Blog #15, Worksheet #5.1, Distance Ladder, Problem #2

2) Okay, let’s use your new-found knowledge of magnitudes. 

(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?

Using our chart from question #1, we can see that for a star with a difference in magnitude \(\delta m\) of 3, we have a difference of flux of about 16. That means we must observe star A 16 times longer to obtain the same amount of energy as star B.

(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?

We know that flux is related to distance through a relationship of F ~ d^2. Thus we take our flux equations from question #1 and come up with the relationship, \[\frac{F_1}{F_2} = (\frac{d}{10})^2 = 10^{.4(M-m)}\]We take the log of both sides and end up with \[2 log (\frac{d}{10}) = .4(M-m)\]Rearrange to solve isolate the M's and get \[5 log(\frac{d}{10}) = (M-m)\] and finally we solve for little m to get \[m = M - 5 log (\frac{d}{10})\]

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

From part (b) we know that \[\frac{F_1}{F_2} = (\frac{d}{10})^2 = 10^{.4(M-m)}\]From here we can solve that \[d^2 = 100 pc \cdot 10^{.4(M-m)} \]Then to solve for parallax all we do is solve for d, giving \[d = 10 pc \cdot (10^{.4(M-m)})^{\frac{1}{2}} \]Which when simplified gives us a final answer of \[d = 10 pc \cdot 10^{.2(M-m)} = p\]