Chapter 16--Properties of Stars

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Alcor Types of Binary Star Systems About half of all stars orbit a companion star of some kind. These star systems fall into three classes: ● A visual binary is a pair of stars that we can see distinctly (with a telescope) as the stars orbit each other. Mizar, the second star in the handle of the Big Dipper, is one example of a visual binary (Figure 16.6). Sometimes we observe a star slowly shifting position in the Figure 16.8 The apparent brightness of an eclipsing binary system drops when either star eclipses the other. apparent brightness Mizar is a visual binary. Mizar B Mizar Spectroscopy shows that each of the visual “stars” is itself a binary. Figure 16.6 Mizar looks like one star to the naked eye but is actually a system of four stars. Through a telescope Mizar appears to be a visual binary made up of two stars, Mizar A and Mizar B, that gradually change positions, indicating that they orbit every few thousand years. However, each of these two “stars” is actually a spectroscopic binary, making a total of four stars. (The star Alcor appears very close to Mizar to the naked eye but does not orbit it.) A B 1900 1910 1920 Mizar A Figure 16.7 Each frame represents the relative positions of Sirius A and Sirius B at 10-year intervals from 1900 to 1970. The back-and-forth “wobble” of Sirius A allowed astronomers to infer the existence of Sirius B even before the two stars could be resolved in telescopic photos. 530 part V • Stellar Alchemy We see light from both A and B. 1930 1940 1950 1960 1970 B A sky as if it were a member of a visual binary, but its companion is too dim to be seen. For example, slow shifts in the position of Sirius, the brightest star in the sky, revealed it to be a binary star long before its companion was discovered (Figure 16.7). ● An eclipsing binary is a pair of stars that orbit in the plane of our line of sight (Figure 16.8). When neither star is eclipsed, we see the combined light of both stars. When one star eclipses the other, the apparent bright- We see light from all of B, some of A. B A time We see light from both A and B. B We see light only from A. A A
ness of the system drops because some of the light is blocked from our view. A light curve, or graph of apparent brightness against time, reveals the pattern of the eclipses. The most famous example of an eclipsing binary is Algol, the “demon star” in the constellation Perseus (algol is Arabic for “the ghoul”). Algol becomes three times dimmer for a few hours about every 3 days as the brighter of its two stars is eclipsed by its dimmer companion. ● If a binary system is neither visual nor eclipsing, we may be able to detect its binary nature by observing Doppler shifts in its spectral lines [Section 6.5].Such systems are called spectroscopic binary systems. If one star is orbiting another, it periodically moves toward us and away from us in its orbit. Its spectral lines show blueshifts and redshifts as a result of this motion (Figure 16.9). Sometimes we see two sets of lines shifting back and forth—one set from each of the two stars in the system (a double-lined spectroscopic binary). Other times we see a set of shifting lines from only one star because its companion is too dim to be detected (a single-lined spectroscopic binary). Each of the two stars in the visual binary Mizar is itself a spectroscopic binary (see Figure 16.6). Measurements of stellar masses rely on Newton’s version of Kepler’s third law [Section 5.3],for which we need to know the orbital period p and semimajor axis a. As described in the text, it’s generally easy to measure p for binary star systems. We can rarely measure a directly, but we can calculate it in cases in which we can measure the orbital velocity of one star relative to the other. If we assume that the first star traces a circle of radius a around its companion, the circumference of its orbit is 2pa. Because the star makes one circuit of this circumference in one orbital period p, its velocity relative to its companion is: v 2p distance traveled in one orbit a period of one orbit p Solving for a, we find: pv a 2p Mathematical Insight 16.4 Once we know both p and a, we can use Newton’s version of Kepler’s third law to calculate the sum of the masses of the two stars (M1 M2 ). We can then calculate the individual masses of the two stars by taking advantage of the fact that the relative velocities of the two stars around their common center of mass are inversely proportional to their relative masses. Example: The spectral lines of two stars in a particular eclipsing binary system shift back and forth with a period of 2 years (p 6.2 107 seconds). The lines of one star (Star 1) shift twice as far as the lines of the other (Star 2). The amount of Doppler shift indicates an orbital speed of v 100,000 m/s for Star 1 relative to Star B spectrum at time 1: approaching, therefore blueshifted to Earth Star B spectrum at time 2: receding, therefore redshifted Figure 16.9 The spectral lines of a star in a binary system are alternately blueshifted as it comes toward us in its orbit and redshifted as it moves away from us. Measuring Masses in Binary Systems 1 approaching us Even for a binary system, we can apply Newton’s version of Kepler’s third law only if we can measure both the orbital period and the separation of the two stars. Measuring orbital period is fairly easy. In a visual binary, we simply observe how long each orbit takes (or extrapolate from part of an chapter 16 • Properties of Stars 531 B B A 2 receding from us Orbital Separation and Newton’s Version of Kepler’s Third Law Star 2. What are the masses of the two stars? Assume that each of the two stars traces a circular orbit around their center of mass. Solution: We will find the masses by using Newton’s version of Kepler’s third law, solved for the masses: p2 4p2 a G(M1 M2 ) 3 ⇒ (M1 M2 ) 4p G 2 a p 3 2 We are given the orbital period p 6.2 107 s, and we find the semimajor axis a of the system from the given orbital velocity v: pv (6.2 10 a 2p 7 s) (100,000 m/s) 2p 9.9 1011 m Now we calculate the sum of the stellar masses by substituting the values of p, a, and the gravitational constant G [Section 5.3] into the mass equation above: (M1 M2 ) (9.9 1011 m (6.2 107 s) 2 ) 3 4p 2 6.67 1011 m3 kg s2 1.5 1032 kg Because the lines of Star 1 shift twice as far as those of Star 2, we know that Star 1 moves twice as fast as Star 2, and hence that Star 1 is half as massive as Star 2. In other words, Star 2 is twice as massive as Star 1. Using this fact and their combined mass of 1.5 1032 kg, we conclude that the mass of Star 2 is 1.0 1032 kg and the mass of Star 1 is 0.5 1032 kg.
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Chapter 16--Properties of Stars

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