Homework Help: Physics help please

Posted by Jim on Friday, April 6, 2012 at 1:18pm.

Twins who are 19.0 years of age leave the earth and travel to a distant planet 12.0 light-years away. Assume that the planet and earth are at rest with respect to each other. The twins depart at the same time on different spaceships. One twin travels at a speed of 0.854c, and the other twin travels at 0.513c.

(a) According to the theory of special relativity, what is the difference between their ages when they meet again at the earliest possible time?

The answer is 4.3 years, but I'm not sure how they arrived at that...

Physics help please - Elena, Saturday, April 7, 2012 at 10:31am
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)

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