1. II-2b. Magnitude 2015 (Main Ref.: Lecture notes; FK Sec.17-3) b  1 / d 2 Lec 2

2. 2 2b-(i) m naked-eye

3. 3 = 100 1/5 Therefore, six magnitudes must have ratios = 100 1/5 = 2.512 1 2.512 2.512 2 2.512 3 2.512 4 2.512 5 1 2.512 6.310 15.851 39.818 100.023 Note” the smaller the magnitude, the brighter the star! Table II-1 Sun :  26.7 Full Moon:  12.6 Venus:  4.4 Serius (brightest star):  1.4 Pluto: +15.1 Largest telescope: +21 Hubble Space Telescope: +30 EX 7 Modern Magnitude (See Fig. II-5 for more details.)

4. 4 Astronomers often use the magnitude scale to denote brightness The apparent magnitude scale is an alternative way to measure a star’s apparent brightness The absolute magnitude of a star is the apparent magnitude it would have if viewed from a distance of 10 parsecs Fig. II-5: The Apparent Magnitude Scale

5. 5 Fig. II-6: Apparent Magnitudes

6. 6 Math Expression  m = m 2 – m 1 = 2.5 log ( b 1 / b 2 ) Eqn(6) See examples in FK Box 17-3. ******************************************************************* EX 8: Venus m 1 =  4; dimmest star we can see m 2 = + 6. How many times brighter is Venus than the faintest star we can see? Ans: 10,000 times brighter (See class notes, also FK Box 17-3, Example 1)

7. 7 EX 9: RR Lyrae, variable: b peak = 2 b min. What is the magnitude change? Ans: 0.75 ( See class notes, also FK Box 17-3, Example 2) EX 10 (#) (#) Note: If use  m = 1.12, we get 2.8 times as bright. EX 10 2.8

8. 8 EX 11

9. 9 2b-(ii) Absolute Magnitude M Absolute Magnitude M = m a star would have if it were located at 10 pc

10. 10 Math Expression m – M = 2.5 log ( b M / b m ) Eqn(7) m – M = 5 log ( d m / d M ) Eqn(8a) d M = 10 pc; d m = true distance m – M = 5 log d (pc) – 5 Eqn(8b) (See lecture notes for derivation.) Distance Modulus DM = m – M Eqn(9) See FK Box 17-3 for DM(=m – M) vs d(pc). e.g., DM =  4 d = 1.6 +20 10 5

11. 11 EX 12  Note: If we use the exact value of 1pc = 2.066 x 10 5 AU  get M sun = 4.8!

12. 12 EX 13: A Star with m = +6 (faintest we can see by unadied eyes) at d = 20pc. What is the absolute magnitude? Ans: M = + 4.5 (See class notes.) ************************************************************** EX 14: Suppose we are at 100 pc away from Sun. Can we still see Sun with naked eyes? What is m of the sun then? Note: M sun = 4.8 (see Ex 12). Ans: No, too faint to be seen. Reason: m = 9.8 > 6 (See class notes and FK Box 17-3, Example 4.) ********************************************************************* Study more examples in FK Box 17-3. Luminosity Function: The Population of Stars (See FK pp 472-473) M #/vol

13. 13 Stars of relatively low luminosity are more common than more luminous stars Our own Sun is a rather average star of intermediate luminosity Fig. II-7: The Luminosity Function = FK Fig. 17-5