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Motion in Gravitational Field

  1. Mar 11, 2012 #1
    I want to find out the general equations of motion for a particle with an initial velocity [itex]v_0[/itex] in a gravitational field by a point/spherical mass (assuming this is a large mass which doesnt move). Assume that the origin of the coordinate system is the point mass. If the vector equation of the particle's path is [itex]\mathbf{r}(t)[/itex], then the acceleration should be the second derivative,
    [tex]\frac{d^2 \mathbf{r}}{dt^2}[/tex]
    The acceleration is caused by the gravitaional field (acceleration field) given by,
    [tex]A(\mathbf{r}) = GM\frac{\mathbf{r}}{|\mathbf{r}|^3}[/tex]
    But we already have the acceleration of the particle,
    [tex]\frac{d^2 \mathbf{r}}{dt^2} = GM\frac{\mathbf{r}}{|\mathbf{r}|^3}[/tex]
    So the general solution for [itex]\mathbf{r}[/itex] can be found by solving the above differential equation but I couldn't do it. Can anyone show me how it is done?
     
  2. jcsd
  3. Mar 13, 2012 #2
    Does anybody know how to solve it or is there no exact analytic solution?
     
  4. Mar 13, 2012 #3
    Yes, this equation is solvable analytically. It is best to approach it using Lagrangians. The solutions are elliptical orbits first described by Kepler's laws.
     
  5. Mar 13, 2012 #4
    I have seen Lagrangians and Euler-Lagrange equation but I don't know them well. Would someone kindly show me the steps in the solution?
     
  6. Mar 15, 2012 #5
    Does any one know how to do it?
     
  7. Mar 16, 2012 #6
    By Kepler:
    [itex]\frac{d^{2}r}{dt^{2}} = GM\frac{r}{|r|^{2}}[/itex]
    by Newton:
    F=ma, P=mv
    [itex]F=m*GM\frac{r}{|r|^{2}}[/itex]
    Where m is the mass of the orbiting body and M is the point mass (center of orbit).
    We also know that motion for a particle follows the path dictated by:
    [itex] r=r(r,θ,t)=re_{r}[/itex]
    [itex] v=e_{r}\frac{dr}{dt}+rωe_{θ}[/itex]
    [itex] a=e_{r}(\frac{d^{2}r}{dt^{2}}-rω^{2})+(rα+2ω\frac{dr}{dt})e_{θ}[/itex]
    This is using a polar coordinate system. If you substitute your acceleration into this equation and then solve for zero velocity, you will have a suitable equation for two dimensional orbit. For higher dimensions there is considerably more work, and I frankly have too much else to do!

    By the way: This is taken directly from Newton's equations of motion, Under General Planar Motion.
     
  8. Mar 16, 2012 #7
    You can find a detailed derivation in Landau and Lifgarbagez, Mechanics.
     
  9. Nov 10, 2013 #8

    andrewkirk

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    There is no analytic solution to the differential equation, if by 'solution' we mean an explicit formula giving r as a function solely of t and items that are constant over time.

    However, there are explicit formulas for various values that would otherwise be calculated via such a solution. For instance there is an explicit formula for the 'free-fall time'

    Effective potential techniques can be used to answer many questions about the motion.

    However calculation of the radius at a given point in time can generally only be done by numerical techniques.

    The following page from Wolfram alpha confirms that there is no analytic solution to the DE:
    http://www.wolframalpha.com/input/?i=%28x%27%28t%29^2%29%2F2-C%2Fx%3D-1

    or this: http://www.wolframalpha.com/input/?i=x%27%27%28t%29%3D-C%2Fx^2
     
  10. Nov 10, 2013 #9

    phyzguy

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    Or in Goldstein "Classical Mechanics".

    There is also a derivation of the solution at this Wikipedia page

    I think andrewkirk is wrong in saying there is no analytic solution. There is an analytic solution as others have said. In polar coordinates the orbit is given by:

    [tex] r = \frac{p}{1+e \cos(\theta)}[/tex]

    where p and e are constants of the motion. You can derive p and e, as well as the position along the orbit and the orientation of the orbit in space from your initial position and velocity values. The part of the Wikipedia page titled "Determination of the Kepler orbit that corresponds to a given initial state" gives the detailed algorithm for doing this.
     
  11. Nov 10, 2013 #10

    andrewkirk

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    Unfortunately that is not a solution in terms solely of ##t## and items that are constant with respect to ##t##, because ##\theta## varies with ##t##. It is just a new differential equation.
     
  12. Nov 10, 2013 #11

    phyzguy

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    It is not a new differential equation. It is an algebraic equation and it is a solution to the initial problem. And it is possible, as I said in the earlier post, to explicitly calculate r(t) and θ(t) given the initial conditions, and the Wikipedia page gives the specific algorithm for doing this. You can then translate this to x(t), y(t) and z(t) if you wish. This is the Kepler problem, and the solution has been known for centuries. Why do you say no solution exists?
     
  13. Nov 10, 2013 #12

    andrewkirk

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    In the post to which you were replying, I made clear what I meant by 'solution', as follows:
    Nothing has been posted in this thread that meets that criterion.
    If there is such a formula, I will be very pleased to see it, as it would help greatly in a little project I am currently doing. But I suspect there is not.
     
  14. Nov 10, 2013 #13

    phyzguy

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    So what you are looking for is r(t) and θ(t), right? The way you get there is from the following three equations:
    [tex] t = a \sqrt{\frac{a}{G M}}(E - e \sin(E))[/tex]
    [tex]\cos(\theta) = \frac{\cos(E)-e}{1-e \cos(E)}[/tex]
    [tex]r = \frac{p}{1+e \cos(\theta)}[/tex]

    So given t, you calculate E from the first equation, then calculate θ from the second equation, then calculate r from the third equation. G, M, a, p, and e are all constants of the motion. This gives θ(t) and r(t). Perhaps this doesn't satisfy your definition because the first equation is hard to invert. Is that your issue? If so, I think the issue is purely semantic, since it is an easy practical matter to get E(t) by building a look-up table or solving the first equation using a numerical technique like Newton's method.
     
  15. Nov 10, 2013 #14

    andrewkirk

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    You missed the key word 'analytic' in my original statement:
    Of course there are numerical solutions. Indeed, the original DE itself can be solved numerically, given initial conditions, without having to use the Kepler equations.
     
  16. Nov 10, 2013 #15

    phyzguy

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    We're quibbling over semantics, but the usual definition of an analytic solution is:

    "In mathematics, an expression is said to be a closed-form expression if it can be expressed analytically in terms of a finite number of certain "well-known" functions.

    What I posted earlier certainly qualifies. It is not a numerical solution, but a closed-form, analytic expression.
     
  17. Nov 10, 2013 #16

    D H

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    No, it is not quibbling, and what you posted earlier does not qualify.

    There is no closed form analytic solution for the eccentric anomaly as a function of time. There is an analytic solution for mean anomaly as a function of time, but there is no analytic solution to solve for eccentric anomaly as a function of mean anomaly. The problem of finding E given ##M=E-e\sin E## (Kepler's equation) does not have an analytic solution in the elementary functions.
     
  18. Nov 10, 2013 #17

    phyzguy

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    OK, I won't argue the point.
     
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