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 m_{v}P(days) M_{v}m_{v}-M_{v}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: m_{v}- M_{v}= __________

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

In the simplest case, the distance

d(pc) = 10In 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^{ [ 0.2 (mv - Mv + 5) ]}.

We will have to account for interstellar extinction when we calculate
the distance to M100. The absorption in magnitudes at visual
wavelengths is denoted A_{v}. How can we correct the distance
modulus for A_{v}? 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

mAlthough the extinction varies somewhat over the extent of M100, a typical value for the extinction to M100 is A_{v}- M_{v}- A_{v}.

There's only one more bookkeeping item, and this is simply a technical point. After the data in Ferrarese

m_{v}- M_{v}- A_{v}+ 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) = 10Calculate the distance to M100, using the average distance modulus from Part II, Section B for m^{ [0.2 (mv - Mv + 5 - Av + 0.05) ]}.

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) = 10which gives^{ [ 0.2 (30.57 + 5 - 0.30 + 0.05) ]}.

d = 11.6 x 10^{6}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.

On to Hubble's Law and the current Hubble constant

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