Question

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 0.903c, and the other twin travels at 0.470c. (a) According to the theory of special relativity, what is the difference between their ages when they meet again at the earliest possible time?

a) A2 − A1 =

(b) Which twin is older?

Answer 1

Lo= 12 ly
[twin a] La= Lo*sqrt(1-v^2/c^2)= 5.15 ly
[twin b] Lb= Lo*sqrt(1-v^2/c^2)= 10.59 ly

the time to reach the planet:
[twin a] ta=La/va= 5.15ly/0.903c= 5.70 y
[twin b] tb=Lb/vb= 10.59ly/0.470c= 22.53 y

The age of each twin when each arrives at the planet:
[twin a] 19 y + 5.70 y= 24.70 y
[twin b] 19 y + 22.53 y= 41.53 y

Twin1 has to wait for twin 2 to arrive. As seen from Earth
Δt1 = (12 ly)/0.903c = 13.28 years,
Δt2 = (12 ly)/0.470c = 25.53 years.

Twin1 must wait another 25.53 – 13.28 = 12.25 years for Twin2 to arrive.

When Twin2 arrives, twin A has an age of
24.70 + 12.25 =36.95 years.

The difference between their ages when they meet is
41.53 - 36.95 = 4.58