# What are the masses of the two stars (given the information provided)?

If anyone would at least be willing to point me in the direction of finding the answer, or help me solve it I'd appreciate it. I don't know how to go about this, and need to know how to solve this on my own (plus I need the answer for a homework assignment, but just an answer isn't gonna help).

Question:

The spectral lines of two stars in a particular eclipsing binary system shift back and forth with a period of 10 months. The lines of both stars shift by equal amounts, and the amount of the Doppler shift indicates that each star has an orbital speed of $~8.0\cdot 10^4~ \mbox{m}/\mbox{s}.$ What are the masses of the two stars? Assume that each of the two stars traces a circular orbit around their center of mass.

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You need to use the generalized (a la Newton) form of Kepler's third law; see here.

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The equation M1 + M2 = a^3 / p^2 should solve this, correct? Or am I mistaken? – user1026 Feb 14 '14 at 3:32
Yes, but be careful of the units!! Note that equation has the masses in solar masses, the time in years, and the radius in AU. – Scott Griffiths Feb 14 '14 at 3:38
I got 3.71885*10^16 (half of 7.4367*10^16 since it is asking for M1 = M2 = X). Mastering Astronomy says I am incorrect though, do you know where I messed up? I converted ten months to seconds (2.628*10^7) since P is in meters/second. I got 3.3024*10^12 for A also. My equation was ((4π^2)/(6.67*10^-11))*((3.3024*10^12)/(2.628*10^7)). Any idea what I did wrong...? – user1026 Feb 14 '14 at 4:02
Just realized I forgot to divide my equation ((4π^2)/(6.67*10^-11))*((3.3024*10^12)/(2.628*10^7)) by the Sun's mass (2*10^30). The answer is still coming out wrong though... now I got 1.859*10^-14. Ahh :( – user1026 Feb 14 '14 at 4:06
Fixed some errors, still coming out incorrect though... ((4π^2)/(6.67*10^-11))*(((3.3460735*10^11)^3)/((2.628*10^7)^2)) – user1026 Feb 14 '14 at 4:36

We can assume that the stars are equal in mass, and their orbits are circular

The orbital speed is 80000 m/s and at an orbital period of 10 months (or $2.628\times 10^7$s) the length of the orbit is $2.1024\times 10^{12}$ m or 14.05 AU The radius of the orbit therefore is $14.05/\tau$ = 2.237AU.

The version of Keplers law given is $$T^2 =\frac{a^3}{m_1+m_2}$$

substituting $T^2 =(10/12)^2 = 0.6944$ (divide by 12 to convert to years) and $a^3= 11.19$ gives $$m_1+m_2 = \frac{11.19}{0.6944} = 16.12\ \mathrm{solar\ mass}.$$

Since $m_1=m_2$, the mass of each star is 8.06 solar masses, or $1.6\times 10^{31}$kg

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