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Circular-elliptic orbit

  1. Jan 16, 2009 #1
    1. The problem statement, all variables and given/known data
    A particle in circular orbit moves in a force field given by F(r)=-k/r^2. If k is halved what will the new orbit be?


    2. Relevant equations
    The answer is the orbit becomes parabolic but how? how does the total energy become zero as for parabolic orbits total energy=0? help please


    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 17, 2009 #2

    D H

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    The particle was in a circular orbit in an inverse square central force system, so you should be able to calculate the particle's pre-halving orbital velocity. Now assume that the particle's momentum remains unchanged when k is halved to get the final answer.
     
  4. Jan 17, 2009 #3
    I equated the given force to the centrifugal force as the orbit is circular. Then i get the orbital velocity. Now after halving k and using the fact that angular momentum is constant in central force motion am still unable to get the final answer. Help me see what am missing. How do I get equation of orbit as that of parabola??
     
  5. Jan 17, 2009 #4

    D H

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    You suggested in your original post an approach you didn't use in your second post: Energy. The total energy for a parabolic trajectory is zero. Beyond that, I find it rather difficult to comment since you have not shown your work.

    In the future, please follow our template when asking for homework help and show your work in the form of mathematical expressions.
     
  6. Jan 18, 2009 #5
    I tried like this: when k is halved, potential energy goes from -k/r to -k/2r (an
    increase of k/2r) so that total energy is also increased by k/2r
    (kinetic energy is unaffected). total energy before k halving K was
    -k/2r. So it is now 0. As total energy goes to zero tha path is parabolic. How correct am I??
     
  7. Jan 18, 2009 #6

    D H

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    There you go. There is no need to use angular momentum at all for this problem.
     
  8. Jan 18, 2009 #7
    Yes indeed, all it needed was a couple of bottles of beer to me,lol. Here's another problem am working on: Two particles moving under the influence of their mutual gravitational force describe circular orbits about one another with a period of T. If they are suddenly stopped and allowed to gravitate each other, show that they will collide with each other after a time T/4squareroot2. I believe that when the particles are stopped the angular momentum goes to zero and the head towards each other for collision. Maybe some energy conservation should lead me to the result. I will cry for help if I ciould not get it done by myself though hints are always welcome.
     
  9. Jan 18, 2009 #8

    D H

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    The resulting integral is doable but is rather nasty. A much easier approach: How long is it until closest approach if you cancel some, but not all, of the angular momentum? What happens when in the limit that angular momentum goes to zero?
     
  10. Jan 24, 2009 #9
    I am now working on the fourier transforms. Please help me how to find the fourier transform of 1/(x square+ 4x + 13). The limit is minus negative to positive infinity so when I applied the formula the limit makes it all zero, which i do not think is the right answer. Shed some light where i made the mistake.
     
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