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Velocity Transforms

  1. May 21, 2009 #1
    1. The problem statement, all variables and given/known data
    A spaceship moves away from a point at speed v and launches a smaller ship at speed v relative to the first ship. The second ship shoots a missile at speed v relative to the second ship
    1. what is the speed of the second ship relative to Earth
    2. what is the speed of the missile relative to Earth

    2. Relevant equations


    3. The attempt at a solution
    I assume that in order to get the speed of the first ship relative to Earth i just use the velocity transform u=(u'+v)/(1+[vu'/c^2]), which gives u= (v+v)/(1+[v^v/c^2]) or u= 2v/(1+[v^2/c^2])

    Is this correct?
    If so, to find the speed of the missile relative to earth would i again use the same equation using the 2v/(1+[v^2/c^2]) as u'? Would this give the answer to part 2?
    Im confused.
     
    Last edited: May 21, 2009
  2. jcsd
  3. May 21, 2009 #2

    CompuChip

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    I think that's what you are supposed to do, yes.
     
  4. May 21, 2009 #3
    hi can i ask, since i am doing same question.. i understand part a.. just simple transforms, but what did u mean by using u=2v/(1+v^2/c^2) for b?? then the answer for b would be.... solving for u'???
     
  5. May 21, 2009 #4
    Well my thinking was that because the speeds are not a number value (they are given as 'v'), to find the answer for part 1 you must find u and simplify. This gives u= 2v/(1+[v^2/c^2]).
    For the second part the u that was found in 1. becomes u' in part 2. (Sorry i forgot to add the ' to the u)

    Now im just confusing myself and im not sure if it is right or not.
     
  6. May 21, 2009 #5
    ya thats wat i was thinking... btu when u do that calculation, u get a weird answer..... but anyways thx for ur help
     
  7. May 21, 2009 #6

    CompuChip

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    Since you both seem to be on the right track, the answer I get (for you to compare to) is

    [tex]\frac{ 3 v + (v^2 / c^2) v }{ 1 + 3 (v^2 / c^2) }[/tex]

    Yes, it looks a little weird.

    Writing it in terms of [itex]\beta = v / c[/itex] gives the still a bit weirdly looking

    [tex]\frac{ \beta(3 + \beta^2) }{ 1 + 3 \beta^2 } c[/tex]
     
  8. May 21, 2009 #7
    Thanks CompuChip. but how did you derive that answer? I tried doing it myself but got really lost.
     
  9. May 21, 2009 #8

    CompuChip

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    I just plugged u = 2v/(1+v^2/c^2) back into the addition formula and simplified it (ok, maybe I cheated a bit by letting the computer simplify it).

    Where did you get stuck?
     
  10. May 21, 2009 #9
    I got stuck in the algebra. After plugging it into the formula as you did.
     
  11. May 21, 2009 #10
    sorry didnt see the c.... = (
     
  12. May 22, 2009 #11

    chiro

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    Formulas in relativity are related by using hyperbolic tangent identities.

    That is to say that we relate

    tanh(v_final/c) = tanh(u/c) + tanh(v/c)

    where v_final is the final velocity and u and v are the initial velocities. Note that the velocity addition formula means that everything is relative to c so that we can not travel faster than light but only relative to it. Think more or less like the speed of light being some kind of infinity where we approach but never reach c. This should help you hopefully understand this relationship.
     
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