# chap33 - RELATIVITY 33 EXERCISES Section 33.2 Matter,...

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RELATIVITY 33 EXERCISES Section 33.2 Matter, Motion, and Ether 13. INTERPRET In this problem we are asked to take wind speed into consideration to calculate the travel time of an airplane. NTERPRET DEVELOP Since the velocities are small compared to c , we can use the non-relativistic Galilean transformation of velocities in Equation 3.7, where u , uuv =+ GGG G is the velocity relative to the ground ( S ), u is that relative to the air G () , S and v is that of relative to S (in this case, the wind velocity). We used a notation consistent with that in S Equations 33.5a and 33.5b. EVALUATE (a) If (no wind), then 0 v = G uu = G G (ground speed equals air speed), and the round-trip travel time is 2 2(1800 km) 4.5 h 800 km/h a d t u == = (b) If is perpendicular to then v G , u G 222 , v = + or 22 2 2 (800 km/h) (130 km/h) 789 km/h =− = = and the round-trip travel time is 2 / 4.56 h. b td u (c) If is parallel or anti-parallel to u on alternate legs of the round-trip, then v G G = ± 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 dd t uvuv =+= + = ′′ +− + ASSESS We find as mentioned in the paragraph following Equation 33.2. , abc ttt << 14. The difference between Equations 33.2 and 33.1 is || 2 2 2 2 2 1 1 1/ cL L L c cv v c v c ⎛⎞ ⎜⎟ Δ= − = = −− ⎝⎠ and for the given interferometer. If 8 2 / 2(11 m)/(3 10 m/s) 73.3 ns Lc = / vc 1 , we can expand the denominators (Appendix A) to obtain (a) For the Earth’s orbital speed (30 km/s), so that 2 2 (2 / )[1 / (1 /2 )] (2 / )( /2 ). tL c v c L c v c Δ≈ + − + = 8 /1 0 , = 8 1 2 73.3 ns( 10 ) 3.67 10 s t Δ= × = × 1 6 (a fraction of the period of visible light). (b) If (a few thousand periods of visible light). (c) At relativistic speeds, we use the exact expression for Δ If 4 1 2 / 10 , 3.67 10 s t = × . t 1 4 /, 1 / 1 / 2 / 3 , γ = and 4 3 73.3 ns( 2/ 3) 13.1 ns. t = (d) If /0 . 9 9 , 1/ 1 0.98 7.09, = = and (2 / ) ( 1) 3.17 s. c γμ Δ = Section 33.4 Space and Time in Relativity 15. INTERPRET This problem is about the distance between two stars, as seen by a spaceship moving at relativistic speed. DEVELOP The distance between stars at rest in system S appears Lorentz-contracted in the spaceship’s system , S according to Equation 33.4: 33.1

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33.2 Chapter 33 22 1/ x xv c Δ= Δ − EVALUATE With and we get 50 ly x 0.75 , v = c 2 1 / (50 ly) 1 (0.75) 33.1 ly xxv c = = ASSESS The distance appears to be shortened or “contracted” as observed by the spaceship. Note that length contraction occurs only along the direction of motion. 16. (a) On Earth (system S , where the velocity of Pluto is much less than 0.65 c ) 9 / (5.8 10 km)/ tx v Δ=Δ = × 84 (0.65 3 10 m/s) 2.94 10 s 8.26 h. ×× = × = (b) Time dilation (Equation 33.3) means that the time interval measured in the spacecraft (system is ) S 4 2 4 1 / (2.94 10 s) 1 (0.65) 2.26 10 s 6.28 h. ttv c Δ=Δ − = × = × = (Note that (/ )1 / / , vv cx Δ=Δ v so Lorentz contraction could have been used to reach the same result.) 17. INTERPRET This is a problem about the length of the spaceship measured in its rest frame.
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## This note was uploaded on 10/21/2010 for the course PHYSICS 2131441 taught by Professor Pheong during the Fall '10 term at University of California, Berkeley.

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chap33 - RELATIVITY 33 EXERCISES Section 33.2 Matter,...

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