concieve the tidal sloshing to be a superposition of tidal modes of the ocean. Making changes in the details by adding some dissipation at certain points, will change the eigenfrequencies and shapes, and one would have to do an analysis of the system plus lunar forcing to figure it out. – Omega Centauri Mar 07 '11 at 00:45
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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.

Lawrence B. Crowell
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I think that the effect of tidal power will be to (very slightly) to increase the tidal drag on the moon. Consider the following thought experiment : Imagine an idealised rigid planet with an idealised rigid moon. This planet has perfectly smooth valleys and hills in which a friction-less ocean resides. Tides will occur but the moon/planet system will not be orbitally slowed because there will be no net tidal energy loss - the moon will simply drag the friction-less ocean around the planet causing the moon's orbital speed to fluctuate up and down slightly as energy is exchanged back and forth slightly with the ocean tides. Now imagine a tidal power plant being installed - this will extract energy from the moon/planet system as the falling water from the power plant must now lag in order to have a height difference between input and output water levels in order to produce power. This will have an overall drag on the moon since the water flowing out from the power plant will be displaced from the correct position to return all the energy back to the moon - thus affecting the orbital speed. David B

David Bowskill
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