<p>It is minute and inconsequential.  Without tidal energy harvesting, the motion of the water is ultimately converted into heat, it has a very small heating effect (that is swamped by the heating caused by the sun). With tidal energy harvesting, instead of heating the water, a little of the energy is converted to electricity (which is then converted into light or heat,  ultimately the energy becomes heat)</p>
<p>The tidal dams can change local tidal patterns, which could have a minute effect on the moon's motion, but this effect is minuscule.</p>

CC BY-SA 4.0
2020-07-30T10:33:00.773Z
I can see that much, but in the extremes of a perfectly frictionless sphere (covered in water) where the tidal forces allow that water to move (with zero viscosity) vs the other extreme where the bulging water pushes against a movable surface (acting like a piston). 

It feels like there should be a difference between these two idealized systems.
orathaic
CC BY-SA 4.0
2020-07-30T11:53:02.847Z
I'm not sure I see your point.  I'm saying that the same work is done by the moon whether or not you build a power station, the only difference (except for small changes in tides) is whether the energy is harvested in the form of electricity or "wasted" in friction etc. Just like putting up a solar panel doesn't make the sun burn faster.
James K
CC BY-SA 4.0
2020-07-30T11:54:41.070Z
I suspect in a perfectly frictionless model, the "bulge" points directly towards the moon, there is no asymmetry, no work done and no transfer of angular momentum.  Whether or not the tide does work against a turbine, or the sea bed is makes no difference to the recession of the moon.
James K
CC BY-SA 4.0
2020-07-29T23:46:22.440Z
2020-07-29T23:46:22.440Z
James K
<p>Reading another question <a href="https://astronomy.stackexchange.com/q/36925/7982">Where do we have it from that the Moon is migrating away from Earth?</a>
I makes me ask what effect does harvesting tidal energy for electricity production have on the rate the moon gains angular momentum?</p>
<p>Presuming the most common harvesting methods involve underwater turbines in tidal zones with water flowing regularly in both directions (twice per day).</p>

<the-moon><tidal-forces>
How does using Tidal energy affect the moon?
orathaic
# How does using Tidal energy affect the moon?

(Asked by: username_0 on 2020-07-29)

Reading another question [Where do we have it from that the Moon is migrating away from Earth?](https://astronomy.stackexchange.com/q/36925/7982)
I makes me ask what effect does harvesting tidal energy for electricity production have on the rate the moon gains angular momentum?

Presuming the most common harvesting methods involve underwater turbines in tidal zones with water flowing regularly in both directions (twice per day).

**Tags:** the-moon, tidal-forces

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## 1 answers

### Answer by username_1 on 2020-07-29. Score: 2

It is minute and inconsequential. Without tidal energy harvesting, the motion of the water is ultimately converted into heat, it has a very small heating effect (that is swamped by the heating caused by the sun). With tidal energy harvesting, instead of heating the water, a little of the energy is converted to electricity (which is then converted into light or heat, ultimately the energy becomes heat)

The tidal dams can change local tidal patterns, which could have a minute effect on the moon's motion, but this effect is minuscule.

> **Comments:**
> **username_0**: I can see that much, but in the extremes of a perfectly frictionless sphere (covered in water) where the tidal forces allow that water to move (with zero viscosity) vs the other extreme where the bulging water pushes against a movable surface (acting like a piston).   It feels like there should be a difference between these two idealized systems.  
> **username_1**: I'm not sure I see your point.  I'm saying that the same work is done by the moon whether or not you build a power station, the only difference (except for small changes in tides) is whether the energy is harvested in the form of electricity or "wasted" in friction etc. Just like putting up a solar panel doesn't make the sun burn faster.  
> **username_1**: I suspect in a perfectly frictionless model, the "bulge" points directly towards the moon, there is no asymmetry, no work done and no transfer of angular momentum.  Whether or not the tide does work against a turbine, or the sea bed is makes no difference to the recession of the moon.  

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