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| Properties of Stars |
1. The reasons are somewhat different. In the star the colors are influenced by the possible electronic excitations of the atoms and ions, not primarily by the types of atoms and molecules present. In the case of Jupiter and Saturn, we think the temperature more directly influences the types of compounds present (and thus the color) because it controls the rates of chemical reactions producing those compounds. They are not completely different; for example, in cool stars molecules are present that are not there in hotter stars.
3. The essential point is the amount of spectral light emitted in the visible relative to other wavelength regions. This is determined by Planck's law and depends only on the temperature, since this determines the intensity curve. For intermediate temperature stars like the Sun most light is in the visible, so the correction is small. For very hot stars most light is in the UV and for very cool stars most light is in the IR, so the corrections are large in that case.
5. Using Earth as a guideline, evolution of intelligence has taken several billion years. Presumably this would have not been very likely unless the Sun were stable over periods of billions of years, which being on the main sequence ensures. Thus, let us assume that life is most likely to have evolved for stars that can spend at least of order a billion years on the main sequence. This eliminates O and B stars for sure, and probably A stars. Thus, we may expect that spectral classes F-M are a good starting point.
9. 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.
11. In a million years the apparent magnitude will be about -0.6. The radial velocity is about -14 km/s if it must cover 63 light years in about a million years.
13. 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
15. 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.
17. 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.
21. 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.
23. 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,
25. 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.
29. 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.
31. 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.
33. 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.
35. 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.