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Calculate the Masses of two binary stars

  1. Mar 16, 2009 #1
    1. The problem statement, all variables and given/known data
    I am trying to understand the following example question in my course book, The answer is actually given but i dont understand how they got to it, i would like someone to give me an idea how this has been worked out

    In the Sirius Binary system the orbital period is 50 years and the semi major axis of the relative orbit is 20AU. Calculate the masses of sirius A and B expressing your answer in solar masses

    2. Relevant equations
    M+m = 4 Pi^2 a^3 / GP^2


    3. The attempt at a solution
    M+m = 4Pi^2 (20 x 1.50 x 10^11 m) ^3 / (6.67 x 10 ^-11 N m^2 Kg^-1) x (50 x 3.16 x 10^7 s) ^2

    I really dont understand as i thought from the equation that it would be 4 Pi^2 x 20^3 / 6.67 x 10^-11 x 50 ^2

    What am i missing ?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 16, 2009 #2

    tiny-tim

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    Homework Helper

    Welcome to PF!

    Hi victoriafello ! Welcome to PF! :smile:

    (have a pi: π and try using the X2 tag just above the Reply box :wink:)
    hmm … they seem to have converted everything to SI units …

    AU to m, and year to s …

    which gives them an answer in kg.

    But the problem asks for an answer in solar masses, so you should be able to use AU and year, and get the answer from the given equation just by dimensional considerations (and without using G) :confused:
     
  4. Mar 16, 2009 #3
    Thanks, I thought that if the units were in AU and Yrs then i could use the equation in the form

    (M+m) / Msun = (a/AU)^3/(P/yr)^2

    So this would make the equation

    20^3/50^2 = 3.2

    Is this correct ? Im still a bit unsure about the first part of the equation is this answer now in solar masses or so i need to convert ?
     
  5. Mar 16, 2009 #4

    tiny-tim

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    Yes, that looks right to me …

    mass is proportional to axis cubed and inversely proportional to period squared, so the result is a multiple of the mass of the Sun+Earth, in other words in solar masses :smile:
     
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