1. Stars Part One: Brightness and Distance

2. Concept -1 – Temperature λ max (metres) = 2.90 x 10 -3 m k T (Kelvin) λ max = Peak black body wavelength T = The star’s surface temperature in Kelvins

3. λmax = 2.90 x 10 -3 m K T From Jay Pasachoff’s Contemporary Astronomy λ max

4. Put this in your notes: A star has a surface temperature of 5787 K, what is its λ max ? λ max = (2.90 x 10 -3 m K)/T = (2.90 x 10 -3 m K) /5787 = 501 nm 501 nm

5. A star has a λ max of 940 nm, what is its surface temperature? λ max = (2.90 x 10 -3 m K)/T, T = (2.90 x 10 -3 m K)/ λ max = (2.90 x 10 -3 m K)/ (940E-9) = 3100 K 3100 K

7. Concept 0 – Total power output Luminosity L = σAT 4 L = The star’s power output in Watts σ = Stefan Boltzmann constant = 5.67 x 10 -8 W/m 2 K 4 A = The star’s surface area = 4πr 2 T = The star’s surface temperature in Kelvins

8. Put this in your notes: Our Sun has a surface temp of about 5787 K, and a radius of 6.96 x 10 8 m. What is its Luminosity? L = σAT 4 = 5.67x10 -8 (4π(7x10 8 ) 2 )(5787) 4 = 3.87x10 26 Watts 3.87x10 26 Watts

9. A star has a radius of 5.0 x 10 8 m, and Luminosity of 4.2 x 10 26 Watts, What is its surface temperature? Luminosity L = σAT 4, T =(L/(σ4  (5x10 8 ) 2 )).25 =6968 K = 7.0x10 3 K 7.0x10 3 K

11. Concept 1 – Apparent Brightness Apparent Brightness b = L 4πd 2 b = The apparent brightness in W/m 2 L = The star’s Luminosity (in Watts) d = The distance to the star L is spread out over a sphere..

12. Apparent Brightness b = L/4πd 2 The inverse square relationship From Jay Pasachoff’s Contemporary Astronomy

13. Put this in your notes: Our Sun puts out about 3.87 x 10 26 Watts of power, and we are 1.50 x 10 11 m from it. What is the Apparent brightness of the Sun from the Earth? b = L/4πd 2 = 3.87E26/ 4π1.50E11 2 = 1370 W/m 2 1370 W/m 2

14. Another star has a luminosity of 3.2 x 10 26 Watts. We measure an apparent brightness of 1.4 x 10 -9 W/m 2. How far are we from it? b = L/4πd 2, d = (L/4πb).5 = 1.3x10 17 m 1.3x10 17 m

16. Concept 2 – Apparent Magnitude Apparent Magnitude: b = The apparent brightness in W/m 2 m = The star’s Apparent Magnitude Note that the smaller b is, the bigger m is. Logarithmic scale: x100 in b = -5 in m

17. Put this in your notes: What is the apparent magnitude of a star with an apparent brightness of 7.2x10 -10 Wm -2 ? What is that of a star with an apparent brightness of 7.2x10 -12 Wm -2 ? 3.9, 8.9

18. From Jay Pasachoff’s Contemporary Astronomy

19. So apparent magnitude is a counter-intuitive scale -27 is burn your eyes out, 1 is a normal bright star, and 6 is barely visible to a naked, or unclothed eye. (You can’t see anything if you clothe your eyes!!) A magnitude 1 star delivers 100 times more W/m 2 than a magnitude 6 star.

