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**Geo-engineering Using Terrestrial Mirrors**

Let's imagine we can reduce global carbon emissions to net zero: we still need to make the planet colder to stop the positive feedback in the Arctic. There are various geo-engineering solutions that might be deployed to reduce the temperature of our overheating Earth. None of them are easy or cheap but, given the positive feedback we've unleashed, we desperately need to get these technologies working.

One idea (the converse of the Arctic Death Spiral) is to compensate for the reduction in sunlight reflected by Arctic sea ice with man-made mirrors. The most sensible place to locate these mirrors would be near the equator in an area of low population density.

The calculations below estimate the size of mirror required in order to change the Earth's average temperature.

As per A Farewell to Ice p47-49, let's approximate the Earth as a sphere with constant temperature over its whole surface area and use the Stefan-Boltzmann equation for the radiation emitted per unit of surface area of any body. As the Earth is in equilibrium, equate it to the total power the Earth receives from the Sun:

\begin{align}
\text{Power emitted as radiation} & = \text{Power received from the Sun} \\
4{\pi}R^2{\epsilon}{\sigma}{T_0}^4 & = {\pi}R^2S(1-\alpha_{av})
\end{align}
where

- \(R\) is the average radius of the Earth, 6.371 × 10
^{6}m - \(\epsilon\) is the Earth's emissivity; for a planet with no atmosphere \(\epsilon\) = 1; according to https://en.wikipedia.org/wiki/Climate_model, with our current atmosphere the Earth's emissivity \(\epsilon\) = 0.612
- \(\sigma\) is the Boltzmann constant, 5.67 × 10
^{-8}W/m^{2}/K^{4} - \(T_0\) is the Earth's average temperature, around 15°C or 288K
- \(S\) is the solar constant, 1,370W/m
^{2}, the amount of radiation reaching the Earth per unit area measured at right angles to the Sun's rays - \(\alpha_{av}\) is the Earth's (average) albedo, about 0.30

Now let's suppose we cover area A of the Earth's surface near the equator with mirrors. For simplicity of visualisation, let's imagine the mirrors in a band all the away around the Earth near the equator.

Looking at the Earth from the direction of the Sun, the projected area of mirror visible at any one time is $$A_{proj} = \frac{A}{\pi}$$

This area of mirrors will cause additional power to be reflected/radiated from the Earth, as follows:

\begin{align}
\Delta_P & = \text{power reflected with mirrors} - \text{power reflected without mirrors} \\
& = {A_{proj}}{P_s}\alpha_m - {A_{proj}}{P_s}\alpha_s
\end{align}
where

- \(P_s\) is the power of raw sunshine reaching the Earth's surface at midday on a cloudless day, 1,000W/m
^{2}, according to https://www.withouthotair.com/c6/page_38.shtml - less than \(S\) because some is absorbed by the atmosphere - \(\alpha_m\) is the albedo of a mirror (let's aim high, choosing the best mirror possible), 0.85, according to https://forum.corona-renderer.com/index.php?topic=2359.0
- \(\alpha_s\) is the albedo of desert sand, 0.4, ibid
- NB1: for simplicity assume the mirrors are located somewhere with near-zero cloud cover
- NB2: this equilibrium overestimates the cooling effect, because it overlooks the reflected sunlight that is absorbed on its way back out of the atmosphere

This simplifies to: $$\Delta_P = \frac{A{P_s}(\alpha_m - \alpha_s)}{\pi}$$

Let's insert this delta into our initial equilibrium equation:

\begin{align}
\text{Power emitted as radiation at temperature } T_m + \text{extra power reflected} & = \text{Power received from the Sun} \\
4{\pi}R^2{\epsilon}{\sigma}{T_m}^4 + \frac{A{P_s}(\alpha_m - \alpha_s)}{\pi} & = {\pi}R^2S(1-\alpha_{av})
\end{align}
where

- \(T_m\) is the new (lower) equilibrium, average temperature of the Earth with the mirrors deployed

We can substitute the right side of the equation from our initial equilibrium equation: $$4{\pi}R^2{\epsilon}{\sigma}{T_m}^4 + \frac{A{P_s}(\alpha_m - \alpha_s)}{\pi} = 4{\pi}R^2{\epsilon}{\sigma}{T_0}^4$$

Rearranging we get $$A = \frac{4{{\pi}^2}R^2{\epsilon}{\sigma}({T_0}^4 - {T_m}^4)}{{P_s}(\alpha_m - \alpha_s)}$$

Suppose we want to reduce the temperature of the Earth by 1°C, ie \(T_m\) = 287K then \begin{align} A & = 1.17 \times 10^{13}m^2 \\ & = 11,700,000km^2 \end{align}

That's a square of mirrors measuring 3,400km on each side, an area larger than the Sahara desert, 9,200,000km^{2}, and equal to 2.3% of the surface of the globe. That's a terrifyingly massive project! Let's hope we find more feasible geo-engineering solutions.

Covering an area larger than the Sahara in mirrors is not at all a sensible idea, in fact it's a terrible idea. There are big risks of unexpected side-effects eg regional climate-change, drought and flooding. This calculation is simply an attempt to get an idea of how large our geo-engineering projects need to be. Dangerous as this idea might be, the alternative of not geo-engineering will lead eventually to even worse outcomes.

*Question: what if we attached mirrors to the roofs of all existing buildings?*

https://ourworldindata.org/land-use reports that "total built-up land (villages, towns, cities and infrastructure) would fit into an area the size of Libya". That's 1,760,000km^{2} and may also include areas such as parks and roads which are unroofed. That's significant but a long way short of the required total.

*Question: what if we painted all roads white?*

The calculation on https://www.quora.com/What-percentage-of-land-area-in-the-world-is-covered-with-pavement looks credible, suggesting that there are around 200 billion m^{2} of road. That's 200,000km^{2}, so relatively insignificant.

See also Geo-engineering Using Mirrors In Space and Geo-engineering By Reforesting

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