A straight forward answer would be, we do not really know. There are a couple of issues which make it hard to simply measure emissivity, and that is not just true for Earth, but also for the Moon for instance. Just like within the visible light, the optical properties of the surface within the LWIR range can be very diverse. Just think of the different colors you see when you look outside. What we perceive as colors are effectively different wavelengths of electromagnetic radiation within the relatively tight spectrum of visible light. The fact that surfaces have all these different colors means their optical properties vary substantially and individually over that narrow spectrum. But also the perspective may play a distinct role.
The total LWIR emission spectrum is much wider by comparison, complicating things even further. With the technology we have today it might yet seem easy to solve the riddle. Let us simply shoot a satellite into space which is going to scan the globe and we should have the answer. And indeed that is exactly being done. However the results are underwhelming. The problem has largely to do with GHGs.
For the largest part of the LWIR spectrum the atmosphere is opaque, even with clear skies. You just can not see through it with IR sensors. Also your best view is straight down onto the surface, because it is the shortest path through the atmosphere. The more you look sideways the stronger the atmosphere will impair your vision. So effectively satellites are restricted both in terms of perspective and spectrum. They can only give a partial answer which is a) relatively useless and b) can even be badly misleading.
We can discuss the problems based on a map provided by NASA1, showing emissivity at the wavelength of 8.3µm. 8.3µm btw. is still within the atmospheric window, so there is a good view onto the surface.
What we see are substantial deviations from 1, pointing out that the surface is not a perfect emitter. Some strongly vegetated areas at least come close 1, while arid regions show up to 40% less emissivity. Sand is indeed a very poor emitter at this specific wavelength, but it is not quite as bad for the whole emission spectrum, which roughly ranges from 3µm to 100µm. Equally we can not generalize the overall map.
The more interesting part is actually what the map does not show. There is no information on water nor ice (actually rather snow). For both the angle of emergence is a critical factor and there is not much use in measuring emissivity with a satellite looking more or less only straight down. In a nutshell satellites can do some auxiliary work, but the main job remains undone. It is something to bear in mind when "satellite data" on emissivity, or even emissions respectively, are presented as top notch state of the art "facts". Circumstantially these are rather low quality data products which need to be taken with a lot of salt and must not be generalized. There is one specific instance where this flaw has been exploited to exaggerate the significance of GHGs and everyone fell for it. I am going to feature this problem in an upcoming article.
For now we are still left with the basic question what is surface emissivity, and how we could find out. The best possible approach is water!? Water covers ~71% of the surface, though if we allow for ice shields it will a bit less, and thus is the monolithic dominant surface type. If we know the emissivity of water we can make some educated guesses on the remainder and we should have a good enough approximation.
The optical properties of water are well known physics, even in the LWIR or far-infrared range respectively. It is all there, we only need to look it up. And again, I can not express my disregard for "climate scientists" enough, who are simply too lazy for doing this, or whatever reason one might name. There is no excuse.
At this point I would also like to thank scienceofdoom for their article "Emissivity of the Ocean"2 which I found quite helpful some years ago. They discuss a lot of things we are going to deal with here, but for some reason they did not really go all the way. Also it seems the content of this article does not quite resonate elsewhere on the site. And definitely the article derails in the "Conclusion".
The “unofficial” result, calculating the average emissivity from the ratio: ε = 0.96.
This result is valid for 0-30°C. But I suspect the actual value will be modified slightly by the solid angle calculations. That is, the total flux from the surface (the Stefan-Boltzmann equation) is the spectral intensity integrated over all wavelengths, and integrated over all solid angles. So the reduced emissivity closer to the horizon will affect this measurement.
Right, emissivity to the surface normal is about 0.96 AND the reduced emissivity towards the horizon WILL modify the result slightly. Actually a lot more than just slightly. So let us pull up sleeves and get started.
With Fresnel equations we can calculate (among others..) the reflectivity of water, that is how much light is reflected on the surface of water relative to the angle of incidence. If the water is deep enough (like oceans, but unlike a swimming pool) we can assume the light not reflected is getting absorbed. So absorption = 1 - reflection. And because of Kirchhoff's law absorptivity = emissivity for identical wavelengths. If we consider absorptivity and emissivity to differ, than it is because both are relevant with different wavelengths. Either way, what we need to know is reflectivity and the rest is a no-brainer.
