Blog #14: WS5.1, #2

2. a) Suppose you are observing two stars, Star A and Star B. Star A is 3 magnitudes fainter than Star B. How much longer do you need to observe Star A to collect the same amount of energy in your detector as you do for Star B? 

If the fluxes of Star A and Star B are \(F_A\) and \(F_B\), their magnitudes (\(m_A\) and (\m_B\)) should obey the relationship \(\frac{F_A}{F_B} = 2.5^{m_B-m_A}\). Since Star B is 3 magnitudes brighter, \(\frac{F_A}{F_B} = 2.5^3 = 15.6\). We'd need to observe Star A for about 16 times longer to get the same amount of energy.

b) Stars have both an apparent magnitude, m, which is how bright they appear from the Earth. They also have an absolute magnitude, M, which is the apparent magnitude a star would have at D = 10pc. How does the apparent magnitude, m, of a star with absolute magnitude M, depend on its distance, d, from you? 

The flux of the star at distance d would be \(F_d = \frac{L}{4\pi d^2}\), so the flux at the standard distance of 10 pc would be \(F_D = \frac{L}{4\pi (10pc)^2}\). We can then take the ratio of the two and use that to solve for m.
\(10^{0.4(m-M)} = \frac{F_D}{F_d}\) which simplifies to \(\frac{d^2}{D^2}\)
\(0.4(m-M) = log(\frac{d^2}{D^2})\)
\(m-M = 2.5log(\frac{d^2}{D^2})\)
\(m = M+5log(\frac{d}{D})\) where D = 10pc.

c) What is the star's parallax in terms of its apparent and absolute magnitude? 

We can rewrite the equation above to solve for d instead of m.
\(d = D*10^{0.2(m-M)}\)
We can then plug this into the parallax equation, \(\theta = \frac{1AU}{d}\).
\(\theta = \frac{1AU}{D*10^{0.2(m-M)}}\) where D = 10pc.