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Twins Adam and Becky, who are 19.0 years of age, leave Earth and travel to a distant planet 12.0 light years away. Assume that the planet and earth are at rest to each other. The twins depart at the same time on different spaceships. Adam travels at a speed of 0.831c, and Becky travels at 0.523c.   When they next meet, who is older and by how much?
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Added Mon, 20 Jul '15
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Let:
d = distance to planet (= 12.0 light years)
Va = Adam's speed (0.831c)
Vb = Becky's speed (0.523c)

Clearly Adam arrives first (because he's going faster). From earth's point of view, it takes Adam this long to get there:

Ta = d/Va

But Adam's time is dilated, so the amount of travel time Adam experiences ("Ta-prime") is:

Ta? = Ta?(1?(Va/c)?)
= d?(1?(Va/c)?)/Va

So when Adam arrives, he is this old:

19.0 yrs + d?(1?(Va/c)?)/Va

But then he has to wait for Becky. Since he's now at rest with respect to earth, the "wait time" he experiences is just the same as the difference in arrival times as measured from earth. That is:

wait time = Tb ? Ta = d/Vb ? d/Va

So, by the time Becky gets there, Adam is this old:

Adam's age = 19.0 yrs + d?(1?(Va/c)?)/Va + d/Vb ? d/Va

Now for Becky's age:

By reasoning similar to the above, the amount of travel time that Becky experiences is:

Tb? = d?(1?(Vb/c)?)/Vb

So, by the time Becky arrives, she is this old:

Becky's age = 19.0 yrs + d?(1?(Vb/c)?)/Vb

That's it: Now just get out your calculator and plug in the numbers for Adam's Age and Becky's Age.
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Added Mon, 20 Jul '15
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