# Cooling film area to undo global warming

It is estimated that only 100 square miles of solar panels could completely satisfy the United States demand for energy.

This got me thinking: A cooling film developed at UC Boulder claims it can eject 93 W/m2 into space under direct sunshine, whereas the sun puts 1120 W/m2 onto the earth.

The cooling power of the film is a significant fraction of the total irradiance of the sun. So how many square kilometers of cooling film would be required to return temperatures to (say) 1950 levels assuming current CO2 levels?

• Isn't the solar constant closer to 1366W/m^2 as in my answer? – Communisty Aug 21 '17 at 12:34
• part of the problem is that the Earth's albedo will be changing, with less snow/ice on the surface. So, the amount of light absorbed will be increasing too. That being said, rooftop modification in urban areas is an ongoing field of research... perhaps this cooling film would be useful there? – farrenthorpe Aug 21 '17 at 14:51
• @Communisty: I was looking for an order of magnitude, so the precise number isn't hugely relevant. – user14717 Aug 21 '17 at 18:32
• While perhaps obvious, it's worth pointing out that 100 square miles of solar panels might provide enough energy but we have no means to store that energy, so it would be quite unproductive without enormous batteries/capacitors/other methods of energy storage and, lets not forget, most cars and all planes/trucks don't run on electricity. Enormously more infrastructure is actually required than just 100 square miles of solar panels to supply the nation's needs. I still support the idea though. – userLTK Aug 22 '17 at 5:33

I will answer the question with some simplifying assumptions that might then provide a coarse guess of the area needed. First lets assume that temperatures have risen $1^{o}C$ and that the cooling film acts as increasing surface albedo effectively reflecting that an additional fraction of sunlight (93/1366) from earth. By following the footsteps of the idealized greenhouse model we can derive the surface temperature as: $$T_{s} = \bigg(\frac{S_{0}(1 - \alpha_{p})}{4 \sigma (1 - \frac{\epsilon}{2})}\bigg)^{1/4}$$ Where $T{_s}$ is surface temperature, $S_{0}$ is the solar constant ($1366W/m^ 2$), $\alpha_{p}$ is the planetary albedo, $\sigma$ is the Stefan Boltzmann constant and $\epsilon$ is the emissivity. By approximating the planetary albedo as 0.3 and emissivity as 0.78 we get that the surface temperature of is 288K. We can now set that the surface temperature is one degree smaller and there is an addition to the planetary albedo due to the cooling films. Solving for albedo: $$\alpha_{p} = -\bigg(\frac{T_{s}^{4}4\sigma (1-\frac{\epsilon}{2})}{S_{0}} - 1\bigg)$$ By substituting the values we find that planetary albedo has to be about 0.31, so cooling of one degree has been reached by a 0.01 additional albedo. The ratio of the cooling film absorbency and solar constant $\frac{93}{1366} \approx 0.07$ can be seen as the additional planetary albedo if the cooling film is installed globally. This being seven times the needed change of albedo implies that a seventh of earths surface should be covered.