Problem Set 7 Solutions

# Problem Set 7 - Physics 341 Problem Set#7 Solutions 1 Recall that the surface brightness of an exponential disk has the form I R = I e R/h R where

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Unformatted text preview: Physics 341: Problem Set #7 Solutions 1. Recall that the surface brightness of an exponential disk has the form I ( R ) = I e- R/h R where I is the central surface brightness and h R is the disk scale length. The total brightness is given by integrating this profile from R = 0 to R = 1 : I total = Z 1 I ( R ) 2 ⇡ R dR (a) Show that the total brightness of the exponential disk is I total = 2 ⇡ I h 2 R Show your work! Hint: Let x = R/h R and rewrite the integral in terms of x and dx . Note that from integration by parts, R xe- x dx =- ( x + 1) e- x plus a constant. We want I total = Z 1 I ( R ) 2 ⇡ R dR = Z 1 2 ⇡ I e- R/h R R dR Now if we make the substitution x = R/h R , that means that R = h R x and dR/dx = h R , so dR = h R dx . The limits of integration don’t change, because when R = 0, x = 0, and when R = 1 , x = 1 . So we have I total = 2 ⇡ I Z x = 1 x =0 e- x ( h R x ) h R dx = 2 ⇡ I h 2 R Z 1 xe- x dx = 2 ⇡ I h 2 R ⇥- ( x + 1) e- x ⇤ 1 = 2 ⇡ I h 2 R [(0)- (- 1)] = 2 ⇡ I h 2 R where I used the hint for the integration. To show that- ( x + 1) e- x goes to zero when x goes to 1 , you can either just try out larger and larger values of x , or prove it using l’Hˆopital’s rule. (b) What fraction of the total light is within one disk scale length ( R h R )? What fraction of the total light is within three disk scale lengths ( R 3 h R )? To figure out the amount of light within some radius R , we just integrate the surface brightness as above, but now going from 0 to R , rather than from 0 to 1 (which gives the total light). The fraction of light f within a radius R is f ( R ) = I ( R ) I total = R R I ( R ) 2 ⇡ R dR R 1 I ( R ) 2 ⇡ R dR = 2 ⇡ I h 2 R R R/h R xe- x dx 2 ⇡ I h 2 R R 1 xe- x dx 1 The constants out front all cancel, and the integral in the denominator is just 1 from part (a), so we are left with f ( R ) = Z R/h R xe- x dx =- ( x + 1) e- x R/h R = 1- ✓ R h R + 1 ◆ e- R/h R where I again used the result from integration by parts. So the fraction of light within one scale length ( R h R ) is f ( R h R ) = 1- (1 + 1) e- 1 = 1- 2 /e = 0 . 264 = 26 . 4% Similarly, the fraction of the total light within R 3 h R is f ( R 3 h R ) = 1- (3 + 1) e- 3 = 1- 4 /e 3 = 0 . 801 = 80 . 1% 2. Using the UGC 5166 rotation curve......
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## This note was uploaded on 11/13/2011 for the course PHY 341 taught by Professor Gawsier during the Spring '11 term at Rutgers.

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Problem Set 7 - Physics 341 Problem Set#7 Solutions 1 Recall that the surface brightness of an exponential disk has the form I R = I e R/h R where

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