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Announcements •Homework: Chapter 13 # 45, 48, 49 & 50 •Exam 2 is next time. Will cover telescope and lens stuff from last week and distance-luminosity-brightness stuff from today. Stars! The fundamental question: How far away is that star The most direct method of determining stellar distances is parallax. The parallax angle is half the total displacement a star makes over the course of a year The distance in parsecs is simply one over the parallax angle in arcseconds 1 d (parsecs) = θ (arcseconds) 1parsec = 3.261ly = 3.0857 ×1016 m Examples The star Tau Ceti has a parallax of 273.96 milliarcsecs. How far away is Tau Ceti in parsecs and lightyears? The star Epsilon Eridanus is the third closest star to Earth with a parallax of 310.94 milliarcseconds. How far away is it in parsecs and lightyears Example Solution: Tau Ceti 1 1 d= (parsec) = 3.6502parsec = −3 θ (arcsec) 273.96 ×10 arcsec Converting to lightyears using 1 parsec = 3.26156 lightyears 3.6502 pc × 3.26156 ly pc = 11.905ly Example Solution: Epsilon Eridanus 1 1 d= (parsec) = = 3.2161parsec −3 θ (arcsec) 310.94 ×10 arcsec Converting to lightyears using 1 parsec = 3.26156 lightyears 3.2161 pc × 3.26156 ly pc 10.489ly = The distance-luminosity relationship can also be used to find distance Luminosity Brightness = 2 4π r Luminosity depends on the size of the star L ∝ 4π R 2 Luminosity also depends on Temperature L ∝ σT 4 Putting it all together L = 4π R σ T 2 4 R is the radius of the star in meters T is the temperature of the star in Kelvin σ is the Stephan-Boltzmann constant = σ 5.67 ×10−8 W m2 ⋅ K 4 Example The bright star in the top left corner of Orion, Betelgeuse, has a radius 936 times that of the Sun and a surface temperature of 3500 K. What is the luminosity of this star? If Betelgeuse is 640 ly from Earth, what is the brightness of the light from Betelgeuse that reaches Earth? Example Solution L = 4π R 2σ T 4 ( = 4π 936 RSun × 6.9599 ×10 m 8 )( 2 5.67 ×10−8 W m2 K 4 ) ( 3500K ) = 4.54 ×1031W The Sun has a luminosity of 3.9 x 1026 W so this is about 116,000 times more luminous than the Sun 4 Example Solution 2 Now that we know the luminosity we can use the distance-brightness-luminosity relationship to find the brightness We want the answer to be in Watts per square meter so we need to convert lightyears to meters L 4.54 ×1031W −8 W = B = = 9.86 × 10 2 m2 2 15 4π r 4π 640ly × 9.46 ×10 m ly ( )