Today we went over an example of the Brower-Cousteau model, which can be found in chapter 7 of the text.

The Sun’s Power on Earth Measured in Paper

We have discussed for some time the magnitude of solar power reaching the surface of the earth, and the point of today’s lecture is to put that in perspective a bit. The question is: We know that the earth receives roughly 81,000 terawatts of energy from the sun, that is, $8.1\times 10^{16}$ joules of energy every single second … but how much energy is that?

100 joules per second (100 watts) will power a standard incandescent light bulb. If we were to measure each watt as a single piece of paper, which averages about 0.1mm thick, then we’d have a stack of paper 1cm thick representing the power being used by that light bulb, since 100 times 0.1mm is 10mm, or 1cm. If we were to view the total energy coming in from the sun, that 81,000 terawatts, how tall would the corresponding stack of paper be? Here, we’re looking at $8.1\times 10^{16}$ times $1.0\times 10^{-3}$, where the first number represents the number of joules hitting the earth each second and the second is the thickness of a single sheet of paper in meters. The result is in meters, $m$. We calculate that as follows.

\[\begin{align}\left(8.1\times 10^{16}\right)\times\left(1.0\times 10^{-3}m\right)&=8.1\times 1.0\times 10^{16}\times 10^{-3}m\cr &=8.1\times 10^{13}m\cr &=8.1\times 10^{10}km\end{align}\]

where now the answer has been reduced to kilometers. How big is that? It’s significantly bigger than the distance from the earth to the sun.

The Brower-Cousteau Model

The Brower-Cousteau model says that we can make something too big or too small easier to intuitively grasp by scaling it appropriately to a human scale. The example we looks at in class was looking at how thick the earth’s atmosphere is. If we were to scale to earth down to the size of a basketball, how thick would the atmosphere be?

The earth has a radius of $6.38\times 10^6$ meters (at the equator anyway), while as basketball has a radius of $1.193\times 10^{-1}$ meters, as we calculated in class from the circumference. That means that proportionally speaking, the ration of the basketball to the earth is roughly

\[\frac{1.193\times 10^{-1}m}{6.38\times 10^6m}\approx 1.8699\times 10^{-8}.\]

This means that if we know a distance on the earth, we can get the corresponding distance on the basketball by multiplying it by $1.8699\times 10^{-8}$. For instance, according to NASA, roughly 50% of the earth’s atmosphere is located below 5.6km (18,000ft) in elevation. How big is that on the basketball? Let’s compute it.

\[\begin{align}5.6km \times 1.8699\times 10^{-8} &= 5.6\times 10^3m\times 1.8699\times 10^{-8}\cr &= 5.6\times 1.8699\times 10^3\times 10^{-8}m\cr &=10.47 \times 10^{-5}m \cr &= 10.47 \times 10^{-2}mm\cr &=1.47\times 10^{-1}mm\end{align}\]

In other words, if the earth were scaled down to the size of a basketball, 50% of the atmosphere would be within the first 2 tenths of a millimeter from the surface of the basketball.


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