NOTES: Star Characteristics: How far (d in parsecs)? Distance to nearby star determined from stellar parallax, p, which is ½ the maximum angular difference.

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Presentation transcript:

NOTES: Star Characteristics: How far (d in parsecs)? Distance to nearby star determined from stellar parallax, p, which is ½ the maximum angular difference in position: d (in parsecs) = 1/p (p in arc seconds) 1 parsec is the distance at which the parallax of a star is 1 arcsec. Parallax method works for stars closer than about 100 parsecs. (1 parsec = 3.26 LY.) How bright (L in watts)? Luminosity at the source is determined from apparent brightness and distance (d). Apparent magnitude (old way). We can see about 1,000 stars in Northern Hemisphere with naked eye. Hipparchus rated them from 1 to 6. A '1' is 2.52 x brighter than a '2', etc. Range in brightness from the sun at '-26' magnitude to the faintest objects seen at about '26' magnitude. Flux (new 'apparent brightness'): b (watts/m 2 ) = L/4πd 2 = Power/unit area of sphere. From d, the distance, we get L, the luminosity (watts of source).

How far (d in parsecs)? Distance to nearby star determined from stellar parallax, p, which is ½ the maximum angular difference in position (seen 6 months later).

distance (in parsecs) = 1/p (p in arc seconds) 1 parsec is the distance at which the parallax of a star is 1 arc second. Parallax method works for stars closer than about 100 parsecs = 326 LY. (1 parsec = 3.26 LY.)

How bright (L in watts)? Luminosity at the source is determined from apparent brightness (flux, f) and distance (R). For the math oriented: f = L/A, A = area of surface of sphere or A = 4πR 2. f = L/(4πR 2 )  L = f(4πR 2 ).

Apparent magnitude (old apparent brightness). Hipparchus rated stars he could see from 1 to 6. A '1' is 2.52 x brighter than a '2', etc.

We can see about 1,000 stars in Northern Hemisphere with the ‘naked eye’.

Again: Flux (new 'apparent brightness'): f(watts/m2) = L/4πR 2 = Power/unit area of sphere. From R, the distance, we get L, the luminosity (watts of source). L = 4πR 2 f

How Big (radius of star in meters)? We get the temperature, T, of the photosphere of a star From the peak wavelength of the black body spectrum. This we called ‘Wien’s Law’.

We then can plug T into: The Stefan-Boltzmann Law which gives the surface flux from surface temperature, T. f(surface) = constant x T 4 for a black body. Ludwig Boltzmann

We can use this flux, b, to find the radius of a star R from: f(surface) = L/4πR 2.

b is flux = f = brightness

At this point you may be really CONFUSED!$%*?!

Aren’t you glad I don’t require you to learn all that math?