Class Activity: Stellar Properties from Afar Problem 1

Given that the angular width of the sun is .5° and that 1 AU is 1.5 * 10¹³ cm, it's possible to find the radius of the Sun in centimeters. If we consider the angle subtended by the Sun from a certain point on Earth, we have a right triangle with one leg equal to 1 AU and one angle equal to half of the angular width of the sun, .25° or 15 arcminutes.

By using a trigonometric relation, we can find a value for R_s, the radius of the Sun.

So the radius of the Sun is 6.55 * 10¹⁰ cm. Next, we can get 1 AU in solar diameters by simply dividing 1 AU in centimeters by the radius of the Sun in centimeters:

So 1 AU is approximately 100 solar diameters. Now, using Newton's version of Kepler's third law, we can get the mass of the Sun.

This is Newton's version of Kepler's third law, if we assume that the mass of the Sun is significantly greater than the mass of the Earth, which it is. Then P is the period of the Earth, a is 1 AU, G is the gravitational constant, and M is the mass of the sun. Rearranging and plugging in the correct numbers:

We see that the mass of the sun from Kepler's third law is 6.4*10³² g.


Secondary author: Joanna Robaszewski