The scenario:

  1. An astronaut, in a space ship capable of accelerating from zero to C instantly ( to reduce complex math to get an easy to understand answer below ).

  2. The space ship will only travel around earth, meaning, it won't get further and further away, but basically always be at a constant distance from earths center point.

  3. To keep example simple, the astronaut will only travel a quarter of a spin around earth.

Ready, steady, go.

The observers with their high tech equipment on earth can calculate in advance what time the astronaut is expected to arrive after a quarter of a spin. Lets call it Xearth.

Now, so can the astronat. In fact, I believe their calculations show the same time. Lets call it Xspace.

Time dilation states that after this experiment, and when both stop to compare watches, earth, since they are the observers (although who is actually spinning, earth or the astronaut is a good question), that Xearth will be lets say 2050, while Xspace still 2019.fraction.of.a.nano.second.

At the same time. Calculations prior showed that the astronaut can expect his watch to be 2019.fraction.of.a.nano.second while on earth, since they considered not themselves being the observers, also expected the astronaut to be there 2019.fraction.of.a.nano.second

How do we address this paradox? What time is it on earth after the astronaut makes this trip?

  • Addressing in advance a response I might get immediately. Namely that since he is spinning around earth, there is no observer and their time should be equal indeed, but that would go against my understanding of the special relativity. It cares not where or what trajectory of the speeding astronaut is, it only cares about what speed he is traveling.

Please try to stick to this example when answering the question.

  • $\begingroup$ Fatal flaws with this question (v1): (a) The spacecraft cannot accelerate "to $c$" because it is massive. Time dilation depends strongly on how closely a moving object approaches $c$. (b) To travel once around the earth at nearly $c$ takes only some milliseconds, so any clock disagreement will be in the millisecond range. (c) As an answer points out, the GPS satellite constellation performs this experiment continuously, including a much smaller correction from general relativity. $\endgroup$ – rob Nov 18 '19 at 19:44

There is no paradox. The physics involved is well understood and is extremely well validated through our experience of deploying global positioning systems, for example. Routinely correction are made to the time settings for GPS satellites to account for the relativistic effects associated with the speed of their motion relative to the surface of the Earth. You can look up the calculations on a variety of sites- Wikipedia for instance.

  • $\begingroup$ Thank you, but you did not address the question or scenario. What will happen here? $\endgroup$ – momomo Nov 13 '19 at 23:41
  • $\begingroup$ Hi Momomo. Your scenario is unphysical, and ambiguous. A body travelling on a circular path around the earth will loose time as a result of its motion and gravitational effects. To calculate the time loss, one would need to know the ship's radial distance from Earth, which you don't specify, so I cannot address the scenario as it is too vague. $\endgroup$ – Marco Ocram Nov 14 '19 at 6:56
  • $\begingroup$ The time on Earth will have advanced by around 0.03 seconds. The time on the watch of the astronaut will not have advanced at all if the ship travelled at c. $\endgroup$ – Marco Ocram Nov 15 '19 at 14:20

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