Is the Earth heated up by the tides?

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?

• 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 Mar 10 '17 at 2:12

A good estimate of the lunar tidal energy dissipated into the oceans is 2.5 Terawatts (Munk, 1997; Le Provost & Lyard, 1997). The value estimated comes mainly from two different sources: from harmonic calculations and from altimetry estimates using satellite observations (e.g., 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).

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

Egbert & Ray (2000) 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). 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.

The repercussion of this fact are given by Wunsch (2000):

...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.

• 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 is inside the mantle, after all. – Spencer Mar 4 '17 at 15:13
• @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 Mar 5 '17 at 10:25
• 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 Mar 3 '17 at 13:04
• (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 Mar 3 '17 at 13:07