P38_006 - (b The equations do not show a dependence on acceleration(or on the direction of the velocity vector which suggests that a circular

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6. (a) The round-trip (discounting the time needed to “turn around”) should be one year according to the clock you are carrying (this is your proper time interval ∆ t 0 ) and 1000 years according to the clocks on Earth which measure ∆ t . We solve Eq. 38-7 for v and then plug in: v = c s 1 µ t 0 t 2 = (299792458 m / s) s 1 µ 1y 1000 y 2 = 299792308 m / s which may also be expressed as v = c p 1 (1000) 2 =0 . 999 999 50 c . The discussion in Sample Problem 38-7 dealing with these sorts of values may prove helpful for those whose calculators do not yield this answer.
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Unformatted text preview: (b) The equations do not show a dependence on acceleration (or on the direction of the velocity vector), which suggests that a circular journey (with its constant magnitude centripetal acceleration) would give the same result (if the speed is the same) as the one described in the problem. A more careful argument can be given to support this, but it should be admitted that this is a fairly subtle question which has occasionally precipitated debates among professional physicists....
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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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