I do understand that open water and open ground cools by the means of convection — lower air takes the heat and goes up, where it cools.

But why does the Earth lose energy and where does it go? Does it emit heat radiation to space? Does the heat go underground? What does happen to it?

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    $\begingroup$ Hint: why are clear nights cooler than cloudy nights? $\endgroup$ Commented Jun 9, 2017 at 13:59
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    $\begingroup$ "During the night"... it's always night somewhere on the planet! $\endgroup$
    – cfx
    Commented Jun 10, 2017 at 13:53
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    $\begingroup$ First, this is a loaded question, it assumes that the Earth DOES cool at night time. Is that so? Don't think so. Second thing that's wrong with it, is that it's imprecise. The Earth doesn't cool of at night, PARTS of the Earth cool off at the NIGHT side. Third thing, this question is almost insultingly naive, since it seems to asssume that Earth goes somewhere at night where the Sun can't find it. Earth temparature in relation to the Sun seems to be a stable system. External influences like Climate Change might change that, but NOT the day-night cycle. $\endgroup$ Commented Jun 12, 2017 at 18:49

3 Answers 3


It emits infrared radiation to space.

This happens constantly, not only during the night, but during the day the net energy flux is positive because the amount energy coming from the Sun is much higher.

Things are complicated a little bit by the atmosphere, meaning that most of the radiation emitted from the ground does not reach space immediately, but instead gets first absorbed by the atmosphere and then re-emitted to space.

Also, some of the energy is first transported to the atmosphere by evaporation and condensation before being radiated away by it.

If you want to know more, this introductory explanation by NASA looks good.

In the picture (taken from the NASA website I linked above): satellite map showing the distribution of thermal infrared radiation emitted by Earth in September 2008).

enter image description here

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    $\begingroup$ fascinating that there's a strip of less-outgoing-hear along the equator. $\endgroup$
    – Fattie
    Commented Jun 8, 2017 at 13:59
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    $\begingroup$ @Fattie Very good observation! That is the intertropical convergence zone, where the trade winds converge, causing the formation of a cloud band :-) $\endgroup$
    – valerio
    Commented Jun 8, 2017 at 14:25
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    $\begingroup$ @Fattie you can also see that the emission is highest in hot desert areas (northern and southern Africa, Arabic Peninsula, Afghanistan/Pakistan, northern Australia, Western North America) where you have very little humidity and little vegetation and lowest where there is tropical rainforest (northern South America, Central Africa, India, South East Asia) where you have very high humidity and a lot of vegetation. You can also easily make out the southern end of the Himalayas from the border of that very strong emission zone. $\endgroup$
    – Adwaenyth
    Commented Jun 9, 2017 at 6:07

The Earth's surface, and the atmosphere, is in thermal contact with outer space, with the heat transfer mechanism being radiation. The Earth-atmosphere system is constantly losing heat by this mechanism. Indeed, this is the predominant mechanism that sets an atmosphereless planet's surface temperature: a planet of radius $R_\oplus$ without atmosphere and lit by a star with light of intensity $I$ will reach thermal steady at temperature $T$ state when incoming and outgoing radiative heat fluxes are equal:

$$\pi\,R_\oplus^2\,I = 4\,\pi\,R_\oplus^2\,\sigma\,(T^4 -T_{CMBR}^4)\tag{1}$$

where the right hand side expresses radiative transfer described by the Stefan-Boltzmann law (neglecting the solid angle subtended by the Sun), $\sigma$ is the Stefan-Boltzmann constant the $T_{CMBR}\approx 2.7{\rm K}$ is the cosmic background microwave radiation temperature, the "temperature of outer space". On solving (1) for $I = 1200{\rm W\,m^{-2}}$ (the power whereby the Sun lights the Earth) we get $T = 270{\rm K}$. For Earth, a more careful calculation takes account of the Earth's albedo (i.e. the proportion of the incident radiation that is reflected straight off in to space and is never converted to heat - the temperature of a perfectly reflecting orb in space would be completely unaffected by incoming radiation and woulf neither absorb nor radiate, staying at whatever its initial temperature is). This lowers the result. Another source of discrepancy is the greenhouse effect, which more than cancels the effect of the albedo of an atmosphereless Earth. But you can see that the result is in the right neighborhood for Earth's temperatures.

During the night, when there is no sunlight warming the atmosphere and the ground, the Earth's surface and atmosphere tends towards the temperature of the outerspace $T_{CMBR}$. The thermal inertia of the air and Earth's surface means that of course the nighttime is not long enough for the temperature fall to anything like all the way to $T_{CMBR}$ but, as you note, it falls some of the way and our nights are thus cooler.

