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| Multiple Star Systems and Star Clusters |
1. There are 60 x 60 = 3600 seconds of arc in a degree. The Moon subtends about 1/2 degree ~ 1800 seconds of arc. Barnard's Star moves by 10.3 seconds of arc per year on the celestical sphere. Thus, it takes about 1800 / 10.3 = 175 years to move the diameter of the Moon on the celestial sphere.
3. The solution is given by the Java Center of Mass Calculator. Alternatively, let R1 be the distance of Sirius A from the center of mass and let R2 be the corresponding distance for Sirius B. The associated masses are given as M1 = 2.1 solar masses and M2 = 1.0 solar masses. Then, since
and inserting this in the first equation,
0.476 x R2 + R2 = 20 AU
R2 = 20 AU / 1.476 = 13.6 AU
R1 = 20 AU - 13.6 AU = 6.4 AU
5. One key point is that if the true orbit is circular the low mass star revolves around a point in the center of the projected ellipse, but if the true orbit is elliptical, the low mass star executes an orbit around the focus of the ellipse. These two situations can be distinguished with sufficient observation of the motion of the two stars.
7. For a visual binary the stars have to be separated enough on their orbits for us to resolve them into two stars at the large distance from which we view them. By Kepler's third law, large orbital separation implies long orbital periods.
9. Stars late in their lives tend to lose large amounts of mass because of very strong stellar winds blowing from their surface. According to our understanding of stellar evolution, Sirius B became a red giant after leaving the main sequence and lost much of its mass in a strong stellar wind that probably produced a planetary nebula. The remaining core of the star, containing only a fraction of the original stellar mass, then became the white dwarf Sirius B. Since Sirius B has evolved faster than Sirius A, this implies that the original main sequence mass of Sirius B was probably considerably larger than the present mass of Sirius A.
13. From the mass-luminosity relation for main sequence stars, the luminosity is proportional to the mass raised to the power 3.5. Therefore a 20 solar mass main sequence star would be approximately (20)3.5 ~ 35,777 times more luminous than the Sun.
15. The solution is given by the Kepler's Law Calculator. Alternatively, from Kepler's modified third law we have
where P is the period in years, a is the average separation in AU, and M1 and M 2 are the masses of the stars. Thus,
17. Better observing techniques may allow the formerly unseen companion to become visible, as happened in the case of Sirius.
19. Using the generalized form of Kepler's third law,
(m1 + m2) P2 =
(a1 + a2) 3
it is clear that the period must become shorter if one of the masses is increased. The ratio of the distances from the center of mass is the same ratio as that of the masses, with the larger mass star being closer to the cm and having the smaller orbit and the smaller mass star having the larger orbit because of its greater distance from the cm.
23. The lines will merge when the two stars emitting the light have the same radial velocity relative to the observer. This happens twice for each orbit, when the stars are moving at right angles to the line of sight.
25. From Kepler's Laws we may conclude that in this case,
The gravitational constant is G = 6.6726 x 10-8 cm3 g-1 s-2 . Thus, the masses will be in grams if the velocities are in cm/s and the period in seconds. Inserting
P = 3 days = 3 x 24 x 60 x 60 = 2.592 x 105 seconds,
we obtain M1 + M2 = 3.31 x 1033 g, which is 1.67 solar masses.
27. The time period is 7 hours = 25,200 seconds. Proceeding as above,
The radius is half of this (1.26 x 106 km), which is 1.8 solar radii.
29. Eclipsing binaries only occur if the geometry is right to cause the stars to pass in front of each other as viewed from Earth. Eclipsing binaries are usually spectroscopic binaries because the orbital motion causes shifts of the spectral lines, but most spectroscopic binaries are not eclipsing as viewed from the Earth because the geometry isn't right to cause eclipses.
31. Since angular momentum L = mvr and is the same throughout the star's orbit, if the r is halved then the v must be doubled at periastron relative to apastron.