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Introductory Physics Homework Help
Speed of rocket as viewed from Earth
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[QUOTE="Rococo, post: 5470185, member: 424318"] Yes, I forgot to say the question mentions that the equations ## z(\tau) = \frac{cosh(g\tau) - 1}{g} ## and ## t = \frac{sinh(g\tau)}{g} ## are in units where ##c=1##. So reinserting factors of ##c##, would these equations become: ##gt/c = sinh(g\tau /c)## ##z(\tau) = \frac{cosh(g\tau /c) - 1}{g/c^2} ##? Then for ##t=40 years=1.26 \times 10^9 s##, we have ##g\tau /c = sinh^{-1}(gt/c) = sinh^{-1}[\frac{(9.8)(1.26 \times 10^9)}{3 \times 10^8}]= 4.41## Giving a distance measured in the Earth frame: ## z= \frac{cosh(g\tau /c) - 1}{g/c^2} = \frac{(3 \times 10^8)^2}{9.8}[cosh(4.41) -1] = 3.69 \times 10^{17} m##. The speed in the Earth frame would then be: ##U^z = \frac{dz}{d\tau}\frac{d\tau}{dt} = c sinh(g\tau /c) \frac{1}{cosh(gt/c)} = c tanh(gt/c) = c tanh(4.41) =0.9997c## Is this the right way to do it? [/QUOTE]
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Speed of rocket as viewed from Earth
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