Unformatted text preview: 998 Ã— 10 8 m / s)(0 . 050) (0 . 0193 m / s Â· ly)(1 + 0 . 050) = 7 . 4 Ã— 10 8 ly . (g) Â±rom the result of part (a), âˆ† t = r c âˆ’ Î±r = (7 . 4 Ã— 10 8 ly)(9 . 46 Ã— 10 15 m / ly) 2 . 998 Ã— 10 8 m / s âˆ’ (0 . 0193 m / s Â· ly)(7 . 4 Ã— 10 8 ly) = 2 . 5 Ã— 10 16 s , which is equivalent to 7 . 8 Ã— 10 8 y. (h) Letting r = c âˆ† t , we solve for âˆ† t : âˆ† t = r c = 7 . 4 Ã— 10 8 ly c = 7 . 4 Ã— 10 8 y . (i) The distance is given by r = c âˆ† t = c (7 . 8 Ã— 10 8 y) = 7 . 8 Ã— 10 8 ly . (j) Â±rom the result of part (f), r B = c (âˆ† Î»/Î» ) Î± (1 + âˆ† Î»/Î» ) = (2 . 998 Ã— 10 8 m / s)(0 . 080) (0 . 0193 mm / s Â· ly)(1 + 0 . 080) = 1 . 15 Ã— 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.
 Fall '08
 SPRUNGER
 Physics, Light

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