The Leavitt Law (period luminosity relation) for Cepheids is roughly M = -2.5 lo

Question

The Leavitt Law (period luminosity relation) for Cepheids is roughly M = -2.5 log(P) - 1.5, where P is the period in days and M is their absolute magnitude. (a) What is the absolute magnitude of a Cepheid with a 10-day period? (b) If a 10-day period Cepheid has an apparent magnitude of m = 16, how far away is the Cepheid (in parsecs)? (c) We want to determine the most distant galaxy for which we can measure a Cepheid distance. Let's say that the longest-period Cepheid has a period of 63 days. If the Hubble Space Telescope can observe Cepheids with an average absolute magnitude of m = 28, how far can Hubble measure Cepheid distances?

Explanation / Answer

A) M = -2.5log(10)-1.5 = -(2.5*1)-1.4 = -4

b)Average distance module m-M = 16-(-4) = 20

d(parasecs) = 10[0.2(m-M+5)] = 10[0.2 x 25]= 105 = 0.1Mpc

c) M = -2.5log(65)-1.5 = (-2.5*1.81291) -1.4 = -4.5322-1.4 = -5.9322

m-M = 28-(-5.9322) = 33.9322

d(parasecs) = 10[0.2(m-M+5)] = 10[0.2( 33.9322+5)]= 107.7 = 50.1187 Mpc

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