This is a thought experiment question on time dilation/ relativity etc.

'Jack McCormick' leaves the Earth at 99.99% the speed of light. Later, he looks at his wrist watch and his wrist watch tells him that 12 hours have passed since he left Earth. Jack turns around and comes back to Earth. According to his wrist watch, only 24 hours passed since he first left Earth. But, the people he knew are all dead because 180 years passed for the people on Earth, while only 24 hours passed for Jack.

For his next trip, Jack wants to leave Earth and return to Earth in such a way that his wrist watch tells him that 24 hours have passed for him, but only 1 hour passes for the people on Earth.

At what Velocity does Jack need to travel, for this to be possible?

  • $\begingroup$ Notice that in order to "turn around and come back to Earth" Jack would need to accelerate, so outbound Jack and inbound Jack would not agree on what time it is for them/on Earth. $\endgroup$ Feb 9 at 21:24
  • 2
    $\begingroup$ The title to this question (v2) is not useful. Consider editing your question so that the title is a complete sentence ending in a question mark; this will make people more likely to read the full question. $\endgroup$
    – rob
    Feb 9 at 22:23

1 Answer 1


He must tie giant rocket motors to Earth. Then send Earth on a similar trip as he made, but much shorter. By remote control. If he carefully plans he can make Earth return so on Earth only 1 hour has passed, while 24 hours have passed for him. He can stay on Mars for a day. If the people on Earth survive the acceleration remains to be seen. They have to get from zero to near c pretty fast.

He could also drag the Earth to a black hole. Let Earth circle it, with him circling at greater distance, and after 24 hours go to Earth.

So, his time, relative to Earth, speeds up.

  • 1
    $\begingroup$ watch out for the tidal waves ... $\endgroup$ Feb 10 at 2:25
  • $\begingroup$ @AndrewSteane Ha! There will be a nice water trail... $\endgroup$ Feb 10 at 7:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.