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Introductory Physics Homework Help
Speed of rocket as viewed from Earth
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[QUOTE="Rococo, post: 5468529, member: 424318"] [h2]Homework Statement [/h2] A rocket starts at rest from the Earth, moving in the z direction so that in its own instantaneous rest frame the acceleration is always g = 9.8 m/s^2. To an observer on Earth, its position is given by ## z(\tau) = \frac{cosh(g\tau) - 1}{g} ## ## t = \frac{sinh(g\tau)}{g} ## where ##\tau## is the rocket's proper time. Measured in a reference frame at rest on the Earth, how far from the Earth will the rocket be in 40 years? How fast will it be going? [h2]Homework Equations[/h2] Four-velocity: ## U^\mu = (\gamma, \frac{d\underline{x}}{d \tau}) = (\gamma, 0, 0, sinh(g\tau)) ## [h2]The Attempt at a Solution[/h2] t = 40 years = ##1.26\times10^9 s## g = ##9.8m/s^2## ##g\tau = sinh^{-1}(gt) = sinh^{-1}((9.8)(1.26\times10^9)) = 23.93## ##z = \frac{cosh(g\tau)-1}{g} = \frac{cosh(23.93)-1}{9.8} = 1.26\times10^9 m## ## U^z = sinh(g\tau) = sinh(23.93) = 1.24\times10^{10} m/s## But this is greater than the speed of light so it can't be right. [/QUOTE]
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Speed of rocket as viewed from Earth
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