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Review Lecture 15 The Central Problem in astronomy is distance. What we see is basically a twodimensional picture of the sky. To interpret many pieces of information available to the astronomer we need to know how far away a star or galaxy is. Example: If you look at the sky, Sirius is brighter than Betelgeuse. But Betelgeuse is actually 300 times more luminous or brighter than Sirius. That is, Betelgeuse actually emits far more energy than Sirius. The reason is that Betelgeuse is just much further away from us than Sirius; thus, apparently less bright or dimmer. To get at these quantities we need three definitions. Luminosity = L, measured in Watts, is the power output(at all wavelengths) of the star, L = σAT4 where σ = 5.67 x10 −8 in Watts W 2 m K 4 , A is the surface area of the star, and T is the temperature measured in Kelvin. This power is spread out into the area of a sphere, where spherical area = 4 πd according to the Inverse Square Law. 2 Apparent brightness and the Inverse Square Law, measured in Watts/meter2, is what we appear to see. luminosity L brightness = b = = spherical area 4 π d 2 (1) Thus, we can only find the luminosity of the star if we know the apparent brightness of the star and the distance to the star, d. The brightness at the surface of a star is called the flux lu min osity L flux = σT4 = = in W/m2. surface area 4 πd 2 [Note: This looks like the Stefan-Boltzmann Law and is similar. If you will, flux is the brightness of a star measured at the surface of the star.] Brightness - unfortunately, Hipparchus (130 B.C.) measured brightness in terms of apparent magnitude, m, with a scale such that m = 1 the brightest star in the sky visible to the eye m = 6 the dimmest “ ” ∆ m = 5 represents 100 times brighter by instrument in a logarithmic fashion. Thus, if x represents the increase in brightness for ∆ m = 1, then x5 = 100 ⇒ x = 2.512 ≅ 2.5 So, going from the brightest to the dimmest brightness magnitude 5 (2.5) = 100 m=1 brightest 4 (2.5) = 39.1 m=2 3 (2.5) = 12.17 m = 3 (2.5)2 = 6.25 m=4 2.5 m=5 1 m=6 dimmest Note: The eye has a different response to light than mechanical devices. Thus: mV = visual apparent magnitude mB = bolometric apparent magnitude (magnitude including all wavelengths) Relationship between apparent magnitude and apparent brightness What this says is that if I have two stars with magnitudes m1 and m2, where 1 is brighter than 2, then the ratio of brightnesses is given by ( 2.5) m − m = ( 2.5) ∆m = 2 1 b1 b2 (2) This can be rewritten to solve for the difference in magnitudes by taking the log of both sides, i.e. ⎛b ⎞ log10 2.5∆m = ∆ m log10 2.5 = 0.4 ∆ m = log10 ⎜ 1 ⎟ or ⎝ b2 ⎠ (3) ⎛ b1 ⎞ ∆ m = 2.5 log10 ⎜ ⎟ ⎝ b2 ⎠ As an example, consider α Centauri(m1 = 0) and Bernard’s Star(m2 = 10), then ∆ m = 10 and the ratio of brightnesses is (2.5)10 = (2.5)5(2.5)5 = 1002 = 10,000. Thus α Centauri has an apparent brightness 10,000 greater than Bernard’s Star. The inverse square law of equation 1 says that equation 3 can be rewritten as 2 ⎛b ⎞ ⎛d ⎞ ⎛d ⎞ ∆ m = 2.5 log10 ⎜ 1 ⎟ = 2.5 log10 ⎜ 2 ⎟ = 5 log10 ⎜ 2 ⎟ ⎝ b2 ⎠ ⎝ d1 ⎠ ⎝ d1 ⎠ (4) Absolute magnitude measures the apparent brightness of a star at 10 parsecs. By implication, a star at a distance of 10 parsecs would have the same apparent magnitude as absolute magnitude. Let the standard distance then be dstandard = 10 parsecs = 3 x 1017 m Example: The Sun mSun = -26.5, the Sun is the brightest star in the sky dsun = 1 AU = 1.5 x 1011 m The Inverse Square Law says that brightness ~ ⇒ b sun b s tan dard 1 dis tan ce 2 = 1 d2 2 d 2s tan dard ⎛ 3 x 1017 m ⎞ 6 2 ⎜ ⎟ 2 