I am finding lots of information about Earth's energy budget, about how objects emit infrared radiation based on heat, and lots of related things, but nothing helping me figure out how much energy is actually being emitted by the ground at night.

At night, the ground will still be emitting electromagnetic energy. The amount of energy being emitted will vary based on many factors. Information that helps with many different types of land in various conditions would be great, but for my case I'm focusing on a stereotypical grassy plain in an area which has been between 5C-40C (40F-100F) for the past weeks.

Ultimately I'd like to have a rough ballpark figure of watts per square meter.

In response to @Michael's comment:

The amount of energy emitted by night depends on how much it receives during the day, water content, length and type of grass and other vegetation, how many "past weeks" have there been, and what was the temperature of the ground when the relevant period began.

During a sunny day the ground is said to receive approximately 1 kilowatt per square meter.

Assume grass of 10cm-50cm (about a half foot to a couple feet). There have been thousands of past weeks. Since we need to get picky let's say the temperature has been a constant 10C (50F) every hour of every night and 15C (60F) every hour of every day for the last thousand years except for the 1 or 2 hours day/night transition where the temperature adjusts linearly back and forth if that makes the exercise easier. And the average annual rainfall of this location is 100cm (40in), rain was coming regularly at that rate until 4 days ago. In the last 4 days, it has rained once, yesterday afternoon dropping about 1cm (half-inch), and the water table is generally about 5 feet under ground level.

The dirt is completely covered and appears as you would expect if you Googled an image of a grassy meadow. For example, the picture I've added at the bottom of the question.

In my opinion, however, the best answer would not just say "The ground emits 10 watts per square meter under those exact conditions!" (even that would be useful though) but would instead say "The type of grass actually makes a significant difference that can throw off even a ballpark estimate, as the difference between grass type A and grass type B produces a whopping half-order-of-magnitude difference in final results due to the way their outer material insulates them! And the difference between wet grass and dry grass is actually 2 orders of magnitude! If I were to simply pick a random, reasonable scenario of ABC it would be within the order of magnitude of 0.1 to 1 watts per square meter, and that's on the high end."

For anything else that matters feel free to simply pick a reasonable value and mention (even if only in a few words) why it matters and if your pick produces a high, low, or mid-range estimate.

enter image description here

  • $\begingroup$ You'll have to be much more specific than that. The amount of energy emitted by night depends on how much it receives during the day, water content, length and type of grass and other vegetation, how many "past weeks" have there been, and what was the temperature of the ground when the relevant period began. $\endgroup$ Feb 5, 2020 at 17:11
  • $\begingroup$ This article might be interesting for you: agupubs.onlinelibrary.wiley.com/doi/pdf/10.1002/2014JD022216 $\endgroup$ Feb 5, 2020 at 17:48
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    $\begingroup$ To get some rough starting point maybe take the temperatures from neo.sci.gsfc.nasa.gov/view.php?datasetId=MOD_LSTN_M and assume a black body spectrum? $\endgroup$ Feb 5, 2020 at 17:55
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    $\begingroup$ If you want a 'rough ballpark figure': treat it as a black-body and you know $T$. $\endgroup$
    – user18801
    Feb 5, 2020 at 19:19
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    $\begingroup$ To clarify: do you mean radiation emitted by the Earth surface, or radiation as observed at the top of the atmosphere (TOA) which is what actually matters for the energy budget? Most radiation emitted by the Earth surface is immediately absorbed by the atmosphere, then re-radiated by layers colder than the surface; this is exactly what the greenhouse effect is. Earth surface radiation is impossible to measure from space, but not immediately relevant for the energy budget. Do you want surface radiation or top of atmosphere radiation? $\endgroup$
    – gerrit
    Feb 5, 2020 at 21:01

1 Answer 1


I'm not sure how accurate you need, but beginning NASA's global 398.2 watts per square meter estimate from their energy budget a rounding down based on temperature variation should give a pretty good estimate. A lot of people have a problem with that number because it's more energy than we get from the sun, but that's because our atmosphere is a kind of blanket that traps heat, which warms the Earth above it's planetary eqilibrium temperature.

