<p>A good estimate of the lunar tidal energy dissipated into the oceans is 2.5 Terawatts (<a href="http://www.sciencedirect.com/science/article/pii/S0079661197000219?via%3Dihub" rel="nofollow noreferrer">Munk, 1997</a>; <a href="http://www.sciencedirect.com/science/article/pii/S0079661197000220" rel="nofollow noreferrer">Le Provost &amp; Lyard, 1997</a>). The value estimated comes mainly from two different sources: from harmonic calculations and from altimetry estimates using satellite observations (e.g., <a href="https://en.wikipedia.org/wiki/TOPEX/Poseidon" rel="nofollow noreferrer">Topex/Poseidon</a>).</p>

<p>The input of energy into the coastal ocean is not uniform and has a peak in the South Atlantic Ocean, while the energy is dissipated by bottom friction in the coastal ocean and mainly in the North Atlantic Ocean (<a href="http://www.sciencedirect.com/science/article/pii/S0079661197000220" rel="nofollow noreferrer">Le Provost &amp; Lyard, 1997</a>).
<a href="https://i.sstatic.net/R7ytb.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/R7ytb.jpg" alt="Egbert &amp; Ray, Nature, 2000"></a> </p>

<p>Source: Nature: Egbert &amp; Ray, 2000. Estimates of tidal energy dissipation.</p>

<p><a href="http://www.nature.com/nature/journal/v405/n6788/full/405775a0.html" rel="nofollow noreferrer">Egbert &amp; Ray (2000)</a> showed that a fraction of the dissipation (1 TW) takes place in the deep ocean in areas of, generally near areas of rough topography. Maintaining ocean stratification and the large scale thermohaline circulation (the commonly known "Conveyor Belt") requires around 2 TW to provide enough mixing (<a href="http://www.sciencedirect.com/science/article/pii/S0967063798000703?via%3Dihub" rel="nofollow noreferrer">Munk &amp; Wunsch, 1998</a>). Therefore, the tides provide about half of that energy with the other half coming from wind forcing.</p>

<p>Here is a summary figure from <a href="http://www.sciencedirect.com/science/article/pii/S0967063798000703?via%3Dihub" rel="nofollow noreferrer">Munk &amp; Wunsch, 1998</a>. <a href="https://i.sstatic.net/vFVHQ.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/vFVHQ.jpg" alt="static.skepticalscience.com"></a></p>

<p>The repercussion of this fact are given by <a href="http://www.nature.com/nature/journal/v405/n6788/full/405743a0.html" rel="nofollow noreferrer">Wunsch (2000)</a>:</p>

<blockquote>
  <p>...it was only recently recognized that the need for an energy source to sustain the vertical mixing (lifting dense water through lighter) has important consequences. The difficulties of driving fluid motions by surface heating and evaporation mean that a mechanical source of energy must control not only the directly wind-driven flows, but also the deep-water components of the meridional overturning circulation. </p>
  
  <p>... changes in tidal distributions and the consequent mixing would need to be understood over geological time. During the Last Glacial Maximum, the sea level was about 130 metres lower than today. This configuration removed much of the present regions of shallow-water energy dissipation and changed the deep-ocean tides, presumably affecting oceanic heat transport. Over longer periods in the past, the entire continental configuration was different, with radically different tidal distributions and mixing. It appears that the tides are, surprisingly, an intricate part of the story of climate change, as is the history of the lunar orbit.</p>
</blockquote>

