This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: AST 3722C  Spring 2008 Homework #8 – solutions 1. (a) What are the units of B ν ? W / m 2 / Hz / sr or dimensionally equivalent. (b) What are the units of S ? W / m 2 or dimensionally equivalent. (c) One way to do the integral is to do a substitution. Let u = hν/kT . Rewrite S by inserting the Planck function and making it an integral over u instead of over ν . Answer : S = 2 πk 4 T 4 h 3 c 2 integraldisplay ∞ u 3 du e u − 1 . (d) I gave you: 1 e x − 1 = ∞ summationdisplay n =1 e − nx . Use this to rewrite the integrand as an infinite sum of integrals. Answer : S = 2 πk 4 T 4 h 3 c 2 integraldisplay ∞ u 3 ∞ summationdisplay n =1 e − nu du. (e) Next do integrationbyparts. Don’t evaluate the summation, integratebyparts on a generic integrand of the form u 3 e − nu . Evaluate the result at infinity and zero (i.e., actually do out the definite integral). You can switch the order of integraltext and ∑ so that instead of taking the integral of a sum you take the sum of the integrals: S = 2 πk 4 T 4 h 3 c 2 ∞ summationdisplay n =1 integraldisplay ∞ u 3 e − nu du. And that’s why you can do the integration parts as the problem states. You do have to do the integration by parts three times though. You should get that the indefinite integral would come out to be integraldisplay u 3 e − nu du = − e − nu parenleftBigg u 3 n + 3 u 2 n 2 + 6 u n 3 + 6 n 4 parenrightBigg . Evaluating that at ∞ and zero is easy if you recall that lim x →∞ e − x x p = 0 for any p . Then you get integraldisplay ∞ u 3 e − nu du = 6 n 4 . So that Answer : S = 12 πk 4 T 4 h 3 c 2 ∞ summationdisplay n =1 1 n 4 . 1 (f) You should wind up with something that depends on ∞ summationdisplay n =1 1 n 4 . It turns out that this equals π 4 / 90. Replace the summation with this expression. At this point you should have an equation of the form S = CT 4 , where C is a constant that depends on a bunch of universal constants. Evaluate C . Make sure you have the right units. Hey hey we did end up with this summation! So we get S = 12 πk 4 T 4 h 3 c 2 π 4 90 = 2 π 5 k 4 15 h 3 c 2 × T 4 . If you calc it out, C = 5 . 67 × 10 − 8 W/m 2 /K 4 . This is the StefanBoltzmann constant and usually represented by σ . In problem 2 I’ll call it σ . (g) S tells you the total amount of power (at all frequencies ) per unit area emitted by a blackbody. Given two stars, one with temperature T and another with temperature 3 T , how much more power per unit area over all frequencies does the hotter star emit? Since it depends on the fourth power of temperature, the power per unit area is 3 4 = 81 times higher. 2. (5 points.) In this problem you will use the result from problem 1....
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.
 Spring '09
 Fernandez
 Astronomy

Click to edit the document details