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If an astronaut leaves planet $A$ for planet $B$ at speed $v$, will the time (measured by the astronaut's clock) that it takes for the astronaut to reach planet $B$ be less than the distance between the planets divided by the speed because of length contraction?

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Yes.

More specifically, if $d$ is the distance between the planets in their rest frame, then in the astronaut's frame the distance between the planets will be $\frac{d}{\gamma}$ so the travel time as measured from his frame will be \begin{align} t_\mathrm{astro} = \frac{d/\gamma}{v} = \frac{1}{\gamma}\frac{d}{v} \end{align} Notice that the quantity $d/v$ is precisely the time it would take for him to complete the journey as measured from the planet frame; \begin{align} t_\mathrm{planet} = \frac{d}{v} \end{align} so the two times are related by \begin{align} t_\mathrm{astro} = \frac{1}{\gamma} t_\mathrm{planet} \end{align} This jives with the effect of time dilation; the astronaut time is less than the planet time by a factor of $\frac{1}{\gamma}$.

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