W11HW6 - 1 Problem 11.6 1.1 Part a We begin by di...

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Unformatted text preview: 1 Problem 11.6 1.1 Part a We begin by di erentiating the velocity addition formula (we will assume the space ship is traveling parallel to the Earth): u = u + v 1 + u v c 2 du dt = 1 + u v c 2 du dt + 7 dv dt ( u + v ) 1 c 2 du dt v + u 7 dv dt 1 + u v c 2 2 = 1 v 2 c 2 1 + u v c 2 2 du dt dt dt |{z} 1 = Letting u = 0 because the ship is, by de nition, at rest in its own instantaneous reference frame: du dt = 1 v 2 c 2 3 = 2 du dt Letting u = v because the space ship's velocity (according to Earth) is equal to the velocity of the space ship's reference frame relative to Earth's: dv dt = 1 v 2 c 2 3 = 2 dv dt Z dv 1 v 2 c 2 3 = 2 = Z dv dt dt Given that dv =dt = g = constant, v q 1 v 2 v 2 = gt Solving for v yields: v = gt q 1 + g 2 t 2 c 2 (1) Now, we integrate both sides of the equation for time dilation: Z dt = Z dt t = Z r 1 v 2 c 2 dt 1 and substitute in equation (1): t = Z v u u t 1 g 2 t 2 c 2 1 + g 2 t 2 c 2 dt = Z 1 + g 2 t 2 c 2 1 = 2 dt Looking up this integral in a table, we nd that it is equal to: t = c g arcsinh gt c Solving for t in terms of t yields: t = c g sinh gt c For the rst leg of the journey, t = 5 years. In addition, g = 9 : 86 m/s 2 , and c = 3 10 8 m/s: t = 3 10 8 9 : 86 sinh (9 : 86)(5 3 : 16 10 7 ) 3 10 8 1 3 : 16 10 7 = 86 years The total journey is 4 70 years = 344 years. Hence, it is the year 2100 + 344 =70 years = 344 years....
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W11HW6 - 1 Problem 11.6 1.1 Part a We begin by di...

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