It's also (but less so), higher because Earth is warmed from below as well as above.

enter image description here

Diurnal temperature variation is between highest day temperature and coldest night temperature and you're looking for average. Using that number won't give a correct answer, but using, perhaps 1/2 that number might be in the range and that's over land. The diurnal temperature variation over water is much lower.

So, 288 Kelvin Earth, say 12 degree variation day to night, so 294 / 282. Ratio of 1.0425, to the 4th power means 18% more heat is radiated from land during the day than at night. Adjust the temperature if you think mine is off. That's 18% x 29% of the surface area.

For oceans, 71% of the surface area the temperature variation is smaller, say 2.5 degrees. 289.25 / 286.75, ratio of 1.0087. 4th power, means the oceans radiate about 3.5% more energy during the day than at night.

(18% x 29%) + (3.5% x 79%) = about 7.7%

So if Earth's surface radiation global average is about 398.2 w/m^2, then the day/night numbers work out to about 7.7% variation between them, or +/- 15.5 degrees, about 413.7 watts/m^2 during the day and about 382.7 at night. That's a pretty good estimate I think. I'm open to a better estimate if someone can think of a way to do it.

  • $\begingroup$ This diagram with back-radiation is pretty hard to understand for me. Back radiation could only happen for IR capturing gasses (I hate the expression "greenhouse gas" because Earth has nothing to do with a greenhouse with glasses suppressing convection) and since the atmosphere is almost black for these gasses wavelengths (CO2 captures almost 100% 15um in 10-20m from the ground up), no such radiation can cross the atmosphere. This would mean that all the "backradiation" would happen in a thin layer above ground. Quite troubling and in violation with the laws of physics (energy is moving from h $\endgroup$
    – Backradek
    Feb 9, 2020 at 18:14
  • $\begingroup$ A physical greenhouse does a few things. A house suppresses convection but isn't warmer than the outside during the day. A house with big windows can be warmer by letting light in. A greenhouse is warmer still for suppressing convection, letting sunlight in and trapping IR radiation. As for how back radiation works. I like to think of individual photons. A photon leaves the Earths surface, it flies generally upwards until it hits a molecule that can absorb it (greenhouse gas), it's absorbed, it makes the molecule dance, and then it's fairly quickly released in a random direction. $\endgroup$
    – userLTK
    Feb 10, 2020 at 15:07
  • $\begingroup$ You're thinking of CO2 in the atmosphere as a mirror reflecting back and you're thinking in percentages. That's kind of what the diagram looks like, but if you think of the photon and molecules that can absorb and release it individually, then it's a lot more like a lot of floating billiard balls moving almost at random and the photon bounces around, shot from molecule to molecule until it escapes into space. The amount of time it takes the average IR photon to leave the Earth or the average number of collisions is proportional to the amount of heat storage. $\endgroup$
    – userLTK
    Feb 10, 2020 at 15:12
  • $\begingroup$ So, more greenhouse gas molecules, more photon storage. The same thing (kind of) happens inside the sun. A photon created in the center of the sun by nuclear fusion can take 100,000 years to make it's way to the surface, then just 8 minutes to Earth. This is, ofcourse, oversimplified as the photons created in the sun are gamma and the photons that leave the sun are visible light and there's some energy conversion on the way and perhaps, splitting a photon into more than one, but the idea is the same. The photons inside the sun take a long time to get out, so the Sun heats up over time. $\endgroup$
    – userLTK
    Feb 10, 2020 at 15:16
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    $\begingroup$ @Backradek The NASA chart just oversimplifies, rather than counting the photons and timing their trip out, it converts to energy in, energy up and energy down, perfectly fine for thermodynamics. Not quite as good for imagining the precise process. $\endgroup$
    – userLTK
    Feb 10, 2020 at 15:18

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