Age comparison to observer on Earth (time dilation)

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Homework Statement



Alpha Centauri is about 4 light years from the Earth. If you were to travel to Alpha Centauri and back at the speed of the electron that you calculated in question A (0.780c), how much would you age compared to an observer on Earth?

Homework Equations



There equations were not provided in the actual problem, but so far we've used these 2 (time dilation and Lorentz contraction) equations:

T' = (T) / sqrt[1 - (v^2/c^2) ]

L' = L * sqrt[1 - (v^2/c^2) ]

The Attempt at a Solution



v = 0.780c

Both ways:
L = 8cY

L = v*T

T = L / v

T = (8cY) / (0.780c)

T = 10.3Y

T' = ?

T' = (T) / sqrt[1 - (v^2/c^2) ]

T' = (T) / sqrt[1 - (v/c)^2 ]

T' = (T) / sqrt[1 - (0.780c/c)^2 ]

T' = (T) / sqrt[1 - (0.780)^2 ]

T' = (T) / sqrt[1 - (0.608) ]

T' = (T) / sqrt(0.392)

T' = (T) / sqrt(0.392)

T' = (T) / (0.626)

T' = (10.3Y) / (0.626)

T' = 16.5Y

From what I recall, T' is the change caused by motion of the moving object, relative to the observer.
In this case, relative to the person on Earth, the travel time to Alhpa Centauri and back would be 16.5 years.

There only other time variable left is T, so I assume this applies to the person/partice traveling to Alpha Centauri.

Am I correct? (I want to avoid using Lorentz contraction to ensure I understand the time dilation equation and use it properly.)

Thanks.
 
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Less time passes for the guy in the spaceship. Remember the guy in Heinlein's story who comes home and marries his great-great-grand niece? (Time For The Stars)
The Earth observer says 10.3 years.
The traveler says 10.3 * .626
(ignoring general relativistic effects during the accelerations)
 
OK. Doesn't that mean I did it right so far? Don't I have half of the answer (the time for the observer on Earth)?