Blog #4: Milky Way Rotation

3) You observe a star and you measure its flux to be $F_*$ and it's luminosity to be $L_*$

a) Give an expression for how far away the star is.

The equation that relates a star's flux, luminosity, and distance is given by the equation:

$$\begin{align} F_* = \frac{L_*}{4\pi D^2} \end{align}$$

In order to figure out how far away the star is, we can rearrange the equation to solve for the radius, $D$, which would be the distance the star is from the observer.

$$\begin{align} D = \left(\frac{L_*}{F_* 4 \pi}\right)^{\frac{1}{2}} \end{align}$$

b) What is its parallax?

The relationship between the parallax angle, $\theta$, and the distance between a planet and the observed star, $D$, is given by the equation:

$$\begin{align} \theta = \frac{1 \text{ AU}}{D} \end{align}$$

Since we solved the for the distance of the star in the previous part, we can substitute that distance for $D$ to solve for the parallax angle, $\theta$.

$$\begin{align} \theta &= \frac{1 \text{ AU}}{D}\\ \theta &= \frac{1\text{ AU}}{\left(\frac{L_*}{F_* 4 \pi}\right)^{\frac{1}{2}}}\\ \theta &=  1\text{ AU}\left( \frac{F_* 4 \pi}{L_*}\right)^{\frac{1}{2}} \end{align}$$

c) If the peak wavelength of its emission is at $\lambda_0$, what is the star's temperature? 

A star's temperature, $T$ and maximum wavelength, $\lambda_{max}$ are related by Wien's Displacement Law by a factor, $b = 2.9 \times 10^{-3}$ m K as follows:

$$\begin{align} \lambda_{max} = \frac{b}{T} \end{align}$$

Since the peak wavelength is $\lambda_{0}$, we can rearranging the previous equation and substitute $\lambda_{max} = \lambda_{0}$ to solve for the temperature as follows:

$$\begin{align} T = \frac{b}{\lambda_0} \end{align}$$

d) What is the star's radius, $R_*$?

We can use the relationship between the star's temperature and its radius using the equation for luminosity:

$$\begin{align} L_* = 4\pi R_*^2 \sigma T_*^4 \end{align}$$

Rearranging this equation to solve for the radius, $R_*$, we get:

$$\begin{align} R_* = \left(\frac{L_*}{4 \pi \sigma T_*^4}\right)^{\frac{1}{2}}\\ \end{align}$$

Finally, we can substitute $T_*$ with the expression for temperature we got in part c to get the radius, $R_*$, as follows:

$$\begin{align} R_* &= \left(\frac{L_*}{4 \pi \sigma \left(\frac{b}{\lambda_0}\right)^4}\right)^{\frac{1}{2}}\\ R_* &= \left(\frac{L_* \lambda_0^4}{4 \pi \sigma b^4}\right)^{\frac{1}{2}}\\ \end{align}$$