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Basic SR question?

  1. Feb 8, 2008 #1
    1. The problem statement, all variables and given/known data

    a ufo travels from one planet to another (distance measured 384000km) at v=0.8c.

    1)How long does the trip take according to observer on planet left?

    2)How long does the trip take according to observer in ufo?

    3) What is the two planets distance measusured my aliens in the ufo?

    2. Relevant equations

    I think I have to use the lorrentx transformations.

    t'=alpha(u/c)(t-ux/c^2)

    t=alpha(u/c)(t'+ux'/c^2)

    x=alpha(u/c)(x'+u't)


    3. The attempt at a solution

    This is my first ever SR question I have done. So I wasnt sure.

    I thought that from the original planet the apparent velocity is going to be 0?

    so i said the time takes was = 1.83 s?

    So then i said using the above formula that alien man thought it would be = 2.69s

    As this is wrong I havnt proceeded to the next part. Can you please tell me where i am going wrong and possibly give me some advice as to how to set up these types of questions.

    thanks
     
  2. jcsd
  3. Feb 9, 2008 #2
    come on surely someone can help me with this???
     
  4. Feb 9, 2008 #3
    can someone tell me if im posting this in the wrong place at least lol?
     
  5. Feb 9, 2008 #4
    I can try to help. First, you should convert the distance in km to s, i.e. 384,000 km = ?s.
    I used [tex]c = 3 * 10^8[/tex] and got d = 1.28s.

    Now, I know that the aliens travelling in the ufo will have the shortest time. I used the spacetime equation

    [tex]\Delta s^2 = \Delta t^2 - \Delta d^2 [/tex]

    in this particular case, d = 0, (this has to do with the distance the 'clock' is from the person (or in this case alien) using the 'clock') I calculated t as distance/velocity so t = 1.6s (this is what I got for part B or 2 of your problem). Do you happen to know if this is the correct answer? If this is right, I may be able to get you to the first part.

    Also, let me know if you're using a different value for c (such as 2.998).
     
  6. Feb 9, 2008 #5

    Doc Al

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    Staff: Mentor

    You are told that the ufo travels at 0.8 c.

    How did you arrive at this number?
     
  7. Feb 9, 2008 #6
    kk so I know that speed =distance/time

    so time = distance/speed = 4.8 seconds. Is that right?
     
  8. Feb 9, 2008 #7

    Doc Al

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    Staff: Mentor

    Good.

    No. What's the distance? What's the speed?
     
  9. Feb 9, 2008 #8
    No. What's the distance? What's the speed?

    the distacne is 384000000m
    and speed is 0.8*3ee9

    when i put this into my calulator i get 1.6
     
    Last edited: Feb 9, 2008
  10. Feb 9, 2008 #9

    Doc Al

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    Don't worry about the time according to the ufo until you've nailed down the time according to the planet.
     
  11. Feb 9, 2008 #10
    and for the other time i get 2.67s is that right?
     
  12. Feb 9, 2008 #11

    Doc Al

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    That's better.

    Will the time for the trip as measured by the ufo be shorter or longer? By what factor? (Hint: The ufo can be treated as a single moving clock.)

    Do moving clocks run slow or fast?
     
    Last edited: Feb 9, 2008
  13. Feb 9, 2008 #12
    Simply from the twins paradox thing im gonna say that the time is shorter for the moving ufo.
    So the time will be 1.25s right?
     
  14. Feb 9, 2008 #13

    Doc Al

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    What you need is the formula for the time dilation of a moving clock. But yes, the moving ufo clock runs slow (according to the planet clocks)--so the time measured by the ufo will be shorter than that measured by the planet.
    How did you get this number?
     
  15. Feb 9, 2008 #14
    I plugged numbers into a fomula in my book.
     
  16. Feb 9, 2008 #15

    Doc Al

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    What formula?
     
  17. Feb 9, 2008 #16
    t' = (t-ux/c^2)/root(1+u^2/c^2)


    is it wrong?
     
  18. Feb 10, 2008 #17

    Doc Al

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    That's one of the Lorentz transforms, but with a typo. Here's the corrected version:
    [tex]\Delta t' = \gamma(\Delta t - v\Delta x/c^2)[/tex]

    Where:
    [tex]\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}[/tex]

    But even simpler is to recognize that moving clocks run slow by a factor of [itex]\gamma[/itex]:
    [tex]\Delta T = \gamma \Delta T_0[/tex]

    (Either way will work.)
     
  19. Feb 10, 2008 #18
    but putting all the numbers into the formula you gave me gives 2.67s? what am i doing wrong?
     
  20. Feb 10, 2008 #19

    Doc Al

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    I gave two formulas. Which one are you using? (Not that it matters--both will give the same answer.)

    Looks to me like you are somehow reversing the primed and unprimed coordinates. What are you plugging in for [itex]\Delta t[/itex] and [itex]\Delta x[/itex]? What do you get for [itex]\gamma[/itex]?
     
  21. Feb 10, 2008 #20
    i get 5/3 for lambda. is that wrong?
     
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