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Special Relativity - time paradox question, not sure if it's right
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[QUOTE="daleklama, post: 4459647, member: 415061"] [h2]Homework Statement [/h2] 2 identical clocks are synchronised in a lab on the equator. One clock is carried around the equator in 24 hours at a constant speed. Given that the radius of the Earth is 6.4 x 10 ^ 6 m, find the difference between the times registered by 2 clocks when the traveling clock returns to the lab. [h2]Homework Equations[/h2] t' = t / γ γ = 1 / (√(1 - v^2 / c^2)) [h2]The Attempt at a Solution[/h2] First I found the circumference of the Earth using c = 2 pi r Found the circumference of the Earth to be 40212385 m. The clock takes 24 hours to go around - 86400 seconds so the speed the clock goes around the Earth at is distance / time = circumference / time = 465 m/s. I used the gamma equation to find the Lorentz factor γ = 1 / (√(1 - v^2 / c^2)) = 1. (the v = 465 m/s was too small to impact the equation?) So then, subbing my Lorentz factor into the time dilation formula t' = t/y t' = t/1 = t, so there's no time difference? That doesn't sound right at all, I've done something stupid, if anyone can correct me that would be great :) [/QUOTE]
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Special Relativity - time paradox question, not sure if it's right
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