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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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  • $\begingroup$ The equation M1 + M2 = a^3 / p^2 should solve this, correct? Or am I mistaken? $\endgroup$ – user1026 Feb 14 '14 at 3:32
  • $\begingroup$ 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. $\endgroup$ – Scott Griffiths Feb 14 '14 at 3:38
  • $\begingroup$ 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...? $\endgroup$ – user1026 Feb 14 '14 at 4:02
  • $\begingroup$ 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 :( $\endgroup$ – user1026 Feb 14 '14 at 4:06
  • $\begingroup$ Fixed some errors, still coming out incorrect though... ((4π^2)/(6.67*10^-11))*(((3.3460735*10^11)^3)/((2.628*10^7)^2)) $\endgroup$ – user1026 Feb 14 '14 at 4:36
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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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