The distance to M100

Now that you know the apparent average magnitude of each Cepheid, and the absolute magnitude of each Cepheid, you can estimate the distance to M100. Although you are not required to understand how all the equations in this section were derived, those who would like to see the details can read about the relationship between magnitudes and distance.

The distance to a star is roughly determined by comparing its apparent magnitude to its absolute magnitude. Astronomers define a quantity called the distance modulus of a star to be the difference between the two:

Distance modulus = mv - Mv

For each Cepheid in Part I, Section B of your lab sheet, subtract column 5 from column 3 to obtain the distance modulus, and enter your result in column 6. For Cepheid C46, your lab sheet will now appear as follows:

Cepheid name   Grid #   Avg mv   P(days)    Mv     mv-Mv

   C46           47      25.3     25.3    -5.27   30.57
____________   ______   ______   _______  ______  ______ 
____________   ______   ______   _______  ______  ______ 

Sum all the distance moduli and divide by the number of Cepheids to get the average distance modulus.

    Average distance modulus: mv - Mv =  __________

Enter your result in Part II, Section B of your lab sheet.

In the simplest case, the distance d to a celestial object may be derived from the distance modulus by the equation

    d(pc) = 10 [ 0.2 (mv - Mv + 5) ].
In the case of M100, this isn't quite true. Interstellar gas and dust along the line of sight cause the galaxy to appear dimmer. We call this effect interstellar extinction. Around the year 1920, Harlow Shapley was fooled by this effect when he tried to determine the size of the Milky Way galaxy. Shapley assumed interstellar extinction was negligible when he determined the distance to globular clusters within the Milky Way. (At the time, it was a perfectly reasonable assumption.) Do you think he overestimated or underestimated the size of the galaxy? Why? (Check your answer here).

We will have to account for interstellar extinction when we calculate the distance to M100. The absorption in magnitudes at visual wavelengths is denoted Av. How can we correct the distance modulus for Av? Remember, brighter stars have smaller magnitudes. We want to make the star brighter than it appears to correct for the dimming effect, so the adjusted distance modulus will be

    mv - Mv - Av.
Although the extinction varies somewhat over the extent of M100, a typical value for the extinction to M100 is Av = 0.3.
There's only one more bookkeeping item, and this is simply a technical point. After the data in Ferrarese et al. (1996) were already in press, new calibration information became available for the WFPC images. As a result, the apparent visual magnitudes which appear in all the light curves used in this lab should be increased by approximately 0.05 magnitudes. Accounting for this effect yields a revised distance modulus of
    mv - Mv - Av + 0.05.

Now you're ready to calculate the distance to M100! Putting it all together, the distance to M100 can be found from the equation

    d(pc) = 10 [0.2 (mv - Mv + 5 - Av + 0.05) ].
Calculate the distance to M100, using the average distance modulus from Part II, Section B for mv - Mv, and setting Av = 0.30 in the above equation. Once you have found your answer in parsecs, convert your answer to megaparsecs (1 Mpc = 106 pc).

Enter your result in Section II, Part C of your lab sheet:

    d = __________ pc,      or      d = __________ Mpc.

Does your answer seem reasonable?

To demonstrate the technique, if we used the distance modulus for Cepheid C46 instead of the average distance modulus, we would derive a distance of

    d(pc) = 10 [ 0.2 (30.57 + 5 - 0.30 + 0.05) ].
which gives

    d = 11.6 x 106 pc,      or      d = 11.6 Mpc.

Your numbers will vary, but the order of magnitude should be the same. Your answer should contain no more than three significant figures.

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