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A problem from science fiction - intersecting curves

  1. May 6, 2013 #1
    1. The problem statement, all variables and given/known data[/b]

    A spaceship is traveling on a curved path, f(t) = (t, t2)

    (We'll assume that the path isn't affected by gravity, this is a math problem, not physics :-) )

    It has to release a pod to intersect a space station that has an orbit described by the following:

    g(t) = (4+cos([itex]\frac{πt}{8})[/itex], 8-sin([itex]\frac{πt}{8}[/itex]))

    The pod will travel at a tangent from the curved path f(t) and I want to know at what time s I should release it to get to the space station and how long it will take.

    3. The attempt at a solution

    OK, so I know that since I am releasing the pod at time s, my first vector to add will be (s, s2). And those numbers will be constant.

    I know that f'(t) = (1, 2t)

    So that means that a pod will be zooming off the curve in a path described by (s+t, s2+2t). (correct?)

    So to see if the pod will hit the space station and when, I need

    (s+t, s2+2t) = (4+cos([itex]\frac{π(s+t)}{8})[/itex], 8-sin([itex]\frac{π(s+t)}{8}[/itex]))


    But after that I am stuck, and I feel like there is some simple step I am missing. I know that the time s of release can't be more than s=3 and less than 0 because if s=0 the line is horizontal and never hits the circle described, and if s≥3 it will miss the circle full stop. So there are some constraints here. I am almost there, I feel like.
     
  2. jcsd
  3. May 6, 2013 #2

    Simon Bridge

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    Not quite.

    For the spacecraft, v(t)=(1,2t)
    At time t=s, the speed of the spacecraft, and thus the pod, will be... ?
    ... and thus, the position-time equation of the pod will be... ?
     
  4. May 6, 2013 #3
    A time t=s the speed of the spacecraft is f'(s) or (1,2s) so that's the speed of the pod.

    The position of the pod would have to be (s+t, s2+(2s)t) after that, yes?

    So you'd want (s+t, s2+(2s)t) = (4+cos[itex]\frac{(s+t)π}{8}[/itex], 8-sin[itex]\frac{(s+t)π}{8}[/itex])

    That's still pretty ugly when you want to solve it, no?
     
  5. May 6, 2013 #4

    Simon Bridge

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    It's usually nastier than that because the pod will be travelling along a conic section as well.
    And then - real orbits are seldom that tidy.

    You want to solve for s - you'll find there are lots of solutions from the restraints you have specified.
     
  6. May 6, 2013 #5
    OK, but when I try to do that I get s = 4-t+cos([itex]\frac{(t+s)π}{8}[/itex])

    And I still have a lot of possibilities for both s and t. I do know that -1<s+t-4<1, b/c of the cosine function. And I know that s>0, and t>0 so for this to work 3< s+t < 5. Am I getting it right so far?

    Further, s2+2st = 8-sin([itex]\frac{(t+s)π}{8}[/itex])

    which gets me s(1+2t) = 8-sin([itex]\frac{(t+s)π}{8}[/itex])

    or 8-s(1+2t)=sin([itex]\frac{(t+s)π}{8}[/itex])

    since -1<8-s(1+2t) <1 7<s(1+2t)<9, yes?
     
  7. May 6, 2013 #6

    Simon Bridge

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    Yes you do - why would you expect any different?

    You are using t as the elapsed time between release and capture.
    You should find an equation for this time in terms of the release time and the speed of the pod.
     
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