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34 years have passed since Voyager I took off and it's just crossing the solar system, being approximately at 16.4 light-hours away. How much time have passed for itself, though?

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34 years. (If you're thinking about relativistic time dilation the effect is very small.) –  Michael Brown Mar 21 '13 at 7:26
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I was - I've just found an online calculator for it. So I guess I can end my hopes we will ever be able to build a time slowdown machine that worked by orbiting the sun ridiculously fast, or something like that? A next question might be: why voyager 1 is so slow? What stops us from just accelerating more? –  Dokkat Mar 21 '13 at 7:31
    
@Dokkat If you can't find the answer to that followup on this site, by all means start a new post. –  Chris White Mar 21 '13 at 7:41
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Similar effect from different physics: physics.stackexchange.com/questions/9921/… –  Chris White Mar 21 '13 at 7:42

1 Answer 1

Using those numbers, let's assume it has been traveling in a straight line at constant speed.1 Then the speed is $$ v = \frac{16.4~\mathrm{hr}\cdot c}{34~\mathrm{yr}} = 5.5\times10^{-5} c, $$ so we can see right away it is traveling at a negligible fraction of the speed of light.

Still, we can compute the effects of time dilation. The relevant factor is $$ \frac{1}{\gamma} = \sqrt{1 - \frac{v^2}{c^2}} = 1 - 1.5\times10^{-9}. $$ This is because the proper time experienced by the spacecraft is the time lapse we experience - call it $\Delta t_\text{Earth}$ - divided by gamma: $$ \Delta t_\text{Voyager} = \frac{1}{\gamma} \Delta t_\text{Earth}. $$ The spacecraft's clocks have progressed just a hair under $34$ years - in fact, they are slow by $$ 1.5\times10^{-9} (34~\mathrm{yr}) = 1.6~\mathrm{s}. $$


1 This is not quite true, but the general result is still the same.

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