Page:Lawhead columbia 0054D 12326.pdf/122

Earth receive? Well, just however much of the sun’s energy actually reaches as far as the Earth multiplied by the size of the area of the Earth that the sun is actually shining on. Filling in some values, we can expand that to:

$S={\tfrac {S_{o}}{4}}=\sigma T_{p}^{4}=F$

(4b)

In this expanded equation, $S_{o}$ is the solar constant (the amount of energy radiated by the sun which reaches Earth), which is something like $1367Wm^{-2}$. Why is this value divided by four? Well, consider the fact that only some of the Earth is actually receiving solar radiation at any particular time—the part of the Earth in which it is day time. Without too much loss of accuracy, we can think of the Earth as a whole as being a sphere, with only a single disc facing the sun at any given time. Since all the surface areas we'll be dealing with in what follows are areas of circles and disks, they're all also multiplied by $\pi r^{2}$; for the sake of keeping things as clean-looking as possible, I’ve just factored this out except when necessary, since it is a common multiple of all area terms. That’s the source of the mysterious division by 4 in (4b), though: the area of the Earth as a whole (approximated as a sphere) is $4\pi r^{2}$, while the area of a disk is just $\pi r^{2}$.

On the other side of the balance, we have $\sigma T_{p}^{4}=F$. The value $\sigma T_{p}^{4}$ is obtained by applying the Stefan-Boltzmann law, which gives the total energy radiated by a blackbody ($F$) as a function of its absolute temperature ($T_{p}$), modified by the Stefan-Boltzmann constant ($\sigma$), which itself is derived from other constants of nature (the speed of light in a vacuum and Planck's constant). Filling in actual observed values, we get:

${\tfrac {[(1367Wm^{-}2]}{4}}=[(5.670373\times 10^{-8}Wm^{-2})K^{-4}]255K^{4})$

(4c)

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