homework7solutions - AST 3722C - Spring 2008 Homework #7...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: AST 3722C - Spring 2008 Homework #7 solutions 1. (1 point.) The star Cyg is actually a double star. (a) Let subscripts 1 and 2 refer to the two stars. In B-band, 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 V-band, 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 B-band, but its 6.3 times fainter in V-band. In other words its 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 ( ) such that: F A ( ) = C A parenleftBigg 1 m parenrightBigg 3 . 97 , where 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 ( m) (mag) 1.2 m-10.058 1.6 m-5.730 2.2 m-2.282 3.6 m 1.115 5.0 m 2.498 8.9 m 3.904 9.7 m 4.040 10.6 m 4.166 11.7 m 4.291 12.5 m 4.366 18.7 m 4.714 20.6 m 4.776 1 (Of course there are uncertainties on these magnitudes but ignore those for now.) (a) Calculate F B , object Bs 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 m / ) 3 . 97 10 . 4 ( m A m B ) . What you get out of that is: Wavelength F B ( m) 1.2 m C A 4 . 597 10 5 1.6 m C A 7 . 900 10 4 2.2 m C A 5 . 343 10 3 3.6 m C A 1 . 728 10 2 5.0 m C A 1 . 676 10 2 8.9 m C A 6 . 202 10 3 9.7 m C A 4 . 995 10 3 10.6 m C A 3 . 944 10 3 11.7 m C A 2 . 990 10 3 12.5 m C A 2 . 464 10 3 18.7 m C A 6 . 861 10 4 20.6 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....
View Full Document

This note was uploaded on 11/09/2009 for the course AST 4700 taught by Professor Fernandez during the Spring '09 term at University of Central Florida.

Page1 / 8

homework7solutions - AST 3722C - Spring 2008 Homework #7...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online