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Stars

Stars. Distance. Find d. Getting distances to stars:. Geometry: ‘Stellar Parallax’ > Only direct method! 2. Inverse-Square Law. * Measure F * Know (or guess!) L. Stellar Parallax. Triangulation:. d. p. x. q. Baseline. x and q (or x and p)  d. p. x. d. Earth.

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Stars

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  1. Stars

  2. Distance

  3. Find d Getting distances to stars: • Geometry: ‘Stellar Parallax’ • > Only direct method! • 2. Inverse-Square Law * Measure F * Know (or guess!) L

  4. Stellar Parallax Triangulation: d p x q Baseline x and q (or x and p)  d

  5. p x d Earth For getting distances to stars, we want longest possible baseline: p  parallax angle d  distance x = 1 AU; measure p  d

  6. p d As a practical matter, how do we get p? B A Star appears to shift position against background – the parallax effect. Shift proportional to 2  p A B

  7. Clearly, as d increases, p decreases. Astronomers find: p: arcseconds (1 arcsec = 1/3600o) If p = 1 arcsec, 1 parsec = 3.26 light year = 206,265 AU

  8. Nearest star: Proxima Centauri p = 0.772 arcsec d = 1.295 pc = 4.22 ly

  9. Inverse-Square Law Luminosity: total amount of energy radiated per second (“wattage”) Watt? Watt? 50 W 100 W Twice the luminosity

  10. These stars would appear to be about equally “bright.” Does this mean they’re equally luminous?

  11. (Apparent) Brightness  Luminosity 2 stars – differ in luminosity – may appear equally bright!

  12. On the other hand . . . two stars that differ in brightness need not differ in luminosity.

  13. Brighter Dimmer How much dimmer? How much brighter?

  14. d L Sphere, radius = d Flux (F) amt. of light energy flowing per second through 1 m2 1 m2

  15. = 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 Inverse-square law of light!

  16. d d (ly) F (watt/m2) For a given star (i.e., specified luminosity): 1 2 5 10 100 25 4 1

  17. Flux & distance  Luminosity  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 (!)

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