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:
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 ____________ ______ ______ _______ ______ ______ ____________ ______ ______ _______ ______ ______
Average distance modulus: mv - Mv = __________
Enter your result in Part II, Section B of your lab sheet.
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.
mv - Mv - Av + 0.05.
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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