Stellar Magitude: Blog 14, Worksheet 5.1, Problem 2

Okay, let’s use your new-found knowledge of magnitudes. In the first problem of the wroksheet, we derived a relationship that looks something like this:

Source

Now lets put that knowledge to use: 

(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? 

From question 1, we know: \[ \frac{F_1}{F_2} \approx 10^{0.4( m_2 - m_1)} \] For this problem we have  \[ \frac{F_1}{F_2} \approx 10^{0.4 \cdot 3} \approx 15.6 \] So we would have to observe for about 16 times longer.

(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 = 10 pc. How does the apparent magnitude, m, of a star with absolute magnitude M, depend on its distance, d away from you? 

For this problem, we are going to look back to a few weeks ago and use our definition of flux:
\[F_M  = \frac{L}{4 \pi d^2} \] \[F_m = \frac{L}{4 \pi r^2} \]  and  \[ \frac{F_1}{F_2} \approx 10^{0.4( M - m)} \] \[ \frac{d^2}{r^2} = 10^{0.4(M- m)} \] \[ \frac{2}{5}(M - m) = log_{10} \left( \frac{d^2}{r^2} \right) \] We know d=10pc, so  \[ \frac{2}{5}(M - m) = log_{10} \left( \frac{100}{r^2} \right) \]  \[ \frac{2}{5}(M - m) = log_{10}(100) - log_{10}(r^2) \] \[ \frac{2}{5}(M - m) = 2 - 2log_{10}(r) \] \[ m= M + 5log_{10}(r) -5\]

(c) What is the star’s parallax in terms of its apparent and absolute magnitudes?

We know from blog 2 that, \[ \theta = \frac{1 \: AU}{r}\] And we can find r from our work above \[ r = \left( \frac{100}{10^{0.4(M-m)}} \right)^{\frac{1}{2}}\] So \[ \theta = \frac{1 \: AU}{\left( \frac{100}{10^{0.4(M-m)}} \right)^{\frac{1}{2}}}\] but we need to multiply by 1000 to get our answer in AU/kpc so it is really  \[ \theta = \frac{1 \: AU}{\left( \frac{100}{10^{0.4(M-m)}} \right)^{\frac{1}{2}}}\cdot (1000) \] This simplifies to \[\theta = \frac{1000 \left( 10^{0.4(M-m)} \right)^{\frac{1}{2}}}{10} = 100 \left( 10^{0.4(M-m)} \right)^{\frac{1}{2}} \]