Plotting the log of the period versus the log of the orbital separation
produces a straight line with a slope of 1.5, in agreement with Kepler's Third Law.
Watch the light curve and the "View from Earth".
Is there a bug in this program? (Which star is more luminous?)
The light curve shows a steep drop in luminosity when the red star
is eclipsed, but almost no change in luminosity when the blue star is
eclipsed (see figure below). This is WRONG! The blue star, being
much hotter and of the same size as the red star, should be much more luminous
than the red star. So the light observed from the system should drop when the
blue star is eclipsed, not when the red star is eclipsed.
Forgetting that you know the answers, begin from
the observations (radial velocity curves and orbital period) and derive the
masses of the two stars.
The orbital velocities can be extracted from the radial velocity curves.
Since the orbit is nearly circular, the maximum velocity on the radial velocity
curve is the orbital velocity: v_red = 200 km/s, v_blue = 50 km/s (approximately!).
First, get the distances from the center of mass given the velocities and period...
Next, get the sum of the masses using Kepler's Third law...
Finally, use the ratio of the masses (m1/m2=r2/r1) to get individual masses...
The radial velocity is always zero, so the orbital velocities, and hence stellar
masses, are unknown (unless the orbital separation can be resolved).
Initial system with orbital period = 313 hours.
K5 Supergiant becomes larger than entire binary system!
The companion star would be swallowed up by the supergiant.