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The Properties of Stars Masses Using Newton’s Law of Gravity to Determine the Mass of a Celestial Body Any two particles in the universe attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the distance between them. F GMm r2 The force on a particle outside an object with spherical symmetry is the same as if all of its mass were concentrated at its center. Newton’s law of gravity, combined with his laws of motion enable us to determine the mass of a celestial body by observing its effect on a second celestial body. For example, we can find Jupiter’s mass by measuring the orbital radius and period for each of its Galilean satellites and using Newton’s form of Kepler’s third law of planetary motion. M Jupiter 9 2 3 a Callisto 1.883 10 m 4 a GP 2 M Jupiter 1.900 10 M Jupiter 6 P Callisto 1.442 10 s 27 kg M Jupiter 1.900 1027 5.977 10 4 1.883 10 6.673 10 1.442 10 2 24 9 11 3 6 2 Earth Masses M Jupiter 318 Earth Masses We’ll use a similar method to find the masses of stars in binary star systems. Using Newton’s Form of Kepler’s Third Law to Measure the Masses of Stars In using M Jupiter 4 2a 3 GP 2 to determine the mass of Jupiter, we assumed that the mass of Jupiter is much greater than the mass of any of its moons. However, in binary star systems, the two bodies have comparable masses and the relevant form of Kepler’s third law is M1 M 2 4 2a 3 GP 2 where M1 and M2 are the masses of the two stars. P is the orbital period of the stars and a is the average distance between them. Because the masses of stars are very large, but a relatively small multiple of the mass of the Sun, it is convenient to use solar mass units. In that case, Kepler’s third law is M1 M 2 a3 P 2 M1 and M2 are multiples of the Sun’s mass if a is in AU’s and P is in years. Center of Mass M1 r2 v 2 M 2 r1 v1 The black dot is the center of mass, and the colored disks are two stars at distances r1 and r2 from the center of mass. M1 and M2 are the masses of the two stars and v1 and v2 their orbital speeds. 3 Types of Binary Star System • • • Visual Binaries Both stars are visible, so their orbits can be plotted. Spectroscopic Binaries The stars appear as a single star but, because of the Doppler effect, the spectral lines can be seen to shift as the stars move in their orbits. Eclipsing Binaries The stars appear as a single star, but we see the orbits edge-on, so the stars periodically eclipse each other. In order to use M1 M2 a 3 2 to find M1 and M2, we have to (1) measure a P and P and (2) determine what fraction of the total mass belongs to each star. In order to accomplish (2), we need the concept of center of mass defined below. When two bodies move through space and are acted on only by their mutual gravitational forces, there is a point between them that moves in a straight line. That point is called the center of mass of the two bodies. The center of mass of a pair of bodies satisfies the equation M1r1 = M2r2 where r1 is the distance from M1 and r2 is the distance from M2. Visual Binaries The figures below show the observed positions of the stars in two different visual binary systems). Poor visual binary observations result from (a) the period being so long that few observations have been made or (b) the stars being so close together and/or so far from Earth that their angular separations cannot be measured accurately. The solid lines in the figures represent the ellipse that best fits the data (colored dots and plus signs). Visual binaries have periods between 1 year and thousands of years. Plenty of observations of stars separated by several arcseconds. Few observations of stars separated by a fraction of a arcsecond. http://ad.usno.navy.mil/ad/wds/hmw5text.html#intro Example 1 Consider a visual binary star system in which star 2 is 5 times farther from the center of mass than star 1, the period is 200 years and the semi-major axis is 100 AU. Calculate (a) the total mass and (b) the mass of each star. The equations to be used are We are given P 200 years and 3 So, M M 100 1 2 200 2 We are also told that M 2 M1 r2 M 2 r1 25 4.17 M 6 and M1 M 2 a3 P2 a 100 AU M1 M 2 25 M1 5 , so M2 5M 2 M 2 25 M1 5M 2 20.8 M 6M 2 25 M Spectroscopic Binaries In spectroscopic binary systems, the two stars are too close together to be resolved by a telescope. Because their separations are less than about 1 AU, their periods are as short as a few hours or as long as a few months. In the animation, the black dot represents the center of mass. Actually, the center of mass moves, but we are looking at the system from the viewpoint of someone at rest relative to the center of mass. Finding the Masses of Spectroscopic Binaries Finding the mass of the stars in a binary star system requires observations that give (a) the sum of the masses and (b) the ratio of the masses. This can easily be done if the system is a