x 10 = = = = 4 x 1012 ( ) 2 11 d sun . x 10 m ⎠ ⎝ 15 This says the ratio of brightness is inversely proportional to the squared distances. Thus, the Sun is 4 x 1012 times brighter where it is, as seen from the Earth, than at 10 parsecs. What is the absolute magnitude of the Sun at 10 parsecs. Note: increased distance ⇒ decreased intensity ⇒ ∆ m positive. According to equation ∆m = 4 x101 2 ∆ m1 ( ) 1, we need to determine ∆ m such that ( 2.5) Two methods: 12 1) Estimate 4 x 10 = 4 x (100) (2.5)1 = 2.5 (2.5)2 = 6.2 6 = 2.5 x 2.5 5 ∆ m2 = (2.5) ∆ m1 + 5 ∆ m2 4 is somewhere between 2.5 and 6.2. Guess ∆m1 = 1.5 ( ) 100 = 2.5 6 5 6 = (2.5) 30 ⇒ ∆m2 = 30 ∆m = ∆m1 + 5 ∆m2 = 15 . + 30 = +315 . 2) Calculate by taking logarithms of both sides ( 2.5) ∆m = 4 x 1012 log(4 x 1012 ) = log(2.5m ) = m log( 2.5) log(4 x 1012 ) 12.6 ∆m = = = + 317 . ≅ 315 . estimate log( 2.5) .4 ⇒ The Absolute Visual Magnitude of the Sun is MSun = -26.5 + 31.5 = +5 The Sun would be a dim star just visible in Cortland at 10 parsecs. Distance Distance can only be determined directly from a local group of stars near enough to have their parallax measured. 1 d= (in par sec s) p(in arc sec onds) From Earth the smallest parallax measurable is about .01 arcseconds corresponding to a distance of 100 pc = 326 light years. There are only about 2000 stars within this range. Satellite measurements extend this distance to about 1000 pc and 200,000 stars. In comparison to the Sun, equation (1) can be written as L LSun 2 ⎛ d ⎞ b =⎜ ⎟ ⎝ d Sun ⎠ bSun Absolute Magnitude from distances Equation 3 can be rewritten as ⎛ D⎞ ∆ m = M − m = 5 log⎜ ⎟ = 5 log D − 5 log d ⎝ d⎠ = 5 log 10 − 5 log d = 5 − 5 log d (5) This can be rearranged to say that the absolute magnitude can be determined from the apparent magnitude and the distance from the observer, i.e. (6) M = m + 5 − 5 log10 d This equation can be used to produce the log-log plot shown below. To use the plot you need two of the three quantities: relative magnitude, distance, or absolute magnitude. In the figure below, Sirius has a bolometric apparent magnitude of m = -1.4 (the brightest star in the sky other than the Sun has a visual magnitude of 1), and from parallax measurements d = 2.7 parsecs away. Finding this point on the plot and extrapolating over we find that Sirius has an absolute magnitude of MSirius = +1 and a luminosity relative to the Sun of LSirius = 23Lsun. Conversely, the Sun has an absolute magnitude of +5; thus, at 10 parsecs the relative magnitude would be +5. Stars Within 4 Parsecs Distance (pc) Name Part Apparent Magnitude Absolute Magnitude Spectral Type Mass (MSun) Radius (RSun) 0.0 Sun Jupiter A B -26.6 -2.6 5 26 G2 1.0 0.001 1.0 0.1 1.3 Alpha Centauri A B C 0 1 11 4 6 15 G2 K0 M5 1.1 0.9 0.1 1.2 0.9 ? 1.8 Bernard’s Star 10 13 M5 ? ? 2.3 Wolf 359 14 17 M8 ? ? 8 10 M2 0.35 ? o 2.5 BD +36 2147 2.7 L726-8 A B 12 13 15 16 M6 M6 0.11 0.11 ? ? 2.9 Sirius A B -2 8 1 11 A1 A5 Wh Df 2.3 1.0 1.8 0.02 2.9 Ross 154 11 13 M5 ? ? 3.1 Ross 248 12 15 M6 ? ? 3.2 L789-6 12 15 M6 ? ? 3.3 ε Eridani 4 6 K2 0.9 ? 3.3 Ross 128 11 14 M5 ? ? 3.3 61 Cygni 5 6 8 8 K5 K7 0.63 0.6 ? ? 3.4 ε Indi 5 7 K5 ? ? 3.5 Procyon A B 0 11 3 13 F5 Wh Dwarf 1.8 0.6 1.7 0.01 3.5 Σ 2398 A B 9 10 11 12 M4 M5 0.4 0.4 ? ? 3.5 BD +43o44 A B C 8 11 ? 10 13 ? M1 M6 K? ? ? ? ? ? ? 3.6 τ Ceti 4 6 G8 ? 1.0 7 10 M2 ? ? 10 ? 12 ? M4 ? ? ? ? ? A B o 3.6 CD -36 15693 3.7 BD +5o1668 3.7 G51-15 15 17 ? ? ? 3.8 L725-32 12 14 M5 ? ? 3.8 o CD -39 14192 7 9 M0 ? ? 3.9 Kapteyn’s Star 9 11 M0 ? ? 4.0 Kruger 60 A B C 10 11 ? 12 13 ? M4 M6 ? 0.27 0.16 0.01 0.51 ? ? 4.0 Ross 614 A B 11 15 13 17 M5 ? 0.14 0.08 ? ? 4.0 BD -12o4523 10 12 M5 ? ? A B