What is the temperature of the clear night sky from the surface of Earth? 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?
 A: 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.
A: 
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).
