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The Sunspot Cycle • Caused by cyclical change in Sun’s magnetic field. • field reverses polarity each 11 years 22 year cycle. • reverses polarity of leading spots in pairs. Magnetic field lines of force 1st 11-year cycle Next 11-year cycle • Sunspots are a detail showing how the Sun’s magnetic field is leaking out of the zones just below the Sun’s surface. • Magnetic field produced in outer 30% of Sun’s radius. # sunspots Solar Latitude White areas = no sunspots The Sunspot Cycle • 11 years between maxima. • Currently starting an upswing. • A repeating cycle. • Sunspots are formed far from equator at start. • Later, form closer and closer to equator. 1 Solar activity: flares affect Earth. What causes the solar cycle? • Differential rotation of Sun’s outer layers • 24 days at equator. • 30 days at pole. • reason not understood. • “Winds up” magnetic field • field reverses each 11 yrs, when it gets too wound up. • causes 22-year cycle. • but why this reversal??? We don’t know. [Fig 10.20] Other Stars Look like points of light in the sky. How similar are they to the Sun? 2 Other Stars Clicker question: Which of the following things did we NOT need to know about the Sun in order to compute an accurate model of its interior? A. Chemical composition B. Luminosity C. Mass D. Diameter E. We needed to know all of the above Finding the luminosity [Fig. 11.2] L r F = 4r2 [11.1] L = 4r2 F • Luminosity = Energy/unit time • But we measure flux incident on Earth (apparent brightness) = Energy/unit time /unit area Also must know distance r r • For nearby stars, use parallax: [Fig. 11.3] 3 Stellar masses [11.1] • Binary stars • Use Newton’s form of Kepler’s 3rd law: High mass star Center of mass Low mass star 4π 2 a3 P G m1 m2 2 constant Orbits are about center of mass. So can determine ratio m1/m2 Mass - Luminosity Relation Luminosity (Lsun) • Key observational result for theoretical interpretation of different types of stars. Mass (solar masses) 4 The Observed Properties of Stars To compute an accurate model of the interior of a distant star, we need to know: A. Chemical composition B. Luminosity C. Mass D. Diameter [Fig. 5.11] Emitted energy per unit surface area Finding the star’s diameter 5000o [fig 4.8] 4000o 3000o Wavelength Energy • Total energy emitted per unit surface area Stefan-Boltzmann Law: • Total energy from whole star: E = T4 Surface area of sphere of diameter D L = E x (surface area) = ( T4) x (D2) Luminosity 5 Two Ways: • Thermal radiation curve • Spectroscopy Emitted energy per unit surface area Taking a star’s (surface) temperature [Fig. 5.11] 5000o 4000o 3000o Wavelength Energy Sodium Hydrogen Calcium Mercury Neon Wavelength Ionized helium O B A Excited hydrogen Flux F G K Low ionization & neutral atoms M Wavelength Molecules Spectra of different stars look different. Wavelength 6 Due to the surface temperature! Ionized helium • Hot = blue color, abs. from high ionization states. • Cool = red color, abs. lines from molecules. O B A Excited hydrogen 50,000 K 10,000 K Flux F G 6,000 K K M Emitted energy per unit surface area Low ionization & neutral atoms 3,000 K Wavelength Spectra of different stars look different. Molecules Wavelength Energy Wavelength [Fig. 5.11] Emitted energy per unit surface area Finding the star’s diameter 5000o [fig 4.8] 4000o 3000o Wavelength Energy • Total energy emitted per unit surface area Stefan-Boltzmann Law: • Total energy from whole star: E = T4 Surface area of sphere of diameter D L = E x (surface area) = ( T4) x (D2) Luminosity We measure L and T, then solve for D 7