CC BY-SA 3.0
2017-03-04T15:13:53.577Z
You talk exclusively about the oceans; it's hard to believe that no heat (or even less heat) comes out of  mantle and lithospheric flexing. The [Earth-moon barycentre](http://www.astronomycafe.net/qadir/q665.html) is inside the mantle, after all.
Spencer
CC BY-SA 3.0
2017-03-05T10:25:14.053Z
@Spencer - The Earth tides are significantly smaller than are the oceanic tides, and the solid Earth's tidal quality factor Q is much higher than is the tidal quality factor for the oceans.
David Hammen
CC BY-SA 3.0
2017-03-05T16:55:08.750Z
@DavidHammen you should add some material (with numbers) about that to your answer.
Spencer
CC BY-SA 3.0
2017-03-06T00:30:27.937Z
@Spencer, The paper by Munk & Wunsch (cited in this answer) discusses this a bit. A nice figure from the is at https://static.skepticalscience.com/pics/munkwunsch.jpg .
David Hammen
CC BY-SA 3.0
2017-03-06T00:50:32.577Z
That is a great figure. Also thanks for the corrections on the typo. Do you mind if I add the figure to the answer?
arkaia
CC BY-SA 3.0
2017-03-03T19:30:54.713Z
2017-03-06T14:11:09.433Z
2017-03-06T14:11:09.433Z
arkaia
<p>According to <a href="https://en.wikipedia.org/wiki/Tidal_acceleration#Angular_momentum_and_energy" rel="nofollow noreferrer">this article</a>, the energy lost by Earth in tidal acceleration is about 3.321 TW, most of it (95%) is converted to heat by frictional losses in the oceans and their interaction with the solid Earth, and the rest (0.121 TW) is transferred to the Moon.</p>

CC BY-SA 3.0
2017-03-03T12:54:40.463Z
I assume that 3.321 TW is in a year?
JeopardyTempest
CC BY-SA 3.0
2017-03-03T13:04:19.193Z
If so, I'd think the 95% converted to heat would be the temperature gained in the atmosphere.  That's 9.1GW per day.  510 $Tm^2$ surface area of Earth.  So 0.000018 $W/m^2$?  If that's so, a very trivial impact?  If this thinking is all right, perhaps +1 the comment, and it can then be added into the answer?
JeopardyTempest
CC BY-SA 3.0
2017-03-03T13:07:23.587Z
(In comparison, direct solar energy peaks about 1361 $W/m^2$ at the top of the atmosphere, and if I read Wikipedia right, about 1120 $W/m^2$ average after including curvature, scattering, and atmospheric absorption+reemission)
JeopardyTempest
CC BY-SA 3.0
2017-03-03T22:57:44.847Z
You can see the values of tidal dissipation in the figure with my answer to the question. The maximum is around 25 *10-3 W/m2
arkaia
CC BY-SA 3.0
2017-03-04T00:55:26.457Z
@JeopardyTempest The watt is a measure of power ( energy / time). Multiply it by any period of time you like.
Spencer
CC BY-SA 3.0
2017-03-04T16:09:07.157Z
Indeed, have always made a right mess of power units.  So nix the dividing by 365 (I also forgot to apply the 95% part).  So rebuilt estimate of an average of 0.0062 $W/m^2$.  A bit higher than artxabaleta's value (particularly given this one should be the mean, not max), but similar scale.  No idea where the greuze's Wikipedia article for it's values, but the general agreement is the contribution is rather small.
JeopardyTempest
CC BY-SA 3.0
2017-03-04T16:28:16.973Z
You could make a crude estimate: it would heat the bottom 2m of air by $(0.0062 J/{m^2 s})/((2m)(1 m^2)(1.225 kg/m^2)(1006 J/{kg K})) = 0.01 °C/day$ or about $3°C/yr$ contribution.  But that's really highly dishonest, as that energy would heavily radiate away during that year.  I don't know if it's legitimate way to consider it (assumes radiates processes are fairly similar for the different energy forms), but if $1000 W/m^2$ is about $290K$, then $0.0062 W/m^2$ would effect around a $0.002K$ change.
JeopardyTempest
CC BY-SA 3.0
2017-03-05T10:32:13.733Z
Downvoted because the source of that number is a wikipedia article, and in that article, that figure of 3.321 TW is tagged as "citation needed." That number is far too precise. Estimates in the scientific literature vary from well under 3 to almost 4 TW.
David Hammen
CC BY-SA 3.0
2017-03-06T00:15:41.463Z
@David Hammen I was the one who added the "needs citation" on Wikipedia as it bugged me too.  But a loose 3-4 TW is still good for rough estimation, as hopefully the math I've offered gives a fair attempt to.
JeopardyTempest
CC BY-SA 3.0
2017-03-03T10:57:00.227Z
2017-03-03T10:57:00.227Z
greuze
<p>Reading about moons of other planets they often get heated up by the tides so for example Io and Europe of Jupiter get friction by which eruptions and liquid water can arise. But is there also an effect on Earth due to the tides of the moon and sun? If so how many degrees is the Earth heated up?</p>

