How do you calculate the time dilation effect experienced by a traveler traveling at a relatively low speed?

Specifically, how much time dilation would a traveller moving at $v=0.0007 c$ (speed of light) experience during a the time it would take (about 1 month) to travel from Earth to Jupiter (assuming straight line constant speed) when compared to the observer (on earth)? Would the difference (when the traveller returned and clocks were compared) be measured in minutes, hours, days, or weeks?


1 Answer 1


I'm almost certain there is a duplicate question out there, but I can't quite find it.

Anyway, if observer one sees observer 2 moving away from her at a speed $v$, it is a consequence of special relativity that observer 1 will say that a a time that an event that should last a duration $t_{0}$ in observer 2's frame will take a time $t=\frac{t_{0}}{\sqrt{1-\left(\frac{v}{c}\right)^{2}}}$. Having this formula, it is just a matter of calculation. The minimum Earth-Jupiter distance is $5.89\times 10^{11}\,\,$m, and the maximum Earth-Jupiter distance is $9.69\times 10^{11}\,\,$m, we find that $t-t_{0}$ must lie between 0.69 and 1.13 seconds for the scenario you envision. (At least as long as we ignore general relativistic effects).

  • $\begingroup$ I got about half a second by the same calculation. Trip time = about a month. gamma -1 = 2.5*10^-7 $\endgroup$ Mar 1, 2012 at 19:50
  • $\begingroup$ Yes, you're right. There was a data entry error in my spreadsheet. $\endgroup$ Mar 1, 2012 at 19:52
  • 2
    $\begingroup$ This question basically boils down to "how do I calculate time dilation" which is surely on the site somewhere (and besides, it's a duplicate of every relativity book). But I think it's worth mentioning the fact that in the process of turning around for the return trip necessary to compare clocks, the acceleration will throw things off a little. $\endgroup$
    – David Z
    Mar 2, 2012 at 6:38

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