- #1
09jlk
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Homework Statement
A space traveler takes off from Earth and moves at speed 0.99c toward star Vega, which is 26.00 ly distant. How much time will have elapsed by Earth clocks when the traveler reaches Vega?
2. The attempt at a solution
I looked at the problem solution, which demonstrated this in a much simpler way -- but I want to make sure that I can understand the process of doing it the longer way.
I thought to use the Lorentz transformation equation:
∆t = [∆t' + (v∆x')/c^2)] * 1/[sqrt(1-(v/c)2]
where ∆t is the time as experienced by Earth and ∆t' is the time experienced by the traveler.
In order to find ∆t' I first found the new distance between Earth and Vega as experienced by the traveler according to length contraction, and then divided by the speed of the traveler.
I'll use D° for the proper distance (26 ly) and D for the adjusted distance:
D = D°/gamma = 2.459 x 10^17 meters * sqrt(1-(v/c)2)
D = 3.4699 x 1016 meters.
So now I can find my ∆t' based on this distance and the given velocity:
∆t' = D/v = 3.4699x1016/(.99c) = 1.168 x 108 seconds, which is 3.70 years.
Now, because I have ∆x' (the contracted distance) and ∆t', I should be able to solve for ∆t based on the equation from above:
∆t = [∆t' + (v∆x')/c^2)] * 1/[sqrt(1-(v/c)2]
∆t = (1.168 x 108 + [(0.99c)(3.46989x1016)]/c2)*gamma
For the second term, I calculated 1.145 x 108 seconds, added that to my ∆t' and then multipled by the Lorentz factor 1/sqrt(1-(v/c)2) and ended up getting 51.9 years -- the answer should be 26.26 years.
I'm not totally sure where I went wrong -- maybe it was my algebra/calculator but I assume it's something else. It seems that my ∆t' is correct because I got the second part of this question right (not included). Any thoughts?