Posted by Anonymous on Friday, April 6, 2012 at 4:32pm.
twin #1:
speed .854c
travel time: 12c/.854c = 14.05 yr
twin #2:
speed: .513c
travel time: 12c/.513c = 23.39 yr
so, twin #1 travels for 14.05 yr and waits 9.34 yr for twin #2 to arrive.
However, time dilation slowed aging for the twins:
twin #1: t' = .520t -- sqrt(1-.854^2)
twin #2: t' = .858t -- sqrt(1-.513^2)
so,
twin #1 aged 14.05*.520 = 7.31 yrs in transit
twin #2 aged 23.29*.858 = 20.07 yrs in transit
twin #1 age: 7.31+9.34 = 16.65 yrs
twin #2 age: 20.07 yrs
difference: 3.4 yrs
I also did not get 4.3 years, though maybe I'm dyslexic.
Anyone see where I went astray?
Lo = 12 light years (ly)
β1 =0.854,
β2 = 0.513
Twin1: L1 =Lo•sqrt(1- (β1)^2) = 12• Lo•sqrt(1- (0.854)^2) = 6.24 ly,
Twin2: L2 =Lo•sqrt(1- (β2)^2) = 12• Lo•sqrt(1- (0.513.)^2) = 10.3 ly.
The time to reach the planet:
Twin1: t1 = L1/v1 = 6.24 ly/0.854c = 7.307 years,
Twin2: t2 = L2/v2 = 10.3 ly/0.513c = 20.08 years.
The age of each twin when each arrives at the planet
Twin1: 19 + 7.307 =26.307 years,
Twin2: 19 + 20.08 =39.08 years.
Twin1 has to wait for twin 2 to arrive. As seen from Earth
Δt1 = (12 ly)/0.854 c = 14.05 years,
Δt2 = (12 ly)/0.513 c = 23.39 years.
Twin1 must wait another 23.39 – 14.05 = 9.34 years for Twin2 to arrive.
When Twin2 arrives, twin A has an age of
24.8 + 9.34 =34.14 years.
The difference between their ages when they meet is
39.08 -34.14 = 4.94 years.
(The answer 4.3 years may be obtained if you take β1 =0.900, β2 = 0.500)