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Relativistic kinematics.

  1. Aug 4, 2007 #1
    An observer sees two spaceships flying apart with speed 0.99c, what is the speed of one spaceship as viewed by the other? the answer is: 0.99995c.

    well we have the equation of relative velocity:
    [tex]\frac{v_1+v_2}{1+\frac{v_1v_2}{c^2}}=0.99c[/tex]
    now i think from the question that i need to find v1+v2 here, cause compared to spaceship 1, spaceship two is moving with speed -(v1+v2) and spaceship 1 compared to two is moving with speed v1+v2.

    but how to find it, perhaps i need some algebraic manipulation which i don't see how do, or my above equations are flawed?
     
  2. jcsd
  3. Aug 4, 2007 #2
    If I understand the situation correctly, you're seeing two spaceships moving exactly away from you with 0,99 c in opposite directions. Now you have to use the last equation on the following page: http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html with ux=0.99c and vx=-.99c o yield for the relative velocity (as viewed by either of the two spaceships:

    [tex]w = \frac{|ux - vx| }{ 1 - ux vx/c^2}=\frac{2*0.99}{1-0.99^2}=0.99995[/tex]
     
  4. Aug 4, 2007 #3
    it should be 1+0.99^2 in the denominator.
    thanks, i didn't undersatnd the situation, i thought that the observer measures the relative speed between them, and that speed is 0.99c.
     
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