Blog #15: WS5.1, #3

3. A class of stars named Cepheid Variables reside in a special location in the Hertzsprung-Russell Diagram known as the instability strip. They have properties that make them crucial to the determination of large interstellar distances. 
a) Light curves (i.e. brightness time series) are often useful in studying time-variable phenomena in astrophysics. Below are a set of light curves of Cepheid Variables. Using plotting and fitting programs if necessary, find a rough relation obeyed by all the Cepheids. This relationship should relate the brightness (magnitude) of the Cepheid with another property. In other words, your relationship should have the form $M_V = A*feature+M_0$, where A and $M_0$ are variables you should find. 

We can use these light curves to determine the period and the $M_V$ (mean magnitude) and of each Cepheid. 

Ashley: 

• period: 50 days
• $M_V$: -5.7

Moiya: 

• period: 15 days
• $M_V$: -4.4

J Arbanus: 

• period: 20 days
• $M_V$: -4.8

Yutong: 

• period: 100 days
• $M_V$: -6.2

Graphing $M_V$ vs. period appears to give us a logarithmic relationship: 

So we graph $M_V$ vs. log(period) and get a best-fit line: 

A = -2.17 and $M_0$ = 1.925, where the "feature" is log(period): 

$M_V = -2.17log(period)-1.925$

b) Describe how you would measure the distance to a Cepheid.

Based on the light curve, we should know the apparent magnitude of the Cepheid. We can then plug that into the equation from question 2: 

$d = D*10^{0.2(m-M)}$ where D = 10pc

c) Below is an image of a Cepheid Variable in the Large Magellanic Cloud. Determine its period, mean absolute magnitude, and distance. 

The period is the distance between the two peaks, which is approximately 5 weeks or 35 days. 

We can calculate the mean absolute magnitude using the equation from part a.

$M_V = -2.17log(35)-1.925 = -5.28$ 

Using this absolute magnitude, we can calculate the distance using the equation from part b, where m is the mean apparent magnitude of 15.6 and M is the mean absolute magnitude of -5.28. 

$d = 10*10^{0.2(15.6+5.28)} = 1.5 * 10^5pc \rightarrow 150kpc$