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Stellar Properties From Afar Worksheet

Author: Iryna Butsky Co-Author: Juliette Becker

- 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.
From this, we find that the radius of the sun is about 6.5 * 10^10 cm.
- Therefore, the AU, in solar diameters is
- 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.

- 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 . 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, the Luminosity of the sun is about 10^26 Watts or about 10^33erg/second
- We can find the effective temperature of the sun using the equation: 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.
- Once again, we know that , 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). Therefore, the Solar Unit is 3.5*10^5 ergs/cm^2/s

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