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Stars Distance Getting distances to stars: 1. Geometry: ‘Stellar Parallax’ > Only direct method! 2. Inverse-Square Law L F 2 4 d * Measure F * Know (or guess!) L Find d Stellar Parallax Triangulation: d p x Baseline q x and q (or x and p) d For getting distances to stars, we want longest possible baseline: x d p Earth p parallax angle d distance x = 1 AU; measure p d As a practical matter, how do we get p? B d p A Star appears to shift position against background – the parallax effect. A B Shift proportional to 2p Clearly, as d increases, p decreases. Astronomers find: 1 d p p: arcseconds (1 arcsec = 1/3600o) If p = 1 arcsec, 1 1 d 1 parsec p 1 1 parsec = 3.26 light year = 206,265 AU Nearest star: Proxima Centauri p = 0.772 arcsec 1 1 d p 0.772 d = 1.295 pc = 4.22 ly Inverse-Square Law Luminosity: total amount of energy radiated per second (“wattage”) Watt? Watt? 50 W 100 W Twice the luminosity Star Luminosity Sun 1 Proxima Centauri 0.00082 Alpha Centauri 1.77 Sirius 26.1 Betelgeuse 15,000 Rigel 70,000 These stars would appear to be about equally “bright.” Does this mean they’re equally luminous? (Apparent) Brightness Luminosity 2 stars – differ in luminosity – may appear equally bright! On the other hand . . . two stars that differ in brightness need not differ in luminosity. Brighter Dimmer How much dimmer? How much brighter? Sphere, radius = d d L 1 m2 Flux (F) amt. of light energy flowing per second through 1 m2 All of star’s light must pass through sphere . . . So the energy is spread over the sphere’s surface. Amt. of energy per sec through 1 m2 = Total energy per second flowing Total number of sq meters L F 2 4 d Inverse-square law of light! For a given star (i.e., specified luminosity): d 1 F 2 d d (ly) F (watt/m2) 1 100 2 25 5 4 10 1 100 100 F 2 4 5 25 Flux & distance Luminosity L 2 F L 4 d F 2 4 d Sun’s flux at Earth: F = 1370 W/m2 d = 1 AU = 1.5 x 1011 m L = 4(1.5 x 1011)2 x 1370 = 3.9 x 1026 Watt (!)