Problem: Spectroscopy of a binary system

Question

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:

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:

1:

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 $m_1 = 2499 * m_{sun}$

2:

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?

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.