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Vector drive - faster than light?

  1. Oct 13, 2008 #1
    If a spaceship is traveling at 90% the speed of light on the X axis. And it is also traveling 90% the speed of light on the Y axis. What is its velocity?
     
  2. jcsd
  3. Oct 13, 2008 #2

    Vanadium 50

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    1.27c. But your supposition doesn't describe the behavior of a physical spaceship.
     
  4. Oct 13, 2008 #3

    DaveC426913

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    The solution to the apparent paradox is to model how it could get to those speeds. The ultimate answer is: it couldn't.

    The reason is that speeds near C are not summed. They are the result of the formula 1/(v1 = 1-(v0^2/c^2)^.5).
     
  5. Oct 13, 2008 #4
    Morgan (l982) estimated that the exhaust velocity of a pion- relecting matter/antimatter rocket could be in excess of 0.9c. If the spaceship is a sphere, and one antimatter rocket is aimed in the X axis, and one antimatter at 90 degrees to X 1/4 the way across the sphere is aimed in the Y axis, and both push to 0.9c, is what you say still true?
     
  6. Oct 13, 2008 #5

    ZapperZ

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    You need a better and more complete reference than that!

    Zz.
     
  7. Oct 13, 2008 #6
  8. Oct 13, 2008 #7
  9. Oct 13, 2008 #8

    Dale

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    Hi android34, welcome to PF

    This is actually not very difficult to approximate using the four-momentum.

    We will start with a rocket with 1.0 kg payload/vehicle and 2.0 kg fuel (3.0 kg total). We will consider the entire payload of fuel to be burnt instantaneously with 100% efficiency at an exhaust velocity of 0.9 (using units where c=1). We will consider two cases, the first where all of the exhaust goes in the -x direction and the second where half goes 45º above the -x axis and half goes 45º below. We will simply use the conservation of four-momentum before and after the burn to determine the speed of the rocket.

    Case 1:

    3.0 (1,0,0,0) = 1.0 (γ,γv,0,0) + m (2.29,-2.06,0,0)
    eliminating m and solving for v
    v = 0.78

    Case 2:

    3.0 (1,0,0,0) = 1.0 (γ,γv,0,0) + m (2.29,-1.46,1.46,0) + m (2.29,-1.46,-1.46,0)
    eliminating m and solving for v
    v = 0.71

    So you will go faster by sending your exhaust off in the opposite direction you want to go instead of splitting it up on two orthogonal axes.
     
  10. Oct 13, 2008 #9

    DaveC426913

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    Propulsion systems do not "push to [a velocity]"; they provide (a force which results in) an acceleration.

    If you accelerate for time X to reach v=.9c, and then accelerate for time X again, your final velocity will only be about .99c.

    Yes, it is still true.
     
    Last edited: Oct 13, 2008
  11. Oct 13, 2008 #10

    DaveC426913

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    The choice of propulsion is largely academic, so lack of a reference is not a show-stopper. Any propulsion that can (theoretically) accelerate a craft to .9c will do. There are a few well-known ones that are unlikely to be contested.
     
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