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    $\begingroup$ one should maybe ling to the black body radiation page en.wikipedia.org/wiki/Black-body_radiation where there is also a discussion of planets around the sun. $\endgroup$
    – anna v
    Commented Jun 8, 2017 at 8:47
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    $\begingroup$ Nitpick: A perfectly reflective orb in a vacuum would stay at whatever temperature it is, since it would not neither absorb nor emit anything. My gut feeling is that the albedo in the general case does not change the equilibrium temperature because albedo affects emission and absorption alike (i.e. covered in snow or in coal, the equilibrium is the same). Am I wrong? $\endgroup$ Commented Jun 8, 2017 at 11:00
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    $\begingroup$ @PeterA.Schneider LOL - that's a great point about the orb! I don't think you're right about albedo, though. Although you're absolutely right that emissivity and absorptivity are equal (Kirchoff's Radiation Law), they are equal but frequency / wavelength dependent. Therefore, the albedo is different for blackbody radiation of different temperatures. So the game changes once 6000K sunlight gets absorbed to become 300K heat. This is why albedo is so important for setting climate, and why Earth experienced extreme events such as snowball Earth. $\endgroup$ Commented Jun 8, 2017 at 11:13
  • $\begingroup$ @PeterA.Schneider I also changed my statement about the orb. $\endgroup$ Commented Jun 8, 2017 at 11:16
  • $\begingroup$ Oh, good argument about different wavelengths. A body absorbing radiation should always heat up at least to the lowest temperature where it has low albedo (which may be much lower than that of the radiation it absorbs). $\endgroup$ Commented Jun 8, 2017 at 11:45

Though they aren't quite as high resolution as might be ideal, the meteoblue website appears to give reasonable meteograms for stations around the globe showing the energy components...

enter image description here
(Sample energy balance forecast, copyright meteoblue)

On the radiation section (the bottom of the many graphs), you can see:

  • The incoming solar radiation (SW down)... energy from the sun that will warm the ground. A daytime-only process.
  • The sensible heat... which is the conduction+convection up into atmosphere that you mentioned, and will be a cooling process for the ground... basically in it, the ground directly warms the air just above it (conduction) and that leads to thermals/mixing (convection). Basically a daytime-only process.
  • The latent heat... an often overlooked factor that can be shockingly significant in places like Florida (where I'm at). Energy is expended in evaporating water... and that energy is later released into the atmosphere whenever that vapor is finally condensed (or deposited). So it's a process that also removes energy from the surface (cooling it). Basically a daytime-only process (results in less drastic daytime temperature rises in maritime environments than in deserts).
  • The upwelling longwave radiation (LW up)... the Earth continually emitting radiation as all bodies do with any temperature... most of it in longer wavelengths than the sun due to the cooler temperature here compared to there (thus the name). This is a continual release of energy 24/7 that only varies a bit as the temperature changes... and is a cooling process.
  • The downwelling longwave radiation (LW down)... constituents in the atmosphere absorb energy (most of which is Earth's longwave energy), warm, and then re-emit the energy as their own radiators. A portion of this returns to the Earth. The most predominant re-emitors are greenhouse gases ($H_2O$, $CO_2$, methane, etc) and clouds (small contribution also from aerosols (solid particles in the atmosphere)). This will occur 24/7, though vary somewhat depending on cloud coverage, and is a net warming process to the Earth.

So typically:
During the day we get sunlight bringing in energy, with latent heat (evaporation) and sensible heat (conduction/thermals) also ramping up, dispelling some of the rapid energy gain. Then through the night, we really only have longwave radiation as the factor... a large portion of which is returned to Earth.

In more arid areas, there's less latent heat during the day (so heats up quicker), less LW down at night (due to less water vapor) (so cools down quicker), and so larger fluctuations overall. The opposite is true of maritime regions.

On cloudy days, there's less solar radiation, but more LW down. Daytime warms less, but nighttime cools less. So small fluctuations overall.

Some energy does indeed get conducted (and emitted) downwards into the Earth. But it must be a long-term balanced process (a mix of primarily diurnal and annual oscillations within the first few meters)... otherwise if the underground kept gaining energy overall, it would cause it to continue to warm up. Likewise, a very small amount of energy is actually provided to the surface from conduction (and emission) from energy actually emanating within the Earth (due to radioactive decay). However it's a tiny amount (Wikipedia quotes it as 0.027% of the overall budget).

This graphic and this associated Wikipedia article give a larger depiction of all the avenues of energy exchange going on within Earth's atmosphere and their typical scales. To extend that fully through the surface, you'd just need to add the small components of radiation/conduction into and out of the underground.

Please note that meteoblue's meteograms are forecast meteograms, not observations. Generally I don't think you'll find much full observational data. This is because pyranometers (to measure incoming solar radiation) aren't standard (though are included in some useful datasets, such as the Oklahoma Mesonet [scroll to bottom]). And other values require additional measurements and calculation (here is a great deal more information on how the other fluxes are determined).

  • $\begingroup$ (Towards sunrise, we'll often cool to the dewpoint, which will mostly stunt the low-atmospheric temperature decrease. I'm not sure just how much that impacts ground cooling... but dew/frosting probably actually adds some of the latent heat energy back down to the surface in the condensation process plus slows surface temperature change by changing heat capacity(?). And as the ground cools faster than the air, a very small bit of downward conduction may start up from ground to air as well.) $\endgroup$ Commented Jun 9, 2017 at 9:09

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