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My apologies if my understanding is incorrect, but I believe that as you approach relativistic speeds you experience time dilation as compared to an outside observer.

So taking into account this effect how fast do you have to be travelling in order to reach an object one light year away in one year (subjective time of the traveller). How much normal time would have passed?

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You need to travel at $v=\sqrt{0.5}c=0.707c$. To see this, notice that because of lenght contraction the traveler will see shorter distance:

$L=L_0 \sqrt{1-(v/c)^2}$ (1)

where $L_0$ is one year light. If you travel for a year $t_y$, then $L=v*t_y=(v/c)ct_y=L_0(v/c)$.

Replacing that in (1) results in $v/c=\sqrt{0.5}$

UPDATE: the time at earth will be $t$, where $t'$ is one year:

$t'=t/\sqrt{1-(v/c)^2}=1.84$ years

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  • $\begingroup$ Are you sure your math is right? How are you accounting for the time dilation in the frame of reference of the traveler? $\endgroup$
    – Floris
    Commented Jun 5, 2015 at 19:47
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    $\begingroup$ @Floris the traveler travels one year in its own reference frame, that is what I undesrtood the op asked $\endgroup$
    – user66432
    Commented Jun 5, 2015 at 19:48
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    $\begingroup$ I think the result is $v^2/c^2 = 1/2$, not $v/c = 1/2$. $\endgroup$
    – Michael Seifert
    Commented Jun 5, 2015 at 19:48
  • $\begingroup$ @MichaelSeifert you are right, I solved it too fast, thanks $\endgroup$
    – user66432
    Commented Jun 5, 2015 at 19:49
  • $\begingroup$ @brucesmitherson so 0.25c? that's cool. How much normal time passes in this situation? $\endgroup$
    – Aequitas
    Commented Jun 5, 2015 at 19:55

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