Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    0
    ReputationRep:
    A team of astronauts set off from a stationary space station at one quarter of the speed of light. They are carrying a highly accurate atomic clock. They travel to another space station, stationary with respect to the first one , and which is one quarter of a light year away from the first one. They immediately turn round and travel back to the original space station, at half the speed of light. How much time has elapsed on the atomic clock ? How does this compare to the time elapsed on the space station ?

    Therefore I know for the first journey v=0.25c x=0.25 light years.
    The return journey is v=0.5c and x=0.25 light years.

    I know I have to find the time but Im confused as to which formula to use please help guys exam in afternoon!
    • Thread Starter
    Offline

    0
    ReputationRep:
    Anyone?
    Offline

    8
    ReputationRep:
    Special Relativity states:

     t_{actual} = \dfrac{t_{experienced}}{\sqrt{1-\frac{v^2}{c^2}}}

    Now then for the first part, the time it takes is exactly one year, so:

     t_{actual} = 3.15 \times 10^7

    Put the numbers into the equation:

     t_{experienced} = 3.15 \times 10^7 *\sqrt{1-\frac{0.25^2}{1^2}} = 3.05 \times 10^7 s

    For the second part of the journey, the time it takes is half a year, so:

     t_{actual} = 1.58 \times 10^7

    Again into the equation:

     t_{experienced} = 1.58 \times 10^7 *\sqrt{1-\frac{0.5^2}{1^2}} = 1.37 \times 10^7 s

    Overall time =
     4.42 \times 10^7 \text{seconds}
    • Thread Starter
    Offline

    0
    ReputationRep:
    Is that overtime for the atomic clock?
    (Original post by Piguy)
    Special Relativity states:

     t_{actual} = \dfrac{t_{experienced}}{\sqrt{1-\frac{v^2}{c^2}}}

    Now then for the first part, the time it takes is exactly one year, so:

     t_{actual} = 3.15 \times 10^7

    Put the numbers into the equation:

     t_{experienced} = 3.15 \times 10^7 *\sqrt{1-\frac{0.25^2}{1^2}} = 3.05 \times 10^7 s

    For the second part of the journey, the time it takes is half a year, so:

     t_{actual} = 1.58 \times 10^7

    Again into the equation:

     t_{experienced} = 1.58 \times 10^7 *\sqrt{1-\frac{0.5^2}{1^2}} = 1.37 \times 10^7 s

    Overall time =
     4.42 \times 10^7 \text{seconds}
    • Thread Starter
    Offline

    0
    ReputationRep:
    Also how did you the the T actual value?
    (Original post by Piguy)
    Special Relativity states:

     t_{actual} = \dfrac{t_{experienced}}{\sqrt{1-\frac{v^2}{c^2}}}

    Now then for the first part, the time it takes is exactly one year, so:

     t_{actual} = 3.15 \times 10^7

    Put the numbers into the equation:

     t_{experienced} = 3.15 \times 10^7 *\sqrt{1-\frac{0.25^2}{1^2}} = 3.05 \times 10^7 s

    For the second part of the journey, the time it takes is half a year, so:

     t_{actual} = 1.58 \times 10^7

    Again into the equation:

     t_{experienced} = 1.58 \times 10^7 *\sqrt{1-\frac{0.5^2}{1^2}} = 1.37 \times 10^7 s

    Overall time =
     4.42 \times 10^7 \text{seconds}
    Offline

    8
    ReputationRep:
    (Original post by mathslover786)
    Also how did you the the T actual value?
    Speed = Distance * Time,
    you can just cancel out the speed of light parts, so you end up with a year, and half a year respectively...
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Brussels sprouts
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.