solutions-ps07

# solutions-ps07 - Physics 341: Problem Set #7 Solutions 1....

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Physics 341: Problem Set #7 Solutions 1. Recall that the surface brightness of an exponential disk has the form I ( R ) = I 0 e - R/h R where I 0 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 = : I total = Z 0 I ( R ) 2 πR dR (a) Show that the total brightness of the exponential disk is I total = 2 πI 0 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 0 I ( R ) 2 πR dR = Z 0 2 πI 0 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 = , x = . So we have I total = 2 πI 0 Z x = x =0 e - x ( h R x ) h R dx = 2 πI 0 h 2 R Z 0 xe - x dx = 2 πI 0 h 2 R ± - ( x + 1) e - x ² 0 = 2 πI 0 h 2 R [(0) - ( - 1)] = 2 πI 0 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 , 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 (which gives the total light). The fraction of light f within a radius R is f ( R ) = I ( R ) I total = R R 0 I ( R ) 2 πR dR R 0 I ( R ) 2 πR dR = 2 πI 0 h 2 R R R/h R 0 xe - x dx 2 πI 0 h 2 R R 0 xe - x dx 1

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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 0 = 1 - ² R h R + 1 ³ e - R/h R where I again used the result from integration by parts. So the fraction of light - 2 /e = 0 . 264 = 26 . 4% Similarly, the fraction of the total light within R 3 h R is - 4 /e 3 = 0 . 801 = 80 . 1% 2. observed spiral galaxy rotation curves from Rubin, Ford, & Thonnard (1978), ApJ, 225, L107 5 rotation curve: UGC 5166 from Kristine Spekkens total dark matter halo stars: bulge stars: disk gas 6 (a) What is the mass of dark matter within 10 kpc?
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## This note was uploaded on 10/18/2011 for the course PHYSICS 341 taught by Professor Keeton during the Fall '08 term at Rutgers.

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solutions-ps07 - Physics 341: Problem Set #7 Solutions 1....

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