You observe a star you measure its flux to be F*. If the luminosity of the star is L*,
(a) Give an expression for how far away the star is.
(b) What is its parallax?
(c) If the peak wavelength of its emission is at λ0, what is the star’s temperature?
(c) If the peak wavelength of its emission is at λ0, what is the star’s temperature?
(d) What is the star’s radius, R*?
(a) Give an expression for how far away the star is.
Let's start with the star's distance away. Imagine that the star is emitting a fixed number of photons in all directions (which it is). Now imagine that you have a a sheet of cardboard (red lines). When you are close to the star, this sheet will block a greater number of photons (blue rays) than when you are far away from the star. Since luminosity is the total amount of energy radiated form the star, and flux is the energy received per unit of area, this lines up with our cardboard analogy to get us the equation that we want, where \(d\) is distance away from the star.
\[F_*=\frac{L_*}{4\pi d^2}\]
\[d^2=\frac{L_*}{4\pi F_*}\]
\[\boxed{d=\left(\frac{L_*}{4\pi F_*}\right)^{1/2}}\]
(b) What is its parallax?
Parallax angle (\(P\)), is the angle formed between a star and Earth when the Earth is 3 months further around the sun. Parallax angle relates to the star's distance in the following way:
\[d=\frac{1}{P}\]
\[P=\frac{1}{d}\]
We solved for \(d\) in the previous problem, so let's plug that in to get our answer.
\[\boxed{P=\left(\frac{4\pi F_*}{L_*}\right)^{1/2}}\]
(c) If the peak wavelength of its emission is at λ0, what is the star’s temperature?
Peak wavelength and stellar temperature are related by Wien's Law which states:
\[\lambda_0=\frac{b}{T_*}\]
Where \(b\) = Wien's constant = \(2.9\times 10^{-3}mK\)
Solving for temperature, we get:
\[\boxed{T_*=\frac{b}{\lambda_0}}\]
(d) What is the star’s radius, R*?
The luminosity of a star is dependent upon it's radius and temperature.
\[L_*=4\pi R_*^2 \sigma T_*^4\]
Where \(\sigma\) is Stefan's Constant with a value of \(5.67\times 10^{-5} erg\text{ }cm^{-2}s^{-1}K^{-4}\)
Solving for radius, we get:
\[R_*=\left(\frac{L_*}{4 \pi \sigma T_*^4}\right)^{1/2}\]
Plugging in for \(T_*\), we are left with:
\[R_*=\left(\frac{L_* \lambda_0^4}{4 \pi \sigma b^4}\right)^{1/2}\]
\[\boxed{R_*=\frac{\lambda_0^2}{2b^2}\left(\frac{L_*}{\pi \sigma}\right)^{1/2}}\]
Very nice rearrangement.
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