Main Concept

The Sun's radiation (sunlight) is closely approximated as blackbody radiation. Accordingly, the intensity or flux of solar radiation, $F$, which is a measure of how much power is being radiated from the Sun's surface per unit area, can be expressed in terms of the temperature, $T$, of the sun using the Stefan-Boltzmann law:

$F\={\mathrm{\σT}}^4$,

where $\mathrm{\σ}\=5.67 \times {10}^{-8}\frac{W}{{\mathrm{K}}^4 {\mathrm{m}}^2}$ is the Stefan-Boltzmann constant.

The luminosity of the Sun is a measure of how much energy the Sun produces per unit time. It is obtained by multiplying the solar intensity by the Sun's surface area:

$L\=4 \mathrm{\π} r_s^{2}F$,

where $r_s$ represents the radius of the Sun. To determine the flux passing through any other point in space where the Sun is visible, $r_s$ is replaced by $d$, which is the distance from the center of the Sun to the point in question. So, the formula now becomes:

$F\=\frac{L}{4 \mathrm{\π} d^2}$.

In this formula, the flux is proportional to the inverse square of the distance. This means that if an object's distance from the Sun doubles, the amount of sunlight hitting a given area will drop by a factor of four. This property is an example of the inverse-square law, which affects conserved quantities propagating evenly in all directions through three-dimensional space. Consecutive wavefronts of decreasing intensity can be visualized in the following interactive diagram:

As you can imagine, as the radiation moves farther from the Sun, the same amount of energy is spread out over a larger area (which is proportional to the square of the distance from the source), which makes the flux (power per unit area) correspondingly smaller.

Example

What is the radiant flux emitted into space by a light source with temperature of $1000 \mathrm{K}$? What is the luminosity of the solar radiation given by this surface if its radius is $5\times {10}^8\mathrm{m}$?   Adjust the sliders to change the radius and temperature of the surface providing solar radiation. Click the checkboxes to see the flux and luminosity calculations.   Radius:  Temperature:  Solution Using the formulas introduced in the previous section, you can determine both the flux and the luminosity produced by the specified surface.   To begin, calculate the flux:   $F\=\mathrm{\σ}\cdot T^4$ $F\=5.67 \times {10}^{-8}\frac{W}{{\mathrm{K}}^4 {\mathrm{m}}^2} {\left(1000 K\right)}^4$ $F\=56700 \mathrm{W}\/{\mathrm{m}}^2$.   You can now use this result to determine the luminosity:   $L\=4\cdot \mathrm{\π}\cdot R^2\cdot F$ $L\=4\cdot \mathrm{\π}\cdot {\left(5 \times {10}^8\mathrm{m}\right)}^2\cdot \left(56700 \mathrm{W}\/{\mathrm{m}}^2\right)$ $\mathrm{L}\=1.781 \times {10}^{23} \mathrm{W}$.   Therefore, a surface with a radius of $5 \times {10}^8\mathrm{m}$ and a temperature of $1000 \mathrm{K}$ has a radiant intensity of $56700 \mathrm{W}\/ {\mathrm{m}}^2$ and a luminosity of $1.781 \times {10}^{23 }\mathrm{W}$.

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