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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.

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  • $\begingroup$ The problem might be approached by looking at how much the earth's rotation has been slowed over a number of years. Of course you would also have to consider the sun's influence. $\endgroup$ – Friddy Jul 2 at 23:25
  • $\begingroup$ Take a look at this what if - the interesting part - for your question - is, that moonlight is about only 100°C "hot". So not much IR that is transported there. $\endgroup$ – Erik Jul 3 at 8:06
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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

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  • $\begingroup$ I presume the IR input from moonlight is pretty inconsequential? $\endgroup$ – Andy Jul 3 at 0:45
  • $\begingroup$ @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. $\endgroup$ – Semidiurnal Simon Jul 3 at 4:19
  • $\begingroup$ @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. $\endgroup$ – gerrit Jul 3 at 8:27
  • $\begingroup$ @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. $\endgroup$ – jeffB Jul 3 at 18:23
  • $\begingroup$ @jeffB That fits over on Physics (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). $\endgroup$ – gerrit Jul 3 at 20:32

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