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Homework Help: Spaceship voyage at close to light speeds

  1. Aug 25, 2016 #1
    1. The problem statement, all variables and given/known data
    The last of the human race are leaving the earth, after a total nuclear destruction, to reach the only known planet suitable for lives, 2 million light years away from earth. They are traveling on the spaceship ARK, capable of close to speed of light. There is only enough food and energy to last 10 years for those on board.

    a). How fast must the ship travel so that the human race survive?
    b) If you made any assumption in part a), state it and justify it using your final results.
    c) In the eyes of the ship crew, how far did they travel when they reach the planet?
    d) In your calculation, which reference frame is the “proper frame”?

    Just posting my work to see if anybody can spot any errors with it. Thank you all in advance!
    2. Relevant equations
    Δt = Δt_p*γ

    L = (L_p)/γ

    γ = 1/sqrt(1-β^2 )

    β = v/c

    c = 3 * 10^8 m/s

    L_p = 1.892 *10^16 m

    Δt_p = 3.154 *10^8 s
    3. The attempt at a solution

    a.) since the passengers only have 10 years worth of food, that is by definition the max proper time interval (Δt_p) for them to reach the new planet, as measured by the passengers on the spaceship.

    Δt_p = L/v plugging in the length contraction relation ---> Δt_p = (L_p)/(γ*v)

    v*Δt_p = L_p * sqrt(1-β^2) ...... after some algebra i get

    β = 1/sqrt[ (c*Δt_p/L_p)^2 + 1] .... plugging in the numbers i get

    β = 0.196 ..... hence v = 0.196*c

    b.) since the distance between the earth and the new planet is given to be 2 light years, I assumed that they are at rest with respect to each other. This assumption is used when the passengers measure the time interval Δt_p = L/v. Since the ship is at rest with respect to itself, and the planets are at rest with respect to each other, that means that the planet is approaching the spaceship at the same speed that earth is receding away from the spaceship, speed v.

    c.) L = (L_p)/γ .... plugging in the numbers

    L = 1.86 * 10^16 m

    d.) There is no proper frame of reference, each is equivalent to the other by the first postulate of special relativity. The proper time is measured by the passengers on the spaceship, and the proper length is measured by those at rest on earth with respect to those on the other planet.
  2. jcsd
  3. Aug 25, 2016 #2


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    Your work looks good except that the problem gives a distance of 2 million light years instead of 2 light years.
  4. Aug 25, 2016 #3
    oh woops, thank you for that. Would have sucked to get points taken away for that lol
  5. Aug 25, 2016 #4
    My calculator is giving me β = 1 when I change L_p to be 1.892 * 10^22 m. Am I missing something?
  6. Aug 25, 2016 #5


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    In the equation β = 1/sqrt[ (c*Δt_p/L_p)2 + 1], the quantity (c*Δt_p/L_p)2 is a very small number (but it is not zero). So you can see that β should be very close to 1, but not actually equal to one. Your calculator might not be able to handle the very small numbers.

    But note that you can write β as β = (1 + x)-1/2, where x is the small number (c*Δt_p/L_p)2.
    Do you know how to approximate (1 + x)-1/2 for small x?

    Also, there is no need to convert the distance to meters and the time to seconds. To evaluate the quantity c*Δt_p/L_p you can use distance units of light years and time units of years. What is the value of c in these units?
  7. Aug 25, 2016 #6
    β = 1/sqrt[ (c*Δt_p/L_p)^2 + 1] is approximately 1 + (-1/2)*((c*Δt_p/L_p)^2) for small x = (c*Δt_p/L_p) . and I see what you mean about the units. c = 1 light year/year by definition. funny enough my calculator still gives me β = 1 with this approximation, but I definitely see how its not exactly equal to 1 now. I think its fine to simply say β = 0.99 right?
  8. Aug 25, 2016 #7


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    β is much closer to 1 than 0.99
    Can you figure out how may 9's there are before you get some other digits?
  9. Aug 25, 2016 #8
    1 + (-1/2)*((c*Δt_p/L_p)^2) gives me 1 - 1.25 * 10^-11 after the subtraction is when my calculator gives me β =1 but id assume its more along the lines of β = 0.99999999999
  10. Aug 25, 2016 #9


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    Right, 1 - 1.25 * 10-11.

    Put the calculator aside. You don't quite have the correct number of 9's and you should also be able to include a couple of the other digits beyond the last 9.
  11. Aug 25, 2016 #10
    β = 0.9999999999875? I think I got it right this time
  12. Aug 25, 2016 #11


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    Looks right. Good work! (Actually, I see now that your previous answer with eleven 9's is correct if you round off to the eleventh decimal place.)
  13. Aug 25, 2016 #12
    awesome! thank you very much, you were a great help :)
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