The amount of solar radiation received by the earth can be calculated using the constant known in the physical sciences as the solar constant. The solar constant, \(K\), is a measure of the amount of energy per unit time of all wavelengths in the sun's radiation normal to a unit area at the earth's surface.

\begin{align*} K&=1.353*10^{6}\frac{\text{ergs}}{(\text{cm.}^{2})(\text{sec.})} \\ &=1.048*10^{14}\frac{\text{BTU}}{(\text{mile}^{2})(\text{yr.})} \end{align*}The amount of radiation received by any particular area of the earth, \(A\), can be calculated as follows.

Let the area under consideration have a width \(W\) in the longitudal (NS) direction and length \(L\) in the latitudinal (EW) direction. See Fig. 1. Consider a strip of width \(W\) that encircles the earth as an extension of the \(L\) dimension as shown in Fig. 1. The length of this strip is the circumference of the earth at the latitude of the area \(A\). The area of the strip, \(A_{s}\), is given by

\begin{align*} A_{s}=W\pi D\cos\theta \\ \end{align*}where \(D=\) diameter of the earth

\(\theta=\) latitude at the center of the area \(A\)

The amount of radiation falling on area \(A\) at a particular time will depend on both the NS location of the strip that \(A\) is in and on the EW location of \(A\) within the strip. For example, in the longitudinal direction, an area of width W will receive less radiation if it is located near the poles than if it is located near the equator. In the latitudinal direction, a given length \(L\) of the strip will receive less radiation in the morning and evening hours than at midday. These factors can be taken into consideration by the realization that only that portion of an area that is perpendicular to the sun's rays will receive the full intensity of radiation. Therefore, in order to calculate the solar energy received by this strip, the amount of area normal to the incident radiation must be determined.

It can be shown that the normal component of the width \(W\) of the strip is given by \(W\cos\theta\), where \(\theta\) is the latitude of the center of width \(W\). In the longitudinal direction, it can be shown that the normal component of the length of the strip is given by \(D\cos\theta\), where \(D\) is the diameter of the earth. Thus the area of the strip, \(A_{n}\), normal to the incident radiation is

\begin{align*} A_{n}=WD\cos^{2}\theta \end{align*}The amount of solar energy, \(E_{s}\), received by the strip is

\begin{align*} E_{s}&=KA_{n} \\ &=KWD\cos^{2}\theta \end{align*}It is to be noted that even though a particular area within the strip may receive a large variation of energy over a 24-hour period, the energy received by the entire strip is constant over a 24-hour period.

The density of energy, \(D_{e}\), received by the strip, in terms of its global area, is thus

\begin{align*} D_{e}&=\frac{E_{s}}{A_{s}} \\ &=\frac{KWD\cos^{2}\theta}{W\pi D\cos\theta} \\ &=\frac{K\cos\theta}{\pi} \end{align*}Over a 24-hour period, this density applies to any part of the strip. For an area \(A\) of the strip, the energy \(E\) received by \(A\) over a 24-hour period is

\begin{align} E&=AD_{e}\text{ or} \\ E&=\frac{K}{\pi}A\cos\theta\tag{1}\label{eq:1} \\ \end{align}There will be a summer-winter difference in the solar energy received by area \(A\) that must also be considered. This difference is caused by the slight tilt of the earth's NS axis relative to its orbital plane around the sun. The formulas developed assume no tilt in the earth's axis, which results in the actual winter energy received being less than that calculated, and the actual summer energy received being more. The formulas therefore represent an approximate yearly average of the amount of energy received.

\(\text{Equation }\ref{eq:1}\) can be used to calculate the incident radiation on any area \(A\) on the earth. Note that the energy depends only on the magnitude of \(A\) and the latitude of \(A\) (i.e., the latitude of the center of \(A\)). The formula averages out (approximately) the seasonal and day/night variations in the energy received. In order to avoid confusion on this point, it is best to calculate the energy on a per year basis (rather than a per hour basis).

For the USA (excluding Alaska), \(A = 3.03x10^{6}\text{ miles}^{2}, \theta=37^{\circ}\).

\begin{align*} E&=\frac{(1.048*10^{14})(3.03*10^{6})\cos 37^{\circ}}{\pi} \\ &=0.807*10^{20}\text{ BTU/yr.} \end{align*}For Alaska, \(A=0.586x10^{6}\text{ miles}^{2}, \theta=65^{\circ}\).

\begin{align*} E&=\frac{(1.048*10^{14})(0.586*10^{6})\cos 65^{\circ}}{\pi} \\ &=0.083*10^{20}\text{ BTU/yr.} \end{align*}Thus while Alaska constitutes \(16\%\) of the USA land area, it receives only \(9\%\) of the solar energy.

\(\text{Equation }\ref{eq:1}\) is only approximate and becomes less accurate as the area \(A\) becomes larger. It cannot be used to calculate the amount of solar energy received by the entire earth. This energy, \(E_{a}\), can be shown to be

\begin{align*} E_{a}&=\frac{K\pi D^{2}}{4} \\ &=52*10^{20}\text{ BTU/yr.} \end{align*}where \(D\), the diameter of the earth, is \(7926\text{ miles}\).

It is to be noted that the USA comprises \(1.8\%\) of the area of the earth (\(197x10^{6}\text{ miles}^{2}\)) and receives \(1.7\%\) of the incident radiation. As an efficient receiver of solar energy, therefore, the USA is about average for the world as a whole.