Saturday
March 25, 2017

Post a New Question

Posted by on .

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 - ,

    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)

  • Physics help please - ,

    Lo= 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

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question