The Problem is:

For a binary system (2 Stars) with Orbital Period of $P =4.822 days = 416620.8 second$ and inclination $i=90$ and with speeds very less than $3 .10^8 m/s$. Their orbital planes around Center of Mass is completely circular. we have this lines on their spectroscopy: enter image description here

And we also know that the smaller star has mass $m = m_{sun}$. find the other star mass and their distance.

according to this we can calculate velocity (absolute values):

$\frac{\Delta\lambda_1}{\lambda} = \frac{v_1}{c} \Rightarrow v_1 = 428673 m/s$

$\frac{\Delta\lambda_2}{\lambda} = \frac{v_2}{c} \Rightarrow v_1 = 1286020 m/s$

$r_1 = 428673 * 416620.8/ (2 * \pi) = 2.842 *10 ^{10} m$

$r_2 = 1286020 * 416620.8/ (2 * \pi) = 8.527 *10 ^{10} m$

from know we can solve this question in two ways which gets into a contradiction:


Their distance $r= 1.136 *10^{11} m$

And also we know that $p^2 = \frac{4*\pi^2 r^3}{G(m_1+m_2)}$ so $m_1 +m_2 = 4.999 *10^{33} \approx 2500 * m_{sun}$ so $m_2 = m_{sun}$ and $m_1 = 2499 * m_{sun}$


Center of mass equation:$m_1r_1 = m_2r_2 \Rightarrow \frac{m_1}{m_2}=\frac{r_2}{r_1}$ and also due to that $P_1 = P_2 = P$ in binary stars $\frac{v_1}{v_2} = \frac{\omega r_1}{\omega r_2} = \frac{r_1}{r_2} = \frac{m_2}{m_1}$ according to this we calculate $m_1 = 3 m_2$ which is a contradiction with the another solution. what is the problem? which one is true?

  • $\begingroup$ I don't entirely follow what you're doing in (1), but if both masses were equal, then we would require that both the semi-major axes ($r_1$ and $r_2$) are the same, and the velocities would be the same, just by symmetry. And that's clearly not the case.... $\endgroup$ Commented Mar 31, 2017 at 16:35
  • $\begingroup$ @DilithiumMatrix In (1) , I calculated $r=r_1 + r_2$. and then calculated sum of masses. My problem is with this calculation we get $m_1 + m_2 = 2500 m_{sun}$ and in the other calculation, we get $m_1 + m_2 = 4 m_{sun}$. So there must be a problem in one of solutions. $\endgroup$
    – titansarus
    Commented Mar 31, 2017 at 17:48
  • $\begingroup$ Something is wrong with the numbers you have given (or have been given). Is it possible the period is 4.8 years? $\endgroup$ Commented Mar 31, 2017 at 21:10

1 Answer 1


I assume the picture shows three spectra taken equally spaced in time. In which case, the middle spectrum is taken at quadrature with one star moving towards the observer and the other directly away (with respect to the centre of mass).

The respective relative velocities are given by the two doppler shifts $18c/4199$ and $-6c/4199$.

From this, we can immediately say that the primary star is three times the mass of the Sun, since the mass ratio is given by the reciprocal of the radial velocity amplitudes.

The separation of the objects can then be found from Kepler's third law. Sticking to solar masses, years and astronomical units, then $a^3 = (M_1 + M_2) P^2 = 4P^2$, where $a$ is the separation of the stars.

The problem here is that you have been given a physically impossible situation. The absolute values of the doppler shifts are far too big to be produced by such a system.

In fact if we know the total mass is $4M_{\odot}$ and the period then the maximum doppler shift of either star can be calculated. For example, for the primary star $$ V_{\rm max,1} = \left( \frac{2\pi G M_2^{3}}{(M_1+M_2)^2 P}\right)^{1/3}$$

For a mass ratio of 3, total mass of $4M_{\odot}$ and $P=4.822$d, then $V_{\rm max,1}=50.1$ km/s, corresponding to a wavelength shift of just 0.7 Angstroms for a rest wavelength of 4199 Angstroms.


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