p45_043 - 998 × 10 8 m s(0 050(0 0193 m s ly(1 0 050 = 7...

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43. (a) During the time interval ∆ t , the light emitted from galaxy A has traveled a distance c t .M e an - while, the distance between Earth and the galaxy has expanded from r to r 0 = r + t .L e t c t = r 0 = r + t ,wh ichleadsto t = r c . (b) The detected wavelength λ 0 is longer then λ by λα t due to the expansion of the universe: λ 0 = λ + λα t .Thu s , λ λ = λ 0 λ λ = α t = αr c αr . (c) We use the binomial expansion formula (see Appendix E): (1 ± x ) n =1 ± nx 1! + n ( n 1) x 2 2! + ··· ( x 2 < 1) to obtain λ λ = αr c αr = αr c ³ 1 αr c ´ 1 = αr c · 1+ 1 1! ³ αr c ´ + ( 1)( 2) 2! ³ αr c ´ 2 + ··· ¸ αr c + ³ αr c ´ 2 + ³ αr c ´ 3 . (d) When only the Frst term in the expansion for ∆ λ/λ is retained we have λ λ αr c . (e) We set λ λ = v c = Hr c and compare with the result of part (d) to obtain α = H . (f) We use the formula ∆ λ/λ = αr/ ( c αr )toso lvefor r : r = c (∆
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Unformatted text preview: 998 × 10 8 m / s)(0 . 050) (0 . 0193 m / s · ly)(1 + 0 . 050) = 7 . 4 × 10 8 ly . (g) ±rom the result of part (a), ∆ t = r c − αr = (7 . 4 × 10 8 ly)(9 . 46 × 10 15 m / ly) 2 . 998 × 10 8 m / s − (0 . 0193 m / s · ly)(7 . 4 × 10 8 ly) = 2 . 5 × 10 16 s , which is equivalent to 7 . 8 × 10 8 y. (h) Letting r = c ∆ t , we solve for ∆ t : ∆ t = r c = 7 . 4 × 10 8 ly c = 7 . 4 × 10 8 y . (i) The distance is given by r = c ∆ t = c (7 . 8 × 10 8 y) = 7 . 8 × 10 8 ly . (j) ±rom the result of part (f), r B = c (∆ λ/λ ) α (1 + ∆ λ/λ ) = (2 . 998 × 10 8 m / s)(0 . 080) (0 . 0193 mm / s · ly)(1 + 0 . 080) = 1 . 15 × 10 9 ly ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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