20. What is the Apparent Magnitude of a star that has an apparent brightness of 1.4 x 10 -9 W/m 2 ? m = 2.5log 10 (2.52 x 10 -8W / m 2/b) = 2.5log 10 (2.52 x 10 -8W / m 2/ 1.4 x 10 -9 W/m 2 ) = 3.1 3.1

21. The Hubble can see an object with an apparent magnitude of 28. What is the apparent brightness of such a star or galaxy? m = 2.5log 10 (2.52 x 10 -8W / m 2 /b), b = 2.52 x 10 -8W / m 2 /10 (m/2.5) = 1.6 x10 -19 W/m 2 1.6 x10 -19 W/m 2

22. Concept 3 – Absolute Magnitude Absolute Magnitude: m - M = 5 log 10 (d/10) M = The Absolute Magnitude d = The distance to the star in parsecs m = The star’s Apparent Magnitude The absolute magnitude of a star is defined as what its apparent magnitude would be if you were 10 parsecs from it.

23. Absolute Magnitude: m - M = 5 log 10 (d/10) M = The Absolute Magnitude d = The distance to the star in parsecs m = The star’s Apparent Magnitude Example: 100 pc from an m = 6 star, M = ? (10x closer = 100x the light = -5 for m) M = 6 - 5 log10(100/10) = 1

24. Put this in your notes: The Sun has an apparent magnitude of -26.8, we are 1.5x10 8 km or 4.9 x 10 -6 pc from the sun. What is the sun’s absolute magnitude? M = m - 5 log 10 (d/10) = -26.8 - 5 log 10 (4.86 x 10 -6 /10) = 4.7 4.7

25. You are 320 pc from a star with an absolute magnitude of 6.3. What is its apparent magnitude? M = m - 5 log 10 (d/10), m = M + 5 log 10 (d/10) = 6.3 + 5 log 10 (320/10) = 14 14

27. Concept 4 – H-R diagrams In 1910, Enjar Hertzsprung of Denmark, and Henry Norris Russell at Princeton plotted M vs T in independent research. From Douglas Giancoli’s Physics Main Sequence

28. HotCooler Dim Bright From Douglas Giancoli’s Physics

29. HotCooler Dim Bright From Jay Pasachoff’s Contemporary Astronomy

30. OOh( Hot) Bbe AaAa Ffine Ggirl, Kkiss Mme! ( Cooler)

31. The Star The atmosphere Light from star is filtered by atmosphere

32. From Jay Pasachoff’s Contemporary Astronomy

33. Spectral types: Suffixes 0 (hottest) - 9 (coolest) so O0, O1…O9, then B0, B1… O - B - A - F - G - K - M - 30,000 - 60,000 K, ionized H, weak H lines, spectral lines are spread out. O types are rare and gigantic. 10,000 - 30,000K, H lines are stronger, lines are less spread out (Rigel, Spica are type B stars) 7,500 - 10,000K, strong H lines, Mg, Ca lines appear (H and K) (Sirius, Deneb and Vega are A type stars) 6,000 - 7,500K, weaker H lines than in type A, strong Ca lines (Canopus (S.H.) and Polaris are type F) 5,000 - 6,000K, yellow stars like the sun. Strongest H and K lines of Ca appear in this star. 3,500 - 5,000K, spectrum has many lines from neutral metals. Reddish stars (Arcturus and Aldebaran are type K stars) 3,500 or less, molecular spectra appear. Titanium oxide lines appear. Red stars (Betelgeuse is a prominent type M)

34. SO… Hot stars are: Big, Bright, Brief and Blue Cool stars are: Diminuitive, Dim, and Durable and um… reD More about brief and durable next time…

35. Dim Bright Spectroscopic “parallax” Spectrum = M (from diagram) Measure m Find d From Douglas Giancoli’s Physics

36. Put in your notes: How far is an B0 that has an m of 8? M = m - 5 log 10 (d/10) 5 log 10 (d/10) = m - M log 10 (d/10) = (m - M)/5 d/10 = 10 ((m - M)/5) d = (10 pc)10 ((8 - -4.3)/5) d = 2884 pc 2884 pc

37. How far is an K0 that has an m of 14? d = (10 pc)10 ((14 - 5.8)/5) d = 437 pc 437 pc