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Unformatted text preview: AST 3722C  Spring 2008 Homework #7 – solutions • 1. (1 point.) The star β Cyg is actually a double star. (a) Let subscripts 1 and 2 refer to the two stars. In Bband, the stars have magnitudes m B 1 = 4 . 171 and m B 2 = 5 . 027. Which star is fainter and by what factor in flux density? Star 2 is the fainter star. The ratio of flux densities would be 10 − . 4 × ( m B 1 − m B 2 ) = 2.200 . So that means Star 2 is fainter by this factor (or in other words its flux density is 1/2.200 = 0.455 times that of Star 1). (b) In Vband, the stars have magnitudes m V 1 = 3 . 085 and m V 2 = 5 . 088. Which star is fainter and by what factor in flux density? Star 2 is the fainter star. The ratio of flux densities would be 10 − . 4 × ( m B 1 − m B 2 ) = 6.327 . So that means Star 2 is fainter by this factor (or in other words its flux density is 1/6.327 = 0.158 times that of Star 1). (c) Use these numbers to figure out and explain which is the bluer (hotter) star, 1 or 2? Star 2 is 2.2 times fainter than Star 1 in Bband, but it’s 6.3 times fainter in Vband. In other words it’s even fainter in the longer wavelength. So that means that Star 2 must be bluer . • 2. (3 points.) Suppose you are looking at a star – call it “A” – that provides a flux density F A ( λ ) such that: F A ( λ ) = C A × parenleftBigg 1 μ m λ parenrightBigg 3 . 97 , where λ is the wavelength and C A is a constant with dimension power per area per wavelength interval. Nearby, a new object has appeared in the sky – call it “B” – and very conveniently you can take pictures of B while A is in the same FOV. You observe the two objects at several wavelengths and measure the following differences in magnitudes between A and B: Wavelength m A − m B ( μ m) (mag) 1.2 μ m10.058 1.6 μ m5.730 2.2 μ m2.282 3.6 μ m 1.115 5.0 μ m 2.498 8.9 μ m 3.904 9.7 μ m 4.040 10.6 μ m 4.166 11.7 μ m 4.291 12.5 μ m 4.366 18.7 μ m 4.714 20.6 μ m 4.776 1 (Of course there are uncertainties on these magnitudes but ignore those for now.) (a) Calculate F B , object B’s flux density, at all 12 wavelengths in terms of C A . We know that F A /F B = 10 − . 4 × ( m A − m B ) , so: F B = F A × 10 . 4 × ( m A − m B ) = C A × (1 μ m /λ ) 3 . 97 × 10 . 4 × ( m A − m B ) . What you get out of that is: Wavelength F B ( μ m) 1.2 μ m C A × 4 . 597 × 10 − 5 1.6 μ m C A × 7 . 900 × 10 − 4 2.2 μ m C A × 5 . 343 × 10 − 3 3.6 μ m C A × 1 . 728 × 10 − 2 5.0 μ m C A × 1 . 676 × 10 − 2 8.9 μ m C A × 6 . 202 × 10 − 3 9.7 μ m C A × 4 . 995 × 10 − 3 10.6 μ m C A × 3 . 944 × 10 − 3 11.7 μ m C A × 2 . 990 × 10 − 3 12.5 μ m C A × 2 . 464 × 10 − 3 18.7 μ m C A × 6 . 861 × 10 − 4 20.6 μ m C A × 4 . 947 × 10 − 4 (b) Make a semilog plot of wavelength versus F B . Since C A is in all values of F B , you can assume C A = 1 (in proper units), or anything you want for that matter. You should see that the plotted points more or less trace out a Planck function.that the plotted points more or less trace out a Planck function....
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 Spring '09
 Fernandez
 Astronomy, parsecs, µm µm µm, λu λl, β Cyg

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