Stars

Distance

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

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

For getting distances to stars, we want longest possible baseline: Earth p d x 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. Shift proportional to 2  p A B

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

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

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

they’re equally luminous? 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 Flux (F)  amt. of light energy flowing per second through 1 m2 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 Inverse-square law of light!

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

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 (!)