CC BY-SA 3.0
2017-03-10T02:12:19.510Z
I can think of one thing that was missed in these answers (I didn't read all of the cited articles). You mention tidal effects of Jupiter on its moons. This is caused by the "stretching" of the moons due to elliptical orbit. Our moon's orbit is actually a lot more elliptical than Io and Europa so it definitely will have some effect on the Earth. The orbit is complex though and varies from about .02 to .07 eccentricity, which may negate the effect somewhat. Also, of course, the main difference is the relative masses of the moon and planet (or vice versa).
Jack R. Woods
<geophysics><tides>
Is the Earth heated up by the tides?
Marijn 
# Is the Earth heated up by the tides?

(Asked by: username_0 on 2017-03-02)

Reading about moons of other planets they often get heated up by the tides so for example Io and Europe of Jupiter get friction by which eruptions and liquid water can arise. But is there also an effect on Earth due to the tides of the moon and sun? If so how many degrees is the Earth heated up?

**Tags:** geophysics, tides

> **Comments:**
> **username_1**: I can think of one thing that was missed in these answers (I didn't read all of the cited articles). You mention tidal effects of Jupiter on its moons. This is caused by the "stretching" of the moons due to elliptical orbit. Our moon's orbit is actually a lot more elliptical than Io and Europa so it definitely will have some effect on the Earth. The orbit is complex though and varies from about .02 to .07 eccentricity, which may negate the effect somewhat. Also, of course, the main difference is the relative masses of the moon and planet (or vice versa).  

---

## 2 answers

### Answer by username_2 on 2017-03-03. Score: 6

A good estimate of the lunar tidal energy dissipated into the oceans is 2.5 Terawatts ([Munk, 1997](http://www.sciencedirect.com/science/article/pii/S0079661197000219?via%3Dihub); [Le Provost & Lyard, 1997](http://www.sciencedirect.com/science/article/pii/S0079661197000220)). The value estimated comes mainly from two different sources: from harmonic calculations and from altimetry estimates using satellite observations (e.g., [Topex/Poseidon](https://en.wikipedia.org/wiki/TOPEX/Poseidon)).

The input of energy into the coastal ocean is not uniform and has a peak in the South Atlantic Ocean, while the energy is dissipated by bottom friction in the coastal ocean and mainly in the North Atlantic Ocean ([Le Provost & Lyard, 1997](http://www.sciencedirect.com/science/article/pii/S0079661197000220)).
[![Egbert & Ray, Nature, 2000](https://i.sstatic.net/R7ytb.jpg)](https://i.sstatic.net/R7ytb.jpg)

Source: Nature: Egbert & Ray, 2000. Estimates of tidal energy dissipation.

[Egbert & Ray (2000)](http://www.nature.com/nature/journal/v405/n6788/full/405775a0.html) showed that a fraction of the dissipation (1 TW) takes place in the deep ocean in areas of, generally near areas of rough topography. Maintaining ocean stratification and the large scale thermohaline circulation (the commonly known "Conveyor Belt") requires around 2 TW to provide enough mixing ([Munk & Wunsch, 1998](http://www.sciencedirect.com/science/article/pii/S0967063798000703?via%3Dihub)). Therefore, the tides provide about half of that energy with the other half coming from wind forcing.

Here is a summary figure from [Munk & Wunsch, 1998](http://www.sciencedirect.com/science/article/pii/S0967063798000703?via%3Dihub). [![static.skepticalscience.com](https://i.sstatic.net/vFVHQ.jpg)](https://i.sstatic.net/vFVHQ.jpg)

The repercussion of this fact are given by [Wunsch (2000)](http://www.nature.com/nature/journal/v405/n6788/full/405743a0.html):

> ...it was only recently recognized that the need for an energy source to sustain the vertical mixing (lifting dense water through lighter) has important consequences. The difficulties of driving fluid motions by surface heating and evaporation mean that a mechanical source of energy must control not only the directly wind-driven flows, but also the deep-water components of the meridional overturning circulation.
>
> ... changes in tidal distributions and the consequent mixing would need to be understood over geological time. During the Last Glacial Maximum, the sea level was about 130 metres lower than today. This configuration removed much of the present regions of shallow-water energy dissipation and changed the deep-ocean tides, presumably affecting oceanic heat transport. Over longer periods in the past, the entire continental configuration was different, with radically different tidal distributions and mixing. It appears that the tides are, surprisingly, an intricate part of the story of climate change, as is the history of the lunar orbit.

