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?