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Homework Help: Orbital Dynamics problem help

  1. Jan 25, 2009 #1
    1. The problem statement, all variables and given/known data


    I've found the a general form of equation for the explicit integration of the shape, [itex]r(\theta)[/itex], of a one-body-problem (a particle rotating about a set point, O) for arbitrary [itex]\mathbf{f} = -f(r)\hat{\mathbf{r}}[/itex], and used it to find [itex]r(\theta)[/itex] for an inverse square law [itex]f(r) = \mu / r^2[/itex]. Where [itex]u = (1/r)[/itex].

    I was able to get [tex]r = r(\theta(t)) => \dot{r} = {{dr} \over {d\theta}} {{d\theta} \over {dt}} = {{dr} \over {d\theta}} \left({{h_0} \over {r^2}}\right)[/tex]

    Where [itex]h_0 \neq 0[/itex] is the constant angular momentum.

    Can anyone help me to figure out the shape under an inverse cube law of force, [itex]f(r) = {{\nu} / {r^3}}[/itex], and show that if [itex]\nu \le h^2[/itex], the orbit is unbounded?

    I know that this solution will depend on the relative sizes of [itex]\nu[/itex] and [itex]h[/itex], but I can't seem to get this to work.


    EDIT:: Got LaTex to work
     
    Last edited: Jan 26, 2009
  2. jcsd
  3. Jan 26, 2009 #2
    One way to approach this problem is to use the substitution

    [tex]r=\frac{1}{u}\mbox{ and u =}\frac{1}{r}[/tex]

    then differentiate r with respect to t giving

    [tex]\dot{r}=-\frac{1}{u^2}\frac{du}{d\theta}\frac{d\theta}{dt}[/tex]

    and

    [tex]\dot{\theta}=\frac{L}{mr^2}[/tex]

    This will give

    [tex]\dot{r}=-\frac{L}{m}\frac{\mbox{d}u}{\mbox{d}\theta}[/tex]

    Differentiate again with respect to t to find the second derivative of r in terms of u and theta. Write the general equation of motion for r in terms of F(r) and "centrigugal" force (L2/(mr3), and substitue the second derivative of r with respect to t. You will arrive at an equation of motion in terms of u and theta. This differential equation is easier to solve than straightforward methods. Once u is found substitue 1/r and solve for r.
     
  4. Jan 26, 2009 #3
    Got it figured out. Thanks
     
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