PowerPoint PresentationSlide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18StarsDistanceGetting distances to stars:1. Geometry: ‘Stellar Parallax’> Only direct method!2. Inverse-Square Law d 4L F2* Measure F* Know (or guess!) LFind dpdxqx and q (or x and p) dBaselineStellar ParallaxTriangulation:For getting distances to stars, we want longest possible baseline:p parallax angled distanceEarthpdxx = 1 AU; measure p dAs a practical matter, how do we get p?Star appears to shift position against background – the parallax effect.ABShift proportional to2 pBpAdClearly, as d increases, p decreases. Astronomersfind:p1 d p: arcseconds (1 arcsec = 1/3600o)If p = 1 arcsec, 1 parsec = 3.26 light year = 206,265 AUparsec 1 11 p1 d Nearest star: Proxima Centaurip = 0.772 arcsec 0.7721 p1 d d = 1.295 pc = 4.22 lyLuminosity: total amount of energy radiated per second (“wattage”)Watt? Watt?100 W50 WTwice theluminosityInverse-Square LawStar LuminositySun 1Proxima Centauri 0.00082Alpha Centauri 1.77Sirius 26.1Betelgeuse 15,000Rigel 70,000These stars would appear to be about equally “bright.”Does this mean they’re equally luminous?(Apparent) Brightness Luminosity2 stars – differ in luminosity – may appear equally bright!On the other hand . . . two stars that differin brightness need not differ in luminosity.BrighterDimmerHow much dimmer?How much brighter?LdSphere, radius = d1 m2Flux (F) amt. of light energy flowingper second through1 m2All of star’s light must pass through sphere . . .So the energy is spread over the sphere’s surface.Amt. of energyper sec through1 m2= Total energy per second flowingTotal number of sq meters2d 4L FInverse-squarelaw of light!For a given star (i.e., specified luminosity):2d1 F d (ly) F (watt/m2)125101002541d4 25100 5100 F2Flux & distance Luminosity Fd 4 L d 4L F22Sun’s flux at Earth:F = 1370 W/m2d = 1 AU = 1.5 x 1011 mL = 4(1.5 x 1011)2 x 1370 = 3.9 x 1026 Watt
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