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Distance traveled, v->c

  1. Feb 1, 2006 #1


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    I have this equation here and I don't see how I can isolate v, the speed of the ship...

    [tex]\frac{{2(4ly)(9.5x10^{15} \frac{m}{{ly}}}}{v} = \frac{{16y}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}[/tex]

    y = years
    ly = light years

    The book doesn't tell you how it got v=0.447c, it just jumped to the conclusion after presenting the equation. It seems like no matter how I try the equation, the v-squared will always cancel and i'll lose my velocity equation. Am I not doing this correctly or is there a more complex method of solving for v?
  2. jcsd
  3. Feb 1, 2006 #2
    I will do some cleaning up:

    [tex] a = 2(4ly)(9.5x10^{15}) [/tex]

    [tex] b=16y [/tex]

    so we write it much cleaner as:

    [tex] \frac{a}{v} = \frac{b}{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} } [/tex]

    Now I just multiply by the denominator of the RHS and get:

    [tex] \frac{a}{v} [\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }] = b [/tex]

    Now I just bring the v inside the square root:

    [tex] a[\sqrt {\frac{1}{v^2} - \frac{{1}}{{c^2 }}}}] = b [/tex]

    No v's cancel out.

    You can simply it to get,

    [tex] ( \frac{1}{v^2} - \frac{1}{c^2}) = \frac{b^2}{a^2} [/tex]

    [tex] \frac{1}{v^2} = \frac{b^2}{a^2} + \frac{1}{c^2} [/tex]

    invert both sides and take square root:

    [tex] v = +/- \sqrt{(\frac{b^2}{a^2} + \frac{1}{c^2})^{-1}} [/tex]
    Last edited: Feb 1, 2006
  4. Feb 1, 2006 #3


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    Just from eying the equation, wouldn't squaring both sides, then multiplying both sides by v²(1-(v/c)²) give you a quadratic equation in v²? the uniqueness of the v would then be due to the fact that one of the root is negative, so not a physically acceptable solution.
  5. Feb 1, 2006 #4


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    Ok I have another problem where I can't seem to isolate v correctly... A rocket will go to a galaxy 20 light years away and it can only take 40 years round trip according to the rocket passengers. I figured this is how i could write it.

    T' = \frac{{2L}}{v} = \frac{{T_0 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }} \\
    \sqrt {1 - \frac{{v^2 }}{{c^2 }}} *2L = T_0 *v \\
    \sqrt {4l^2 - \frac{{4l^2 v^2 }}{{c^2 }}} = T_0 v \\
    4l^2 - \frac{{4l^2 v^2 }}{{c^2 }} = T^2 v^2 \\
    4l^2 (1 - \frac{{v^2 }}{{c^2 }}) = T^2 v^2 \\
    \frac{1}{{v^2 }} - \frac{1}{{c^2 }} = \frac{{T_0 ^2 }}{{4l^2 }} \\

    Now im not sure how to isolate v
  6. Feb 1, 2006 #5
    Add 1/c^2 to both sides, take the inverse, then take the square root, exactly as before.
  7. Feb 1, 2006 #6


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    How do i do the inverse of two added quantities though?
  8. Feb 1, 2006 #7
    make them a common fraction and do the addition. Then take the inverse of that new fraction you get.
  9. Feb 1, 2006 #8


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    How would i do that.... man i can't do basic math!!!
  10. Feb 1, 2006 #9


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    Do you do this?

    [tex]v^2 = \frac{{4l^2 + c^2 }}{{T_0 }}
  11. Feb 1, 2006 #10
    In your latter example you would get:

    [tex] \frac{T^2}{4L^2} + \frac{1}{c^2} [/tex]

    [tex] \frac{c^2T^2}{4L^2c^2}+ \frac{4L^2}{4L^2c^2} [/tex]

    [tex] \frac{c^2T^2+4L^2}{4L^2c^2} [/tex]

    The inverse would be:

    [tex] \frac{4L^2c^2}{c^2T^2+4L^2} [/tex]
    Last edited: Feb 1, 2006
  12. Feb 1, 2006 #11


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    Dang that's insane... thats from like 5th grade :( and i forgot!
  13. Feb 1, 2006 #12


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    Ok and since I know the space ships speed to be 0.894c.... does that mean the time measured from the stationary observers on Earth is ~2.236*40 years using gamma*T proper?
  14. Feb 1, 2006 #13
    Sorry, im not a physics major. Thats as much help as I can provide.
    Last edited: Feb 1, 2006
  15. Feb 1, 2006 #14


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    You are great up to this point but remember you want to solve for v- you want to isolate it, not mix it with other things.
    [tex]4l^2= T^2v^2- \frac{4l^2}{c^2}v^2= (T^2- \frac{4l^2}{c^2})v^2[/tex]
    [tex]v^2= \frac{4l^2}{T^2- \frac{4l^2}{c^2}}[/tex]
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