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This question already has an answer here:

Imagine an electric power generator in a form of a floating object moving up and down and utilizing the buoyant force of sea tides. Can anyone explain how it doesn't violate the principle of conservation of energy? Where does this energy come from?

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marked as duplicate by Bob D, Dale, Cleonis, stafusa, Jon Custer Aug 17 at 23:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Where do the tides come from? $\endgroup$ – Aaron Stevens Aug 17 at 15:25
  • $\begingroup$ I mean lunar tide $\endgroup$ – Adam Aug 17 at 15:28
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    $\begingroup$ I didn't ask for clarification. I was asking to get you to think :) $\endgroup$ – Aaron Stevens Aug 17 at 15:29
  • $\begingroup$ Are you suggesting that the electric energy will come from the potential energy of the moon? $\endgroup$ – Adam Aug 17 at 15:32
  • $\begingroup$ What would make you think it violated the principle of conservation of energy? $\endgroup$ – Bob D Aug 17 at 16:27
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Here is a comparison:
This discussion is confined to tidal effect.

Several of Jupiter's Moons are warmer than you would expect them to be, if you go purely by their size.

(In the Earth's interior heat energy is replenished by radioactive decay. (It's not that the Earth interior is more radio-active than the outer layers, but in the interior the heat is - well, on the inside.))

Those Jupiter moons are too small for that. The rate of radio-active decay heating is too small to keep up. Yet they are warmer than you would expect them to be.

In the case of those Jupiter moons the source of the heat is tidal friction. The tidal effect from Jupiter is so strong that the tidal distortion makes the interior of the Jupiter moon rub against itself.

Will that tidal friction go on for eternity?
Well, when a moon is tidally locked there will not be any friction.

Our own Moon is an example of a Moon that is tidally locked; the orbit of the Moon around the Earth and the rotation of the Moon around its own axis have the same period: one month. So the kind of tidal friction that is happening inside those Jupiter moons is not happening inside our Moon.

The Jupiter moons rotate around their own axis faster than their orbit around Jupiter; so their interior is rubbing.

Harvesting energy from tidal motion

There are a number of places in the world where turbines are placed in the path of tidal current, so that energy is harvested from tidal motion.

Here is the comparison that all the above is leading up to:
Both in the case of tidal friction (Jupiter moons interior rubbing) and in the case of using turbines to harvest energy from tidal motion the underlying source of the energy is the same.

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Although tides are caused by the gravity of the moon, waves are most commonly caused by wind. Wind, in turn, is caused by differences in atmospheric pressure. The pressure differences are due to temperature differences caused by variations in solar energy received at the surface of the earth.

So your energy ultimately comes from the sun.

As far as imagining generating electricity from ocean waves, it is already being looked into. See the following: https://www.asme.org/topics-resources/content/harvesting-the-power-of-the-ocean

My question is about the power of rising tide only, so if it make it simpler assume a windless, waveless period of time. I didn't realize I need to spell out all these little assumptions.

Sorry about that. In that case I will vote to close this post since your question is a duplicate of this one which has been answered.

Where does tidal energy come from?

Hope this helps.

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  • $\begingroup$ My question is about the power of rising tide only, so if it make it simpler assume a windless, waveless period of time. I didn't realize I need to spell out all these little assumptions. $\endgroup$ – Adam Aug 17 at 18:15
  • $\begingroup$ @Adam See my revision. $\endgroup$ – Bob D Aug 17 at 18:40
  • $\begingroup$ @Adam If you don't state your assumptions, then how can anyone know what you are assuming? I don't think Bob's answer warrants such condescension. $\endgroup$ – Aaron Stevens Aug 18 at 12:27

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