Solve for an experimenter in the s frame the length

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Solve: For an experimenter in the S frame, the length of the accelerator tube is 3.2 km. This is the proper length A = L because it is at rest and is always there for measurements. The electron measures the tube to be length contracted to ( ) ( ) 2 2 1 1 0.99999997 3200 m 0.78 m L β ′ = = = A
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37.54. Model: Let the earth be frame S and the rocket be frame S´. S´ moves with speed v relative to S. Solve: (a) The round-trip distance is 860 ly. If the rocket takes time t to make the round trip, as measured on earth, its speed (as a fraction of c ) is 860 ly 860 yr v c c t t = = Δ Δ where we used c = 1 ly/yr (1 light year per year). The astronaut’s elapsed time t ´ is the proper time, so τ = 20 yr. The time dilation equation is 2 2 2 2 20 yr 1 (860 yr/ ) (20 yr/ ) 1 ( / ) 1 (860 yr/ ) t t t v c t τ Δ Δ = = Δ = Δ Δ Solving for t gives t = 860.2325 y, and thus 860 y 0.99973 0.99973 860.2325 y v v c c = = = (b) The rocket starts with rest energy E i = mc 2 and accelerates to have energy E f = γ p mc 2 . Thus the energy needed to accelerate the rocket is E = E f E 1 = ( γ p – 1) mc 2 This is just the kinetic energy K gained by the rocket. We know the rocket’s speed, so 8 2 22 2 1 1 (20,000 kg)(3.0 10 m/s) 7.6 10 J 1 (0.99973) E Δ = × = × (c) The total energy used by the United States in 2000 was 1.0 × 10 20 J. To accelerate the rocket would require roughly 760 times the total energy used by the United States.
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37.55. Model: Let S be the earth’s reference frame and S be the rocket’s reference frame. S travels at 0.5 c relative to S. Solve: (a) For the earthlings, the total distance traveled by the rocket is 2 × 4.25 ly = 8.5 ly. The time taken by the rocket for the round trip is 8.50 ly 8.50 ly 17 y 0.5 0.5 ly/y c = = (b) The time interval measured in the rocket’s frame S is the proper time because it can be measured with a single clock at the same position. So, ( ) ( ) 2 2 17 y 14.7 y 15 y 1 1 0.5 t v c τ τ τ Δ Δ Δ = = ⇒ Δ = The distance traveled by the rocket crew is length contracted to ( ) ( ) ( ) 2 2 1 8.50 ly 1 0.5 7.36 ly 7.4 ly L L v c ′ = = = Note that the speed is still the same: 7.36 ly 0.5 c 14.7 y L v τ = = = Δ (c) Both are correct in their own frame of reference.
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37.56. Model: Let S be the earth’s reference frame. Let S , S , and S ″′ be the reference frames of the three spaceships cruising through the galaxy in the direction from Delta to Epsilon at velocities v 1 = 0.3 c , v 2 = 0.5 c , and v 3 = 0.7 c relative to the earth’s frame. Solve: (a) In frame S, x D = 0 ly and t D = 0 y. Also, x E = 2 ly and t E = 1 y. In the moving frames, D D D 0 y. t t t ′′ ′′′ = = = We can use the Lorentz transformation to find the time at which Epsilon explodes. In frame S , ( )( ) ( ) ( )( ) ( ) 2 2 E E 1 E 2 2 2 2 1 1 y 2 ly 0.3 c 1 y 2 y 0.30 0.42 y 1 1 0.3 1 0.3 c t x v c t v c ′ = = = = For t E the terms in the numerator cancel and E 0 y. t ′′ = Lastly, E 0.56 y. t ′′′ = − (b) Spaceship 2 finds that the explosions are simultaneous.
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