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Binary star period calculation

  1. May 18, 2014 #1
    The problem is attached, and the official answer is given.
    The step in the answer does not make a lot of sense to me.
    How does the luminosity L converted directly into distance d, and why is the solar distance is used?
    And where does the dm=0.75 come from.
    It's just not making sense to me...

    Please expand and explain the steps in the answer a bit. Thank you.
     

    Attached Files:

  2. jcsd
  3. May 19, 2014 #2
    m for the binary system is

    m = -2.5log10L + C

    m' for a single star is

    m' = -2.5log10(L/2) + C
    m' = -2.5[log10L - log102] + C
    m' = [-2.5log10L + C] + 2.5log102
    m' = m + 2.5log102 = m + δm

    δm = 2.5log102 = 0.752575... ≈ 0.75
     
    Last edited: May 19, 2014
  4. May 19, 2014 #3
    The relation between magnitude and distance comes from the inverse square law for the radiation intensity. If you double the distance to a star than its radiation intensity drops to (1/2)2 = 1/4 and so forth so if you know the apparent magnitude you can figure the distance.
     
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