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April 19, 2014

April 19, 2014

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

(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 10:31amLo = 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)

- Physics help please -
**edna luz**, Tuesday, July 16, 2013 at 1:43amLo= 12 ly

[twin a] La= Lo*sqrt(1-v^2/c^2)= 5.23 ly

[twin b] Lb= Lo*sqrt(1-v^2/c^2)= 10.39 ly

the time to reach the planet:

[twin a] ta=La/va= 5.23ly/0.9c= 5.81 yr.

[twin b] tb=Lb/vb= 10.39ly/0.5c= 20.78 y

The age of each twin when each arrives at the planet:

[twin a] 19 y + 5.81 y= 24.81 y

[twin b] 19 y + 20.78 y= 39.78 y

Twin1 has to wait for twin 2 to arrive. As seen from Earth

Δt1 = (12 ly)/0.9c = 13.3 years,

Δt2 = (12 ly)/0.5c = 24 years.

Twin1 must wait another 24– 13.3 = 10.7 years for Twin2 to arrive.

When Twin2 arrives, twin A has an age of

24.81 + 10.7 =35.51 years.

The difference between their ages when they meet is

39.78 -35.51 = 4.27 years or 4.3 years

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