<p>According to a slightly old estimate, the total power input from the tides is approx 3.7 TW [1]. Of this about 3.2 TW is from the moon (the rest is from the sun). Most of it goes into moving the oceans around, and some of it into doing the same with the land - but both ultimately end up as heat.</p>

<p>The question of how much this power affects the temperature of the earth is beyond my expertise, but it's a small amount of power at a planetary scale. Some very approximate comparisons:</p>

<ul>
<li>Tidal power from the moon: 3.2 TW [1] </li>
<li>Total global electricity consumption: 3-4 TW</li>
<li>Geothermal heat flow from the earth's interior: 30 TW [1]</li>
<li>Total sunlight hitting Earth (outside the atmosphere): ~175,000 TW [1,2]</li>
</ul>

<p>[1] Munk &amp; Wunsch 1998; <a href="http://www.sciencedirect.com/science/article/pii/S0967063798000703" rel="nofollow noreferrer">http://www.sciencedirect.com/science/article/pii/S0967063798000703</a> - probably paywalled.</p>

<p>[2] <a href="https://www.e-education.psu.edu/earth103/node/1004" rel="nofollow noreferrer">https://www.e-education.psu.edu/earth103/node/1004</a></p>

CC BY-SA 4.0
2019-07-03T00:45:51.490Z
I presume the IR input from moonlight is pretty inconsequential?
Andy
CC BY-SA 4.0
2019-07-03T04:19:14.023Z
@andy hmm, not my area, but from some googling it looks like the intensity of light from a full moon is in the order of 1/100000th of sunlight - so actually possibly comparable to the tidal energy, but still sod all compared to sunlight.
Semidiurnal Simon
CC BY-SA 4.0
2019-07-03T08:27:58.540Z
@Andy On average and on first order, Earth and moon are in radiative equilibrium (both are at same distance from the Sun, after all), so the net IR input from moonlight is zero.
gerrit
CC BY-SA 4.0
2019-07-03T18:23:32.690Z
@gerrit could you expand on this a bit? It seems to me that if I'm surrounded by (in effect) a black body at 4°K, and a small patch of that surrounding surface is replaced by a black body at 373°K, my equilibrium temperature will increase.
jeffB
CC BY-SA 4.0
2019-07-03T20:32:49.000Z
@jeffB That fits over on [Physics.SE] (and would likely be a duplicate).  In absence of anything else, yes, but that's not the equilibrium state.  You can calculate what would be the *eventual* state in this 2 or 3 body situation, and at that point there is radiative equilibrium (neglecting internal heat generation in the Earth at this point).
gerrit
CC BY-SA 4.0
2019-07-02T23:52:55.507Z
2019-07-02T23:52:55.507Z
Semidiurnal Simon
<p>I'm wondering by how much the gravitation effects, tides etc caused by the moon (and other energy such as moonlight IR) increase the surface temperature of the Earth.</p>

<p>So how much warmer than the blackbody temperature is Earth due to energy imparted by the Moon? Or putting it the otherway, how much colder would Earth be if there was no Moon.</p>

CC BY-SA 4.0
2019-07-03T08:06:51.550Z
Take a look at this [what if](https://what-if.xkcd.com/145/) - the interesting part - for your question - is, that moonlight is about only 100°C "hot". So not much IR that is transported there.
Erik
<temperature><moon><gravity>
How much energy does the Moon transfer to the Earth, does this affect the Earth's surface temperature?
Andy
# How much energy does the Moon transfer to the Earth, does this affect the Earth's surface temperature?

(Asked by: username_0 on 2019-07-02)

I'm wondering by how much the gravitation effects, tides etc caused by the moon (and other energy such as moonlight IR) increase the surface temperature of the Earth.

So how much warmer than the blackbody temperature is Earth due to energy imparted by the Moon? Or putting it the otherway, how much colder would Earth be if there was no Moon.

**Tags:** temperature, moon, gravity

> **Comments:**
> **username_1**: Take a look at this [what if](https://what-if.xkcd.com/145/) - the interesting part - for your question - is, that moonlight is about only 100°C "hot". So not much IR that is transported there.  

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## 1 answers

### Answer by username_2 on 2019-07-02. Score: 3

According to a slightly old estimate, the total power input from the tides is approx 3.7 TW [1]. Of this about 3.2 TW is from the moon (the rest is from the sun). Most of it goes into moving the oceans around, and some of it into doing the same with the land - but both ultimately end up as heat.

The question of how much this power affects the temperature of the earth is beyond my expertise, but it's a small amount of power at a planetary scale. Some very approximate comparisons:

* Tidal power from the moon: 3.2 TW [1]
* Total global electricity consumption: 3-4 TW
* Geothermal heat flow from the earth's interior: 30 TW [1]
* Total sunlight hitting Earth (outside the atmosphere): ~175,000 TW [1,2]

[1] Munk & Wunsch 1998; <http://www.sciencedirect.com/science/article/pii/S0967063798000703> - probably paywalled.

[2] <https://www.e-education.psu.edu/earth103/node/1004>

> **Comments:**
> **username_0**: I presume the IR input from moonlight is pretty inconsequential?  
> **username_2**: @username_0 hmm, not my area, but from some googling it looks like the intensity of light from a full moon is in the order of 1/100000th of sunlight - so actually possibly comparable to the tidal energy, but still sod all compared to sunlight.  
> **username_3**: @username_0 On average and on first order, Earth and moon are in radiative equilibrium (both are at same distance from the Sun, after all), so the net IR input from moonlight is zero.  
> **username_4**: @username_3 could you expand on this a bit? It seems to me that if I'm surrounded by (in effect) a black body at 4°K, and a small patch of that surrounding surface is replaced by a black body at 373°K, my equilibrium temperature will increase.  
> **username_3**: @username_4 That fits over on [Physics.SE] (and would likely be a duplicate).  In absence of anything else, yes, but that's not the equilibrium state.  You can calculate what would be the *eventual* state in this 2 or 3 body situation, and at that point there is radiative equilibrium (neglecting internal heat generation in the Earth at this point).  

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