well-observed visual binary. In that case, we can plot the orbit and measure a and P. r1 and r2 can be determined by observing the motion of the system long enough to locate the center of mass. For spectroscopic binaries, it isn’t so easy. In that case, we must extract information from the combined spectra of the stars. Since the Doppler effect only gives the star’s radial velocity (the component along the observer’s line of sight) and most orbits are tilted, we are usually able to only determine a lower limit to the total mass. We’ll just consider the simplest case: the angle between the line of sight and the orbit is 0º, and the orbit is circular. The figure shows the orbits from above the orbital plane. The red circle is the orbit of the red star and the blue circle is the orbit of the blue star. Earth is to the right. vA Toward P Q R S Earth vB The blue star (A) is moving away from us, so its spectrum is red-shifted while that of the red star (B) is blue-shifted. When the stars arrive at the points P, Q, R, and S they are moving across our line of sight so we see no redshift. The Spectrum Usually, the spectrum will show two sets of lines that change positions as the stars move along their orbits. In the following figures, wavelength increases toward the right and only the hydrogen Balmer lines are shown. In each case, the Balmer lines observed in the laboratory are displayed on the bottom for comparison with the binary’s spectrum on the top. The first figure shows the spectrum at a time when the stars are moving across our line of sight so there is no wavelength shift, and spectra of the two stars are superimposed. In practice, the photographs of the stars’ spectra would be black and white, but I’ve used blue for star A and red for star B. Star Lab t = 0 (spectra of A and B are superimposed) Star Lab t = ¼ of the period later. Which star is moving toward us? The Radial Velocity Graphs The following graph shows the radial velocities of the two stars as a function of time. The time scale isn’t the same as that used in the previous slide. Radial Velocity vs. Time 150 Stars moving across our line of sight Radial Velocity (km/s) 100 50 0 0 1 2 3 4 5 6 -50 -100 Star A moving away from Earth. -150 Time (days) Star B moving toward Earth. Note: the radial velocity of the center of mass has been subtracted before the graph was drawn. Calculations From the radial velocity graph, we can read the orbital velocities of the two stars vA and vB as well as the orbital period P. The radii of the orbits are The semi-major axis is The total mass is v P rA A 2 v P rB B 2 a = rA + rB MA MB The ratio of the masses is and a3 P2 MA vB MB vA When the orbits are circular and the angle between them and the line of sight is zero (i.e., we are seeing them edge-on), the last two equations permit us to calculate the masses of the two stars. In that case, the stars will also periodically eclipse each other. Example 2 A spectroscopic-eclipsing binary star system has a period of 3.00 years. The maximum radial velocities of the stars relative to the center of mass are 20 km/s (for star A) and 10 km/s (for star B). Calculate (a) the ratio of their masses and (b) the individual masses, assuming that the orbit is observed edge-on from Earth, (a) vA = 20 km/s, vB = 10 km/s (b) v P rA A 2 M A v B 10 1 M B v A 20 2 v P rB B 2 P = 3.00×(3.16×107 s) = 9.48×107 s rA 20 km/s 9.48 107 3.02 108 km rB 10 km/s 9.48 107 1.51 108 km 2π rA 2π 3.02 108 km 1.50 10 km/AU rB 8 2.01AU 1.51 108 km 1.50 10 km/AU 8 1.00 AU a = rA + rB = (2.01 + 1.00) AU = 3.01 AU 3 3.01 The total mass is M M 3.03M B A 2 3.00 MA 2MA 3.03M 3MA 3.03M MA 1.01M MB 2.02M Flux Eclipsing Binaries MPEG Animation of the Algol System from a Paper by Blondin, Richards, and Malinowski Time t A: eclipse of hotter star begins tB : eclipse of hotter star complete tC : The hotter star is about to emerge from behind the cooler one. tD = eclipse of hotter star ends. v = relative orbital velocity RH = radius of the hotter star 2 RH = v (tB - t A ) RC = radius of the cooler star 2 RC = v (tC - t A ) Example 3 The orbital velocity of an eclipsing binary system is 85 km/s, and the time for the eclipse of the hotter star to be complete is 4.0 hours, what is the radius of the hotter star? 1 1 RH = v (t B - t A ) = v = 85km/s ´ 85km/s ´ 4 ´ 3600 s = 612000km 2 2 Example 4 In the same system, the hotter star is eclipsed for 8 hours, what is the radius if the cooler star? RC = 1 v (tC - t A ) 2 = 1 ´ 85km/s ´ 8´ 3600s = 1.22 ´ 106 km 2 Properties of Stars Some Important Results The Mass-Luminosity Relation A graph of absolute visual magnitude (a measure of luminosity) is plotted as a function of the logarithm of mass, the result is almost a straight line as shown below. Absolute Visual Magnitude -10 -5 L=L M M 3.5 0 L the luminosity of theSun 5 M the mass of the Sun. 10 15 0.1 1 M/M 10 100 Hertzsprung-Russel Diagram