hw1_f11 - Solutions to HW 1: Ch. 1 - 3, 4, 7, 8, 9, 20, 22,...

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Unformatted text preview: Solutions to HW 1: Ch. 1 - 3, 4, 7, 8, 9, 20, 22, 25, 41 The problem uses the Lorentz transform approach. More explicitly, they have done the following: Earth frame: Plane frame: Dlyra = +2500 c ­y = x1 x1’ = ? DOrion =  ­2500 c ­y = x2 x2’ = ? t1 = t2 = 0 (simultaneous) t1’ = ? t2’ = ? Use time Lorentz transform: t’ = γ(t – vx/c2) In earth frame: ∆t = t2 – t1 = 0 In plane frame: ∆t’ = t2’ – t1’ = γ(t2 – v*x2/c2)  ­ γ(t1 – v*x1/c2)  ­>  ­γ(v/c2)(x2 – x1) Note, the x2’ and x1’ in the textbook solution above are NOT correct – it should be x2 and x1! ** Note, the orbit circumference comes from the balance between the gravitational and centripetal forces of an object in a circular orbit about a planet. mv 2 Mm = G e2 where v and r are both unknown r r 2πr but : v = where T is the orbital period T ȹ 2πr ȹ 2 m ȹ ȹ 4π 2mr Mm ȹ T Ⱥ = = G e2 2 r T r 6 So that : r = 6.65 x 10 m 2πr = 4.18 x 107 m v = 7.74 x 103 m/s € ⇒ r3 = G Me T 2 4π 2 ...
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This note was uploaded on 12/07/2011 for the course PHYS 2200 taught by Professor Jjdong during the Fall '08 term at Auburn University.

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