## Friday, October 21, 2011

### Stellar Properties From Afar Worksheet

Author: Iryna Butsky  Co-Author: Juliette Becker

1. We know that the angular diameter of the sun d = 0.5 degrees. We also know that the size of the semi-major axis of the Earth's orbit around the sun, a = 1.5 x 10^13cm.
• We can now set up a simple proportion to find the radius of the sun.
• $\frac{0.5}{360}=\frac{R_{\odot}}{\pi a_{\oplus}}$  $\frac{0.5}{360}=\frac{R_{\odot}}{\pi (1.5\times10^{13})}$ From this, we find that the radius of the sun is about  6.5 * 10^10 cm.
• Therefore, the AU, in solar diameters is  $\frac{1.5\times10^{13}}{2\times6.5\times10^{10}}\simeq 100$
• From Newton's version of Kepler's third law we have the following equation  we know that P =31,556,926 secons , a = 1.5*10^13cm, and G = 6.674*10^-8 cm^3 g^-1s^-2. Plugging the numbers in we find that the mass of the Sun is about 2*10^33g.
2. Using only a lightbulb and our skin, we were able to figure out the luminosity of the Sun. We are given that the power output of the light bulb is 100 W with an efficiency of .1. Therefore, the light bulb emits the equivalence of 10 W. We know that $L=F\times4\pi R^2$. Therefore, we can set up a simple proportion to find the Luminosity of the sun. We decided that the energy given off from the light bulb feels roughly the same as the sun on a warm day at about 5cm distance. Therefore, $\small \frac{L_{\odot}}{10W}=\frac{(1.5\times10^{13})cm}{5cm}$  the Luminosity of the sun is about  10^26 Watts or about 10^33erg/second
3. We can find the effective temperature of the sun using the equation: $\small \lambda_{max}T=0.002897755 \;m\; K$ We know that lambda max is equal to 500 nanometers, we find that the effective temperature of the sun is 5.8*10^3 degrees Kelvin.
4. Once again, we know that $L=F\times4\pi R^2$, and we found the solar luminosity to be 10^33 ergs/second. We can set up a proportion relation to find the bolometric flux at the Earth (the Solar Unit). $\small F_{bol}=\frac{L_{\odot}}{4\pi R_{Au}^2}=\frac{10^{33}}{4\pi(1.5\times 10^{13})^2}$ Therefore, the Solar Unit is 3.5*10^5 ergs/cm^2/s