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Chapter 3

Chapter 3. Stars: Distances & Magnitudes.  Nick Devereux 2006. Revised 8/2012. Astrophysical Units & Constants. In addition to the usual list of physical constants – listed in Appendix A (pg A-2), there is another list of astronomical

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Chapter 3

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  1. Chapter 3 Stars: Distances & Magnitudes  Nick Devereux 2006 Revised 8/2012

  2. Astrophysical Units & Constants In addition to the usual list of physical constants – listed in Appendix A (pg A-2), there is another list of astronomical constants that we must be familiar with and these are listed in on page A-1.  Nick Devereux 2006

  3. The Sun as the “yardstick” Since the distances and masses are so large in astronomy, the basic units of measurement are expressed in terms of the Sun. The Astronomical Unit (AU) is the distance between the Earth and the Sun, 1 AU = 1.496 x 1011 m Which is perhaps more familiar to you as 93 million miles.  Nick Devereux 2006

  4. The parsec for larger distances, there is the parsec (pc) 1 pc = 3.086 x 1016 m  Nick Devereux 2006

  5. Angular units Astronomers can measure the angular extent on the sky for celestial objects, even if they don’t know how far away they are, and therefore unable to attribute a linear size. Radians 360o = 2 radians So, 1o = 2  /360 radians or, 1o =  /180 radians  Nick Devereux 2006

  6. Angular Units (continued) Arc seconds (‘’) and Arc Minutes (‘) 1‘ = 60 ‘’ 1o = 60 ‘ = 3600 ‘’ From previous page… 1o =  /180 radians So,  /180 radians = 3600 ‘’  /(180 x 3600) radians = 1‘’ Or, 1‘’= 4.8 x 10-6 radians and 206265 ‘’ = 1 radian  Nick Devereux 2006

  7. Getting back to the pc…..  Nick Devereux 2006

  8. The pc (radian) = arc /radius 1‘’/ 206265 (‘’/ radian) = 1 AU / 1pc So, 1 pc = 1 AU x 206265 1 pc = 1.496 x 1011 x 206265 m Or, 1 pc = 3.086 x 1016 m  Nick Devereux 2006

  9. In words, a parsec is the distance at which the separation between the Earth and the Sun could be resolved with a medium sized telescope…  Nick Devereux 2006

  10. What does resolved mean?  Nick Devereux 2006

  11. The Resolving Power of a Telescope Depends on both the size of the telescope mirror, D, and the wavelength,  ,of the light under observation. (radian) = 1.22  /D with and D in the same units. For the Hubble Space Telescope D = 2.4m and  = 0.05‘’ @  = 5500 Å .  Nick Devereux 2006

  12. Question: How far away would you have to hold a dime (2cm in diameter) for it to subtend an angle of 1‘’, 0.l‘’?  Nick Devereux 2006

  13. Mass Usually, the masses of stars, galaxies, clusters of galaxies are given in terms of the mass of the sun, 1 M = 1.99 x 1030 kg  Nick Devereux 2006

  14. Measuring Brightness Brightness is measured in a variety of ways; eg. Magnitude, Flux, and Luminosity  Nick Devereux 2006

  15. Luminosity of the Sun The luminosity of the Sun, 1 L  = 3.9 x 1026 W You may recall that Watts = Joules/sec, so the luminosity of the Sun is a measure of the rate of flow of energy through the surface of a star. Concept: Think of luminosity as the rate at which a star emits packets (photons) of energy…  Nick Devereux 2006

  16. Flux The further you move away from the star, the flux of photons, (measured in units of W/m2) passing through a 1m x 1m area goes down as the reciprocal of the distance squared;  Nick Devereux 2006

  17. Flux (continued) Quantitatively, the flux f = L/4D2 Units: W/m2  Nick Devereux 2006

  18. Magnitudes The magnitude scale dates back to the Greek astronomer Hipparchus (200 BC).  Nick Devereux 2006

  19.  Nick Devereux 2006

  20. Definition of Magnitude The human eye perceives, as linear, what are actually logarithmic differences in brightness. m = -2.5log f + c m is the apparent magnitude f is the flux c is a constant related to the flux of a zero magnitude star Note the –2.5, the brighter the star (f increases), the more negative the magnitude (m decreases).  Nick Devereux 2006

  21. Differences in magnitudes are equivalent to ratio’s of fluxes This obviates the need to know the constant c, or, the zero point, of the magnitude scale because; m1 = -2.5log f1 + c m2 = -2.5log f2 + c m1 – m2 = 2.5 log f2 / f1 Note the c’s cancelled (c – c = 0) Also, beware the subscripts are reversed on either side f the equals sign.  Nick Devereux 2006

  22. Question: A binary star system has one star that is 8 times brighter than the other. What is the magnitude difference between the two stars?  Nick Devereux 2006

  23. Absolute Magnitude We are unable to tell just by looking at the night sky if one star is fainter than another because it is intrinsically fainter (ie. lower luminosity) or just further away. To realistically compare stars on an equal basis we introduce the concept of Absolute magnitude (M) which is the magnitude stars have if they are all placed at the same reference distance of 10 pc.  Nick Devereux 2006

  24. --------------------------------------------------------------------- * (2) d(pc) ----------------------------* (1) 10 pc f1 = L/4(10)2 and f2 = L/4 d2 M – absolute magnitude = -2.5log f1 +c m – apparent magnitude = -2.5log f2 +c Then, M-m = 2.5log f2 / f1 M-m = 2.5log L 4 (10)2 /4 d2 .L M-m = 2.5log 100/d2  Nick Devereux 2006

  25. Distance Modulus M-m = 5 – 2.5log d2 M-m = 5 – 5log d So, the absolute magnitude, M = m + 5 – 5log d (remember, the distance d must be in pc) On rearranging, m - M= 5log d – 5 Where the quantity m – M is known as the distance modulus  Nick Devereux 2006

  26. Trigonometric Parallax Of course, to calculate the absolute magnitude of a star, we must know it’s distance. Distances to nearby stars can be found using Trigonometric parallax.  Nick Devereux 2006

  27. Parallax Angle  Question: Given the definitions for angular units provided earlier show that the parallax angle,  measured in arc seconds is equal to the reciprocal of the distance to the star in pc. So that, = 1/d The nearest star a Centauri at a distance of 1.3 pc has a parallax angle = 1/1.3 = 0.77’’. All other stars have even smaller parallaxes.  Nick Devereux 2006

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