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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.
2. The solution is provided by the Star Velocity Calculator.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
4. The solution is given by the Java Kepler's Laws 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 M2 are the masses of the stars. Thus,
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.
8. Even though the companion star cannot be seen, the elliptical motion of the primary star (once parallax and aberration effects have been subtracted) tells us that there is an unseen companion perturbing the motion of the star that we can see. If we can make detailed observations over a period of time, we may be able to infer masses and orbits for the two stars in the system in the most favorable cases.
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.
14. From the mass luminosity relation, L / L (Sun) ~ (0.1 / 1.0)3.5 ~ 0.0003 times as luminous as the Sun. These low mass stars are very common but are difficult to see at any distance because they are so faint.
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,
16. Observation only gives us the angular separation for the components of the binary. We need the distance and simple trigonometry to convert that to a true distance (e.g., in astronomical units) so that we can use it in Kepler's third law to determine masses.
17. Better observing techniques may allow the formerly unseen companion to become visible, as happened in the case of Sirius.
18-19. 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.
22. From the nonrelativistic Doppler formula,
Thus, v = - 0.00166 x c = - 0.00166 x 3 x 105 = - 498 km/s. The value is negative because this is a blue shift.
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.
24. The period corresponds to the time for the velocity curve to repeat itself. From the spectrum below for HD 80715 we can see, for example, that the lines merge at the times marked 0.061 days and 2.038 days. Since, from the preceding question, this time difference must correspond to half of an orbit, the period is approximately 2 x (2.038 - 0.061) = 3.95 days. This is the period in which the velocity curve will repeat itself.
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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.
26. The time period is 2 hours = 7200 seconds. The diameter is thus
The radius is half of this (3.6 x 105
km), and since the solar radius is
6.96 x 105 km, this is 0.52 solar radii.
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.
28. The nature of the dips in the light curve depend on whether the eclipses are partial or total and the relative sizes of the stars. If one star is smaller than the other and the eclipses are total we expect flat bottoms in the light curves. See the adjacent image for an example.
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.
30. Most stars late in their lives will expand to become giant or supergiant stars because of events that take place in their interiors when they begin to exhaust their nuclear fuel. In a close binary, this can cause the star to overflow its Roche lobe, leading to accretion onto the companion.
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.
32. If the white dwarf has the same angular momentum as the Sun does today, and its radius decreases by a factor of 100, then the velocity of a point on the "surface" must increase by a factor of 100. Currently a point at the solar equator moves through a circumference
in 25 days, which is a velocity of 2.2 km/sec. As a white dwarf this would increase to 220 km/sec.