p38_013 - The total time elapsed in the frame of Earth is...

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13. (a) The speed of the traveler is v =0 . 99 c , which may be equivalently expressed as 0 . 99 ly / y. Let d be the distance traveled. Then, the time for the trip as measured in the frame of Earth is t = d/v =(26ly) / (0 . 99 ly / y) = 26 . 3y. (b) The signal, presumed to be a radio wave, travels with speed c and so takes 26 . 0 y to reach Earth.
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Unformatted text preview: The total time elapsed, in the frame of Earth, is 26 . 3 y + 26 . 0 y = 52 . 3 y. (c) The proper time interval is measured by a clock in the spaceship, so ∆ t = ∆ t/γ . Now γ = 1 / p 1 − β 2 = 1 / p 1 − (0 . 99) 2 = 7 . 09. Thus, ∆ t = (26 . 3 y) / (7 . 09) = 3 . 7 y....
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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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