> **Comments:**
> **username_3**: You talk exclusively about the oceans; it's hard to believe that no heat (or even less heat) comes out of  mantle and lithospheric flexing. The [Earth-moon barycentre](http://www.astronomycafe.net/qadir/q665.html) is inside the mantle, after all.  
> **username_4**: @username_3 - The Earth tides are significantly smaller than are the oceanic tides, and the solid Earth's tidal quality factor Q is much higher than is the tidal quality factor for the oceans.  
> **username_3**: @username_4 you should add some material (with numbers) about that to your answer.  
> **username_4**: @username_3, The paper by Munk & Wunsch (cited in this answer) discusses this a bit. A nice figure from the is at https://static.skepticalscience.com/pics/munkwunsch.jpg .  
> **username_2**: That is a great figure. Also thanks for the corrections on the typo. Do you mind if I add the figure to the answer?  

---

### Answer by username_5 on 2017-03-03. Score: 2

According to [this article](https://en.wikipedia.org/wiki/Tidal_acceleration#Angular_momentum_and_energy), the energy lost by Earth in tidal acceleration is about 3.321 TW, most of it (95%) is converted to heat by frictional losses in the oceans and their interaction with the solid Earth, and the rest (0.121 TW) is transferred to the Moon.

> **Comments:**
> **username_6**: I assume that 3.321 TW is in a year?  
> **username_6**: If so, I'd think the 95% converted to heat would be the temperature gained in the atmosphere.  That's 9.1GW per day.  510 $Tm^2$ surface area of Earth.  So 0.000018 $W/m^2$?  If that's so, a very trivial impact?  If this thinking is all right, perhaps +1 the comment, and it can then be added into the answer?  
> **username_6**: (In comparison, direct solar energy peaks about 1361 $W/m^2$ at the top of the atmosphere, and if I read Wikipedia right, about 1120 $W/m^2$ average after including curvature, scattering, and atmospheric absorption+reemission)  
> **username_2**: You can see the values of tidal dissipation in the figure with my answer to the question. The maximum is around 25 *10-3 W/m2  
> **username_3**: @username_6 The watt is a measure of power ( energy / time). Multiply it by any period of time you like.  
> **username_6**: Indeed, have always made a right mess of power units.  So nix the dividing by 365 (I also forgot to apply the 95% part).  So rebuilt estimate of an average of 0.0062 $W/m^2$.  A bit higher than artxabaleta's value (particularly given this one should be the mean, not max), but similar scale.  No idea where the username_5's Wikipedia article for it's values, but the general agreement is the contribution is rather small.  
> **username_6**: You could make a crude estimate: it would heat the bottom 2m of air by $(0.0062 J/{m^2 s})/((2m)(1 m^2)(1.225 kg/m^2)(1006 J/{kg K})) = 0.01 °C/day$ or about $3°C/yr$ contribution.  But that's really highly dishonest, as that energy would heavily radiate away during that year.  I don't know if it's legitimate way to consider it (assumes radiates processes are fairly similar for the different energy forms), but if $1000 W/m^2$ is about $290K$, then $0.0062 W/m^2$ would effect around a $0.002K$ change.  
> **username_4**: Downvoted because the source of that number is a wikipedia article, and in that article, that figure of 3.321 TW is tagged as "citation needed." That number is far too precise. Estimates in the scientific literature vary from well under 3 to almost 4 TW.  
> **username_6**: @username_4 I was the one who added the "needs citation" on Wikipedia as it bugged me too.  But a loose 3-4 TW is still good for rough estimation, as hopefully the math I've offered gives a fair attempt to.  

---

