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Math Help - potential energy function

  1. #1
    C.E
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    potential energy function

    Hi, I was hoping for some help with the following. This is the first question of this type I have attempted so, any help either with answering the parts of the question I could not do or if you notice any mistakes with the parts I have answered would be appreciated.

    A proton of mass m moves in one dimension. Its potential energy function is: U(x)=a/x^2 - b/x where a and b are positive constants. The proton is released from rest at x0=a/b.
    I also showed in earlier questions that that U(x) can be written in the following way:

    U(x)=(a/(x0)^2)[(x0/x)^2-(x0/x)] and that v(x) the speed of the proton as a function of position can be written as:

    V(x)=sqrt((2a/m(x0)^2)[(x0/x)-(x0/x)^2]) with x>x0

    a. Give a short qualitative description of the motion in terms of the classification of Kepler orbits.

    I don't know what a kepler orbit is and my textbook (or the Internet) was not much help so, I could not do this part. What is a Kepler orbit? How are they classified?

    b. Show that at x=2x0 the speed of the proton is maximum and comute that speed.
    My attempt:
    Differentiating the first expression of U(x) with respect to x and setting the derivative equal to zero gives that:

    2a/x=b therefore x=2a/b =2x0 as required.

    Subbing this into v(x) gives v=sqrt(a/2m(x0)^2)

    c. What is the force on the proton at the point x=2x0?
    My answer: F=0.

    d. Instead let the proton be released (from rest) at x1=3a/b. Derive an expression of v(x) for the new release point in terms of a,b and m, also give a qualitative description of the orbital motion.
    My attempt: Initially U(x)=U(3a/b)= -b^2/9a = Total energy
    So at any time: kinetic energy + U(x) = -b^2/9a
    Hence 0.5mv(x)^2 +a/x^2 – b/x = -b^2/9a

    This is as far as I have got for this part of the question; I do not know how to get rid of the x’s. Any ideas? I also don’t know what it means by description of the orbital motion.

    Was it ok for me to use the same potential energy function for a different initial condition?

    e. for each release point, (x0 and x1) find the maximum and minimum x values reached during the motion.
    I did not even know how to start this part of the question, any guidance would be welcome.
    Last edited by mr fantastic; April 14th 2009 at 06:59 AM. Reason: No need for the kinetic energy hyperlink
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  2. #2
    Junior Member qspeechc's Avatar
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    a. Bodies moving under a gravitational force (or any central force) follow trajectories that are conic sections, i.e. hyperbola, parabola, ellipse or circle. Depending on the energy of the body it has on of those orbit. Most physics textbooks have a description, under Kepler's Laws, Central Forces, gravitational force or whatever. Look at Particle Mechanics by Collinson and Roper, or Classical Mechanics by Kleppner and Kolenkow.

    d. If you are unsure, why not simply re-derive the formula for the potential energy from U(x)=a/x^2 - b/x ?
    What do you mean "get rid of the x's"? You want velocity v as a function of x don't you? So re-arrange
    0.5mv(x)^2 +a/x^2 – b/x = -b^2/9a
    to get v(x) = ...whatever...

    The rest looks ok to me.

    e. Draw the potential energy function U(x). Check where the initial condition is on that graph. The particle can only stay in region at the same height, or lower, that that of its initial potential energy. That is, given the initial condition xo for example, the initial potential energy is U(xo), and the particle must always have a potential energy equal to or lower than that. So you can draw a line through (xo, U(xo)) parallel to the x-axis, and the particle is only allowed to move in that region between this straight line and the graph of U(x), by conservation of energy. Then you can find the maximum and minimum x's; it's where the straight line cuts U(x) again, or if it doesn't, +/- infinity.
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  3. #3
    C.E
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    Thanks, that post was really helpful, but there are still a few things I am unsure of, could you please clarify? Firstly Is the fact that the maximum possible value of potential energy is U(x0) because initially the kinetic energy is zero, so the minimum is always where the line cuts U(x) a second time ? Secondly, for b can I just say the Kepler orbit is a hyperbola? Finally for part d what do you think they want me to say when they ask for a qualitative description of the orbital motion?
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  4. #4
    Junior Member qspeechc's Avatar
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    Quote Originally Posted by C.E View Post
    Thanks, that post was really helpful, but there are still a few things I am unsure of, could you please clarify? Firstly Is the fact that the maximum possible value of potential energy is U(x0) because initially the kinetic energy is zero, so the minimum is always where the line cuts U(x) a second time ?
    Yes. Similarly fr the maximum.


    Secondly, for b can I just say the Kepler orbit is a hyperbola? Finally for part d what do you think they want me to say when they ask for a qualitative description of the orbital motion?
    I'm not sure how much detail they require. Maybe they also want you to give the eccentricity of the orbit, and other such stuff. For that you'd have to look into one of those books I mentioned above for more details. But certainly, yes saying it's a hyperbola is required.

    Qualitative description? Dunno. If it's a hyperbola (say), then the particle comes in from infinity, is at nearest approach when the potential energy is a minimum, then flies off to infinity again. The orbit is open (since the total energy is positive for a hyperbola). I really can't tell. But if it's not in your required textbook, then I don't think they need it. Maybe describe how potential and kinetic energy relates to different parts of the orbit, whats happening with the motion etc. If you give a lot of detail, they can't fault you for that, right?
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  5. #5
    C.E
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    Why is the particle between the potential energy function and the line? I understand why it is on or below the line but not why it is not necessarily on the potential energy function curve? (that does tell us its potential energy as a function of position).
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  6. #6
    Junior Member qspeechc's Avatar
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    What I meant is that the positions of the particle, i.e. the x-values that are allowed, that the particle "takes on", must be those where the potential energy function is below, or equal to the straight line. This is a one-dimensional problem, so the particle is only allowed to move on the x-axis, and the x-values it is allow to take on are those I described above. You can think of the particle as actually moving on the potential energy function, with its height U(x) always remaining equal to or below its original height U(x0). At the highest points of the motion U(x0) the speed is zero, and at the lowest points the speed is a maximum etc.

    Btw, the Kepler orbit refers to the motion of the particle, and not the graph of the potential energy function.
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