This is what the Fresnel equation3 looks like. What we will need is the refractive index of air (n1 = 1) and that of water, which in the range of visible light is about n2 = 1.33 on average. With these inputs we can calculate a curve for s-polarized and p-polarized light and the average of the two is a valid approximation for the total.
The black line in the chart gives us the share of sun light reflected by water relative to the angle of incidence. It is about 2% only if the sun is high up, but with the sun close to the horizon most of it is getting reflected. If you imagine a sun set, despite the sun is being dimmed a lot by the atmosphere, there will be an intense reflection in the water. This is the physical description of the phenomenon.
The same thing works for calculating emissivity, though there are a couple of things to consider. First of all I did not allow for the extinction coefficient, which is negligible with SW radiation but becomes significant in the LW range. This is complicated since the refractive index will turn into a complex number with the extinction coefficient as imaginary part. I openly admit I have not mastered that calculation yet. However we have the great site refractiveindex.info4 where we can retrieve these data anyhow. I will also admit I first assumed I could ignore the extinction coefficient altogether, since the results perfectly matched in the SW range. That failing by assumption thing affects me as well.
Then both, refractive index and extinction coefficient differ with wavelength. If it was only the refractive index we could try to calculate a weighted average an go from there. With both variables in play that is not an option. Rather I retrieved the results for individual wavelengths (scroll down to "Reflection Calculator", click the 3 lines right under the chart) and do that so often to get a reasonable resolution over the whole emission spectrum from 3-100µm.
Now we can do something pretty cool. We can add up all the information and draw a 3D chart to show the reflectivity of water relative to solid angle and wavelength. You probably have not seen that before. I had to make a compromise with scaling the wavelength to fit it all into one chart. The interval is 1 for 4-20, 2 for 20-30, 5 for 30-50 and 10 for 50-100µm.
To calculate hemispheric reflectivity (or effectively emissivity) we need to allow for the geometry. The density function of the surface of a hemisphere can be written as 1- cos(x). But of course we have to include Lambert's cosine law5 which turns our density function into sin(x)^2. Note: it is "incidentally" the same geometry as you have with a single source of light shining onto a hemisphere! Doing so will give us the hemispheric reflectivity for every data point on our wavelength curve.
You might want to compare the result with Huang et al (2016), Fig. 3.6 If you invert the curve, as 1- reflectivity = emissivity, and consider the logarithmic wavenumber scale they used, you will find the curves are essentially identical. So far we look good.
The only thing left to do now is to weight this curve according to Planck's law and for simplicity I will just assume a surface temperature of 288K. We would get a bit lower reflectivity with higher temperatures, as the emission spectrum gradually moves towards shorter wavelengths, and vice verse the opposite for lower temperatures. It cancels out, so 288K straight is perfectly fine.
The result is 0.0907. For emissivity we get 1 - 0.0907 = 0.909. The hemispheric surface emissivity of water is 0.909, which is pretty consistent with the 0.91 Baehr, Stefan7 name, referenced on german wikipedia8.
Ok, it was a bit technical but hopefully worth it. With water having an emissivity of only about 0.91 we have a substantial deviation from 1. At a temperature of 288K water does not emit 390W/m2, but only 355W/m2. Something very similar goes for the surface of Earth as a whole. As I have said before there are quite some uncertainties as we just do not have all the data, especially in the far-IR range. At least we know land surface emissivity is deviating substantially from 1 as well, so that global surface emissivity can not deviate a lot from 0.91.
For the GHE that insight is a serious blow. Again, we know clear sky emissions are about 270W/m2 while surface emissions are only about 355W/m2. The difference between the two is 85W/m2. So the GHE attributable to GHGs has shrunk from 150W/m2, which is about the usual erroneous claim, to only 85W/m2. Temperature wise that corresponds to roughly 19K, again not 33K! And of course we will have adept our super sophisticated emission graph anew..
Finally it brings up the question what to think about the "Earth Energy Budget" as presented by NASA9, to name just one instance. What you think might be appropriate to call the "scientists" responsible for it? And what would you call the "critical scientists" who say there is nothing to see here, and who disagrees is stupid? Regrettably I can not give you the answers on these questions. ;)