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Unformatted text preview: RELATIVITY E XERCISES Section 33.2 Matter, Motion, and Ether 13. I NTERPRET In this problem we are asked to take wind speed into consideration to calculate the travel time of an airplane. D EVELOP Since the velocities are small compared to c , we can use the nonrelativistic Galilean transformation of velocities in Equation 3.7, , u u v ′ = + r r r where u r is the velocity relative to the ground ( S ), u ′ r is that relative to the air ( ), S ′ and v is that of S ′ relative to S (in this case, the wind velocity). We used a notation consistent with that in Equations 33.5a and 33.5b. E VALUATE (a) If v = r (no wind), then u u ′ = r r (ground speed equals air speed), and the roundtrip travel time is 2 2(1800 km) 4.5 h 800 km/h a d t u = = = (b) If v r is perpendicular to , u r then 2 2 2 , u u v ′ = + or 2 2 2 2 (800 km/h) (130 km/h) 789 km/h u u v ′ = = = and the roundtrip travel time is 2 / 4.56 h. b t d u = = (c) If v r is parallel or antiparallel to u r on alternate legs of the roundtrip, then u u v ′ = ± and the travel time is (see Equation 33.2, but with c replaced by ) u ′ 1800 km 1800 km 4.62 h 800 km/h 130 km/h 800 km/h 130 km/h c d d t u v u v = + = + = ′ ′ + + A SSESS We find , a b c t t t < < as mentioned in the paragraph following Equation 33.2. 14. The difference between Equations 33.2 and 33.1 is  2 2 2 2 2 2 2 2 2 2 2 1 1 1 / 1 / cL L L t t t c c v v c c v v c ⊥ ∆ = = =  and 8 2 / 2(11 m)/(3 10 m/s) 73.3 ns L c = × = for the given interferometer. If 2 2 / 1, v c << we can expand the denominators (Appendix A) to obtain 2 2 2 2 2 2 (2 / )[1 / (1 /2 )] (2 / )( /2 ). t L c v c v c L c v c ∆ ≈ + + = (a) For the Earth’s orbital speed (30 km/s), 2 2 8 / 10 , v c = so that 8 16 1 2 73.3 ns( 10 ) 3.67 10 s t ∆ = × = × (a fraction of the period of visible light). (b) If 2 2 4 12 / 10 , 3.67 10 s v c t = ∆ = × (a few thousand periods of visible light). (c) At relativistic speeds, we use the exact expression for . t ∆ If 2 2 2 2 1 4 / , 1/ 1 / 2/ 3, v c v c γ = = = and 4 3 73.3 ns( 2/ 3) 13.1 ns. t ∆ = = (d) If / 0.99, v c = 1/ 1 0.98 7.09, γ = = and (2 / ) ( 1) 3.17 s. t L c γ γ μ ∆ = = Section 33.4 Space and Time in Relativity 15. I NTERPRET This problem is about the distance between two stars, as seen by a spaceship moving at relativistic speed. D EVELOP The distance between stars at rest in system S appears Lorentzcontracted in the spaceship’s system , S ′ according to Equation 33.4: 33.1 33 33.2 Chapter 33 2 2 1 / x x v c ′ ∆ = ∆ E VALUATE With 50 ly x ∆ = and 0.75 , v c = we get 2 2 2 1 / (50 ly) 1 (0.75) 33.1 ly x x v c ′ ∆ = ∆ = = A SSESS The distance appears to be shortened or “contracted” as observed by the spaceship. Note that length contraction occurs only along the direction of motion....
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 Spring '09
 Prof.Kim
 Physics, Special Relativity, system S, /C

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