####### PHY 251, Intermediate Astronomy Spring 2012

######## ACTIVITY 4: The Luminosity of the Sun

Now that we've figured out the solar constant S , the amount of solar energy striking a one-meter square at Earth each second, it's straightforward to compute the Sun's luminosity L , the overall amount of energy the Sun emits per second. But first a brief detour.

####### An Earth's Worth of Solar Energy

With the information gained in the previous activity - the solar constant S - it's possible to figure out how much solar energy is intercepted by the entire Earth. The solar constant S is the solar energy illuminating a 1-meter square situated at Earth's distance from the Sun. Imagine placing a second 1 -meter square alongside the first square; together these squares intercept twice the energy of the single square, that is, 2 x S . Adding a third 1- meter square raises the collective solar energy to 3 x S . For 4 such squares, the total solar energy is 4 x S , and so on. Now imagine assembling an immense number N of these squares, enough to cover a circular disk as wide as the Earth itself. (Sure, the corners of some squares will stick out beyond the disk's circular border, but that has minimal effect on our estimation.) The energy striking this big disk - numerically, N x S - is equivalent the solar illumination over our entire planet each second.

The number N is easy to calculate: it is the number of 1-meter squares required to assemble an Earth-sized disk, or equivalently, it is the area A of a circular disk whose radius is equal to Earth's. The area of a circular disk is given by the formula A = R 2  = R , where R is the disk's radius. In this case, R = 6.378 x 10 meters, the radius of Earth. 2 6 

A = R 2 

1. Using the area formula, compute the area A of an Earth-size disk. Since the answer is expressed in units of square meters, it is equivalent to the number N described above.

2. To find the total solar illumination on Earth per second, multiply the accepted value of the solar constant S (1370 W/m ) by your answer for N in part 1, that is, compute N x S x S . Your answer will come out in units of watts. 2

N x S

3. How many times larger is the total solar illumination on Earth than (a) the collective energy generation rate of all the world's nuclear power plants: 4 x 10 watts; and (b) the average energy consumption rate for the entire 11 United States: 3.4 x 10 watts. (Note: These numbers represent the average energy generation or energy 12 consumption per second !)

####### The Sun's Luminosity

Now back to the task at hand: measuring Sun's luminosity L , the overall amount of energy it emits. To solve the problem, we add yet more 1-meter squares to the Earth-sized disk we already created above. We gently curve this growing disk inward so that it increasingly envelops the Sun. Ultimately, we will have built an enormous sphere that completely surrounds the Sun. Any light energy leaving the Sun's surface must necessarily strike this sphere on its way to deep space. (Let's ignore the miniscule fraction of solar energy intercepted by the inner planets Mercury and Venus.)

To find the Sun's luminosity L , we follow the previous procedure, but instead of computing the solar energy swept up by Earth, we compute the solar energy illuminating the Sun-concentric sphere we just created. That means multiplying the solar constant S by the surface area A of the sphere whose radius is equal to the radius of Earth's orbit, 1 AU, as shown in the figure. (Don't confuse the radius of Earth itself with the radius of its orbit, which is more than 20,000 times larger.)

4. The surface area A of a sphere is given by the formula A = 4 R 2  = 4 R , 2  the sphere's radius. In this case, R = 1 AU = 1.50 x 10 meters, the 11

A = 4 R 2 

where R is

average

PHY 251, page 8