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How can I find the velocity of an object given the time elapsed of trip from the frame of reference of the object and the distance from the frame of reference of Earth.

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closed as unclear what you're asking by CuriousOne, Qmechanic Jul 24 '16 at 9:52

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  • $\begingroup$ If s is the distance from the frame of reference of earth, then from the frame of reference of the object, the distance is length contracted $s\sqrt{1-\left(\frac{v}{c}\right)^2}$. If $\tau$ is the elapsed time from the frame of reference of the object, then the velocity of the earth relative to the object v (and vice versa) is the contracted distance divided by the time $\tau$. You then have an equation that you can use to solve for v. $\endgroup$ – Chet Miller Jul 24 '16 at 23:08
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So you have proper time and proper length. Use rapidity, $\eta$. For proper length $x$ and proper time $\tau$-

$\eta=\sinh^{-1}{(\frac{1}{c}{\frac{dx}{d\tau}})}=\tanh^{-1}{(\frac{1}{c}{\frac{dx}{dt}})}$

From this you can calculate the velocity $\equiv dx/dt$

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