How Does Time Dilation Affect the Voyager 1 Space Probe?

AI Thread Summary
The discussion focuses on calculating the effects of time dilation on the Voyager 1 space probe, which travels at 17,000 meters per second. Participants clarify the formula for time dilation, emphasizing the importance of the gamma factor, γ, in the calculations. The correct time dilation formula is confirmed as T = t/√(1 - (v/c)²), where T is the time experienced by the probe and t is the time on Earth. After calculations, it is determined that approximately 30,999,999.95 seconds pass on the probe for every 31,000,000 seconds on Earth, indicating a time difference of about 0.05 seconds. The discussion concludes with the affirmation that moving clocks, like that of the probe, run slower compared to stationary clocks on Earth.
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Homework Statement


The Voyager 1 space probe, launched in 1977, is moving faster
relative to the Earth than any other human-made object, at 17,000
meters per second.
(a) Calculate the probe's
. p
(b) Over the course of one year on earth, slightly less than one year
passes on the probe. How much less?


Homework Equations


\gamma=1/\sqrt{}1-(v/c)^2


The Attempt at a Solution


I calculated the \gamma but i don't know how to get answer (b)
 
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What's the time dilation formula?
 
What it says for the formula. On my first post.\gamma=1/\sqrt{}1-(v/c)^2
 
Stratosphere said:
What it says for the formula. On my first post.\gamma=1/\sqrt{}1-(v/c)^2
That's just the definition of gamma, a factor that appears in many relativistic formulas. What's the formula for time dilation? (It will certainly involve gamma, but also T and T'.)

Look here: https://www.physicsforums.com/showpost.php?p=905669&postcount=3"
 
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\Delta t = \gamma(\Delta + v\Delta /c^2)
By the way what does \Delta mean?
 
Last edited:
Stratosphere said:
\Delta t = \gamma(\Delta + v\Delta /c^2)
That's not the one you want. (Look for "time dilation".)
By the way what does \Delta mean?
Δ means "change"; Δt represents a time interval.
 
T=\frac{}{}t/\sqrt{}1-(v/c)^{}2
This is the one in my textbook.
 
If that's the right formula I think I got the right answer, T= 31000000.05 seconds have passed for the 31000000 seconds on earth.
 
I think you have the values of T and t_0 mixed up. Remember time dilation would mean that slightly less time passes for the probe.
 
  • #10
Stratosphere said:
If that's the right formula I think I got the right answer, T= 31000000.05 seconds have passed for the 31000000 seconds on earth.
Yes, it looks like you mixed things up. The rule to remember is that moving clocks run slow by a factor of gamma. Viewed from earth, the Earth clocks read a longer time than does the moving probe clock.

Since the speed is quite a bit less than c, I suggest using a binomial approximation for gamma.
 
  • #11
So i just redid it and i got 30,999,999.95 seconds passed on the probe. Is that right?
 
  • #12
Figure out how many fewer seconds passed on the probe, not the total number of seconds.
 
  • #13
.05 seconds?
 
  • #14
Stratosphere said:
.05 seconds?
Sounds good.
 
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