Just out of curiosity, based on Einstein's special relativity theory, if Voyager were to return to the Earth now at the same average speed at which it flew to its current position, how much "younger" it would be than its replica displayed at JPL's Von Karman Auditorium?

I'm sorry if my question sounds stupid to any of you. I have little background in astronomy. I just got curious watching a documentary about Voyagers.

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    $\begingroup$ What path does it take to return to Earth? Does it receive a one-time impulse when it reaches a preset distance from the Sun, then coast back home at the same velocity relative to Earth as when it departed? $\endgroup$ Commented Dec 4, 2021 at 12:42
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    $\begingroup$ FWIW, Voyager 1 has been travelling for 16162 days (~44 years), and its current speed (relative to the Sun) is ~16.9 km/s (Lorentz factor $\gamma \approx 1+1.589×10^{-9}$). In your scenario, does it start the return journey today, or 22 years ago? $\endgroup$
    – PM 2Ring
    Commented Dec 4, 2021 at 15:39
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    $\begingroup$ @PM2Ring and on top of that $10^{-9}$ from Lorentz we have roughly $10^{-8}$ with the opposite sign due to Suns gravitation. Voyager should in fact be older than its earthly twin! $\endgroup$
    – asdfex
    Commented Dec 4, 2021 at 16:16
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    $\begingroup$ You are forgetting general relativity. The Voyager clocks are ticking faster than Earth-based clocks. While the velocity effects of relativity theory would suggest that the Voyager clocks are ticking slower than Earth-based clocks, the gravitational effects suggest otherwise. Those gravitational effects on the rate that clocks tick are greater in magnitude than are the velocity effects. $\endgroup$ Commented Dec 4, 2021 at 19:27
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    $\begingroup$ @David Hammen Okay I'll wait to see whether a more correct answer can be made $\endgroup$ Commented Dec 6, 2021 at 4:39

1 Answer 1


Here is the plot of the speed of the Voyager (by its distance from the Sun):

enter image description here

Source: this question

We can assume an about $\rm{20 \frac{km}{s}}$ average speed, what is ${\rm \approx \frac{2}{30000} c}$.

For such low velocities, we can use the approximative time dilation formula: $\rm{\Delta t'=\Delta t \frac{v^2}{2c^2}}$. That gives $\approx 4.44 \cdot 10^{-8}$, meaning a special relativistic time dilation of $\approx$ 1.4 seconds in a year.

Voyager 1 was launched in 1977.09.05, about 45 years ago. Going back on the same path would make an about 90 years travel time, giving the result of about $\rm{\underline{\underline{2 min}}}$ .

We have also general relativistic effects which should be counted, although they do not affect the result too much. The general relativistic time dilation (time slowing due to gravity) caused by the Sun at Jupiter orbit is:

$$\Delta t'=\Delta t \sqrt{1-\frac{v_e^2}{c^2}}$$

$v_e$ is the escape velocity. The orbital speed of the Jupiter is $\rm{\approx 13.6 \frac{km}{s}}$, thus $\rm{v_e \approx 19.3 \frac{km}{s}}$. But the Voyager has spent most if its time far more away as the Jupiter. It might cause maybe some tens of seconds dilation in the result, but not more.

A lesser approximative calculation could be done based on more detailed public data, but it would be a big work and would not affect significantly the result.

  • $\begingroup$ Well... meanwhile, also the Earth went by 30km/s around the Sun and had an about 3 time more gravitational time dilation (due to the Sun). Now I think, in a solar frame of reference, the Earth clone dilated more, making the result reverse and more, maybe even 10 minutes... I am not sure, maybe I should delete this answer $\endgroup$
    – peterh
    Commented Dec 4, 2021 at 22:41
  • $\begingroup$ Thank you so much for doing the math for me. I don't understand all of your calculations but I now at least understand even such a high speed at which Voyager is flying does not affect the time as much as I guessed. I guessed there will be at least a couple of days of difference in time experienced by the two. Thank you again! $\endgroup$ Commented Dec 5, 2021 at 7:18
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    $\begingroup$ You're calculating the time dilation for Voyager with respect to an observer far away at rest with respect to the Sun. This is not what has been asked for - you need to do the same calculation for Earth and look at the difference. Clocks on Earth are running even slower because of its higher speed and position deep down in Suns gravitation well. No need to delete the answer, just amend it with the second set of numbers. Roughly: Sun gravitation: 1E-8, Earth gravitation 1E-9, Earth speed 1E-7. $\endgroup$
    – asdfex
    Commented Dec 5, 2021 at 9:31

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