## A

TEarth a 5270K a 300 K

10 X 103 nm

quantity aT4 is only the energy per second per unit surface area. Therefore, to obtain the luminosity, we must multiply it by the surface area. If the star is a sphere with radius R, the surface area is (4-n-R2 ), so the luminosity is

Scaling results can be useful because they show how different physical parameters are related to each other. It also provides us with a way of using an equation even if we don't remember the constants.

Suppose we are interested in the total energy given off by a blackbody (per unit time per unit surface area) over the whole electromagnetic spectrum. We must add the contributions at all wavelengths. This amounts to taking an integral over blackbody curves, such as those in Fig. 2.5. Since a hotter blackbody gives off more energy at all wavelengths than a cooler one, and is particularly dominant at shorter wavelengths, we would expect a hotter blackbody to give off much more energy than a cooler one. Indeed, this is the case. The total energy per unit time, per unit surface area, E, given off by a blackbody is proportional to the fourth power of the temperature. That is

This relationship is called the Stefan-Boltzmann law. The constant of proportionality, a, is called the Stefan-Boltzmann constant. It has a value of 5.7 X 10-5 erg/(cm2 K4 s). This law was first determined experimentally, but it can also be derived theoretically. The T4 dependence means that E depends strongly on T. If we double the temperature of an object, the rate at which it gives off energy goes up by a factor of 16. If we change the temperature by a factor of ten (say from 300 K to 3000 K), the energy radiated goes up by a factor of 104.

For a star, we are interested in the total luminosity. The luminosity is the total energy per second (i.e. the power) given off by the star. The

Example 2.2 Luminosity of the Sun The surface temperature of the Sun is about 5800 K and its radius is 7 X 105 km (7 X 1010 cm). What is the luminosity of the Sun?

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