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Before you all jump in with 2.73 K or thereabouts, this is more of an experimental question. It will obviously depend on humidity and radiation being scattered back towards the surface of the Earth. Any ideas? Anyone ever pointed a pyrometer or similar at the night sky?

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    $\begingroup$ This is definitely one of my favorite questions ever on this site. Also, what does 2K73 mean? $\endgroup$ – DanielSank Dec 17 '14 at 21:05
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    $\begingroup$ By the way, it might be useful to explicitly state in the question that the sky (like any other object) might not really have a temperature. It's possible that the sky doesn't actually act like a radiation source in thermal equilibrium with a proper temperature. There's cosmic background radiation hitting the atmosphere and doing I-don't-know-what. I don't want to edit your question in case that's not what you're trying to ask, but it's something to think about. $\endgroup$ – DanielSank Dec 17 '14 at 21:08
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    $\begingroup$ Indeed, "2K73" could reasonably be untypod to "273 K" or "2.73 K." I never before realized how close the freezing point of water was to 100 times the CMB temperature. $\endgroup$ – user10851 Dec 17 '14 at 21:19
  • $\begingroup$ Mind you "clear" to the mark 1 eyeball means a wide range of absorbtivity as measured by sensitive instruments. $\endgroup$ – dmckee Dec 17 '14 at 21:21
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    $\begingroup$ Nevertheless, you point a pyrometer at the clear night sky it will give you a temperature reading as a summation over a wide band of IR and optical. Also, 2K73 is just an electronics engineering way of writing the decimal point with the unit in question. For example, 2.73 Ohms is more often written as 2R73, or 2.73 kilo-Ohms as 2K73. Or, 2.73 Volts becomes 2V73 $\endgroup$ – user56903 Dec 18 '14 at 8:59
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What is the temperature of the clear night sky from the surface of Earth?

It's much closer to 273 K than 2.73 K. The answer depends on the surface temperature, the humidity, the temperature gradient through the atmosphere, and what exactly you mean by "the temperature of the clear night sky".

The Swinbank formula provides an ad hoc expression for the power radiated by the night sky. A modified version of this formula from Goforth et al. is $$P_{\text{thermal}} = (1+KC^2)8.78\times 10^{-13}\,T^{5.852}\,{RH}^{0.07195}$$ where

  • $K$ is a scale factor based on cloud height, ranging from 0.34 for very low clouds to 0.06 for very high clouds,
  • $C$ is the fraction of the sky covered by clouds,
  • $T$ is the surface temperature, in kelvins,
  • $RH$ is the surface relative humidity, as a percentage (e.g., $RH$ would be 25 in the case of 25% relative humidity), and
  • $P_{\text{thermal}}$ is the night sky radiation, in watts per square meter.

This can be converted to an effective temperature via the Stefan-Boltzmann law. Now the question arises as to whether you are asking about the effective black body temperature or effective gray body temperature of the night sky. In the first case the Stefan-Boltzmann law yields $T = (P/\sigma)^{1/4}$. Taking emissivity into account yields $T = (P/(\epsilon \sigma))^{1/4}$, where $\epsilon\approx 0.74$ is the emissivity of the atmosphere.

A couple of examples:

  • A cool clear night in the desert, with a temperature of 5°C and a relative humidity of 5%. The modified Swinbank formula yields a flux of 198 w/m2, which in turn corresponds to a black body temperature of -29.9°C or a gray body temperature of -10.9°C.

  • A warm clear night in the countryside, with a temperature of 15°C and a relative humidity of 25%. The modified Swinbank formula in this case yields a flux of 274 w/m2, which in turn corresponds to a black body temperature of -9.5°C or a gray body temperature of 11.1°C.

References

W.C. Swinbank, "Long‐wave radiation from clear skies," Quarterly Journal of the Royal Meteorological Society 89.381 (1963): 339-348.

Mark A. Goforth, George W. Gilchrist, and Joseph D. Sirianni, "Cloud effects on thermal downwelling sky radiance," AeroSense 2002, International Society for Optics and Photonics (2002).

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On 14th Sept 2016 at 20:25 in Woking UK the sky appeared clear to me with a few stars visible, and my infra-red thermometer ( Dr Meter HT550 ) reads -12.9 degrees Celsius when pointed at the sky. OK its un-calibrated, and emissivity of target being measured makes a difference, but the reading may be of interest to some people.

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    $\begingroup$ This answer really only gives a highly localized value & isn't much help without a broader set of data (e.g., whole sky). Contrast this answer with the accepted answer, which provides a mathematical & experimental result. Hence, this really doesn't answer the question. $\endgroup$ – Kyle Kanos Apr 24 at 16:20
  • $\begingroup$ I appreciate a bit of experimental data. Especially as the last part of the question specifically ask for it. $\endgroup$ – Rob Jeffries May 1 at 7:56
  • $\begingroup$ @KyleKanos I think a partial experimental answer of this type is very valuable. It does not address the average situation but it is a pretty good match to what was asked.This sort of direct experimental effort is an equal partner to a theoretical analysis, and adds to the total stock of experimental data. The answer is local but gives sufficient information for us to assess how ordinary or out of the ordinary it may be, and by looking up the instrument we can discover exactly what it measures (acceptance angle etc.) $\endgroup$ – Andrew Steane May 1 at 8:09

protected by Kyle Kanos Apr 24 at 16:10

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