3 ms·
you have to take into account that the closer you get to light speed, the slower time goes. In other words, people on the space ship may age slower, meaning tha
by yohann305 13y ago
you have to take into account that the closer you get to light speed, the slower time goes. In other words, people on the space ship may age slower, meaning that 220 years may be a lot less to them. What is the equivalent of 1 earth year on a spaceship going at 0.1c, that is the question.
- gus_massa 13y agoThe relativistic correction is Time_on_Ship = Time_on_Earth * (1 - (v/c)^2)^(1/2) If v <<c we can use the Taylor approximation Time_on_Ship =~ Time_on_Earth * (1 - (1/2) (v/c)^2) Distributing Time_on_Ship =~ Time_on_Earth - Time_on_Earth (1/2) (v/c)^2 With the current number, assuming instant acceleration to simplify the calculations: Time_on_Earth = (22/.1) Time difference =~ (22/.1) (1/2) * .1^2 years = 1.10 years The important detail is that for small speeds, the correction is quadratic, so the difference is small unless the speed is very close to c. In this case the correction is only 0.5%.
- yohann305 13y agoGood stuff! Thx Gus
- johngalt 13y agoThe Lorentz factor doesn't really make an appreciable difference until you get much closer to c. For instance cutting the trip time in half would require .866c. .99c would cut the trip down to ~30years.
- shawabawa3 13y ago> .99c would cut the trip down to ~30years. Surely you forgot to take into account that increasing the speed shortens the trip as well as dilating time? Travelling at 0.99c would take 22 years to reach that from an observer's point of view, which would be 3 years for the people on board
- johngalt 13y agoYou're completely correct. I didn't take that into account. No more comments about relativity before coffee.