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Scott Kelly just returned from nearly a year of orbiting the Earth in the space station. How much younger is he than his twin, now that he has spent a significant amount of time, traveling at high speeds?

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  • $\begingroup$ Nice question, thanks. +1 for "a significant amount of time traveling" ;) $\endgroup$ Mar 3, 2016 at 9:30
  • $\begingroup$ Do you know the exact time difference between him and his twin on the scale where we could add / remove the time dilation? Cause I would assume that the difference in time is like maybe 1 or 2 seconds. this probably won't change anything on the fact the first born twin keeps staying older :P $\endgroup$
    – Zaibis
    Mar 3, 2016 at 11:26
  • $\begingroup$ Also wasn't his twin brother in space aswell? $\endgroup$
    – Zaibis
    Mar 3, 2016 at 11:28
  • $\begingroup$ Not much younger, but temporarily two inches taller. $\endgroup$ Mar 3, 2016 at 13:21

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Neil DeGrasse Tyson says that Scott Kelly is now 1/100 second younger than he would have been otherwise, which almost certainly isn't enough to alter the birth order of the two; I don't know which of the two was born first.

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    $\begingroup$ Yes, i'd say bearing a second child 1/100th of a second after the first almost certainly can't be done without significant risk to the baby. $\endgroup$
    – kim holder
    Mar 2, 2016 at 17:35
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    $\begingroup$ You say things like that and some people are just going to take it as a challenge. $\endgroup$ Mar 2, 2016 at 17:35
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    $\begingroup$ Scott is most likely considerably older physiologically than is his brother. A year in space takes a significant toll. Understanding that is a big part of the reason for this twin study. $\endgroup$ Mar 2, 2016 at 18:52
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    $\begingroup$ @kimholder Assuming an average length of 20 inches, the first baby would have to be delivered at around 183 kilometers per hour (or 114 miles per hour) to be only 1/100th of a second older. $\endgroup$
    – Nathan K
    Mar 2, 2016 at 21:49
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    $\begingroup$ That's assuming a spherical baby of uniform density, right? $\endgroup$ Mar 2, 2016 at 23:38

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