23
$\begingroup$

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?

|cite|improve this question
$\endgroup$
14
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • 11
    $\begingroup$ But are tidal plants slowing down the rotation more than normal tides would without the plants there? $\endgroup$ – Mark Eichenlaub Mar 5 '11 at 20:12
  • $\begingroup$ @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? $\endgroup$ – daniel.sedlacek Mar 5 '11 at 22:10
  • 1
    $\begingroup$ @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. $\endgroup$ – Ted Bunn Mar 5 '11 at 22:36
  • 3
    $\begingroup$ @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.) $\endgroup$ – Ted Bunn Mar 5 '11 at 22:39
  • 1
    $\begingroup$ 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. $\endgroup$ – Omega Centauri Mar 6 '11 at 2:01
4
$\begingroup$

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.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.