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Path Calculation of point mass in gravitational field

  1. Jan 26, 2013 #1

    VVS

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    Hi,

    I am trying to derive a formula for the path taken of a body in a Gravitational Field using Classical Mechanics.

    I have attached a pdf file with my derivation so far. Basically I used conservation of energy and conservation of angular momentum to derive a diferential equation describing the motion.


    Right now I am stuck. I don't know how to proceed.
     

    Attached Files:

  2. jcsd
  3. Jan 26, 2013 #2

    vanhees71

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    Your derivations are all correct, as far as I can see, and it's in fact a very clever trick to use the conservation laws at hand to get the equations of motion.

    You'll find your way by just another trick! Use just the right coordinates for the symmetry of the problem at hand and you'll succeed. As you've already used, due to the spherical symmetry of the force, angular momentum is conserved and thus the motion goes entirely in a plane, you've choosen to be the [itex]xy[/itex] plane.

    Now it's more convenient to introduce polar coordinates
    [tex]\vec{x}=\begin{pmatrix} x \\ y \end{pmatrix} = r \begin{pmatrix} \cos \varphi \\ \sin \varphi \end{pmatrix}.[/tex]
    Plug this into the expressions for the energy and modulus of the angular momentum. With the latter expression you can eliminate [itex]\varphi[/itex] and [itex]\dot{\varphi}[/itex] completely from the energy expression. Then you get an equation of motion for [itex]r(t)[/itex] alone by taking the time derivative of the energy expression again.

    One final trick is to consider this equation in terms of [itex]u(\varphi)=\frac{1}{r(\varphi)}[/itex]. This leads to an equation for [itex]u(\varphi)[/itex], you can easily solve. The result are Kepler's Laws for the motion of planet around the sun (in the approximation to neglect the center-mass motion of the sun and treat it as fixed central potential).

    If you encounter further problems, let us know!
     
  4. Jan 26, 2013 #3

    VVS

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    Thank you for the quick reply.

    I am going to redo the derivation in polar coordinates again and then I will get back to you.
     
  5. Jan 26, 2013 #4

    VVS

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    Hey vanhees71 I redid the derivation as you suggested with polar coordinates. It really does make life easier. I attached it in the pdf file.

    Do I have to differentiate the energy equation again wrt to t? Wouldn't that give me terms in r'''?

    Moreover I couldn't eliminate theta' and therefore I didn't get an equation only in r, r' an r'' as you suggeste I would.

    How can one solve a differential equation like that?

    Could you please give me a hint?
     

    Attached Files:

  6. Jan 26, 2013 #5

    vanhees71

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    Wait! The trick is to calculate [itex]E[/itex] and [itex]L[/itex] (which you did right!). So you have
    [tex]L=m r^2 \dot{\theta}=\text{const}[/tex]
    Now you use
    [tex]\dot{\theta}=\frac{L}{m r}[/tex]
    to elimininate [itex]\dot{\theta}[/itex] from [itex]E[/itex]. Then you get
    [tex]E=\frac{m}{2} \dot{r}^2 + \frac{L^2}{m r^2}-\frac{G M m}{r}.[/tex]
    In this you substitute
    [tex]r=\frac{1}{u(\theta)}[/tex]
    for the trajectory, expressed in polar coordinates, thinking for the angle as a function of time, [itex]\theta(t)[/itex]. Then you take the time derivative, which leads to a differential equation for [itex]u(\phi)[/itex].

    It is not so easy to get the trajectory as function of time. This leads to Kepler's equation, which is only solvable in implicit form!
     
  7. Jan 26, 2013 #6

    VVS

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    I have done the derivation again as you suggested.

    I did the substitution, but I think I did something wrong the differential equation looks very complicated.
     

    Attached Files:

  8. Jan 27, 2013 #7

    vanhees71

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    Ok, I guess it's time to solve the problem, but I think you already did a good job!

    We start from your equation (3)
    [tex]\frac{\dot{r}^2}{2}+\frac{h^2}{2 r^2} - \frac{G M}{r}=\text{const}.[/tex]
    Here [itex]h=L/m[/itex].

    First we substitute [itex]r=1/u[\theta(t)][/itex]. Using the chain rule this gives
    [tex]\dot{r}=-\frac{u'}{u^2} \dot{\theta},[/tex]
    where the prime refers to differentiation wrt. [itex]\theta[/itex] and the dot to differentiation wrt. time.

    Now use the angular-momentum conservation (it's also on your paper):
    [tex]h=\frac{L}{m}=r^2 \dot{\theta}.[/tex]
    Subsitute this into the equation above, you get
    [tex]\dot{r}=-\frac{u'}{u^2} \frac{h}{r^2}=-h u'.[/tex]
    Put this into the energy-conservation equation above
    [tex]\frac{h^2 u'^2}{2} + \frac{h^2}{2} u^2-G M u=\text{const}.[/tex]
    The time derivative of this gives
    [tex]h^2 u' u'' \dot{\theta}+h^2 u u' \dot{\theta}-G M u' \dot{\theta}=0.[/tex]
    Using again the angular-momentum conservation equation gives after some algebra
    [tex]u''=\frac{G M}{h^2}-u.[/tex]
    This is the equation of an harmonic oscillator subject to an external constant force. The general solution obviously is
    [tex]u(\theta)=\frac{G M}{h^2}+A \cos \theta + B \sin \theta.[/tex]
    Now it is the usual convention to choose the polar coordinate system such that [itex]\vartheta=0[/itex] denotes the perihel, i.e., the point, where the planet comes next to the sun. At this point [itex]u=1/r[/itex] should be maximal, i.e., [itex]u'(0)=0 \; \Rightarrow \; B=0[/itex] and (assuming [itex]A>0[/itex]) yields
    [tex]u(\theta)=\frac{G M}{h^2} + A \cos \theta.[/tex]
    This gives
    [tex]r(\theta)= \frac{1}{G M/h^2 + A \cos \theta}.[/tex]
    Put this a bit in a more familiar shape:
    [tex]r(\theta)=\frac{p}{1+\epsilon \cos \theta},[/tex]
    where
    [tex]p=\frac{h^2}{G M}, \quad \epsilon=\frac{h^2 A}{G M}.[/tex]
    One can show that, if the total energy is negative (bound motion), then [itex]\epsilon<1[/itex]. This means that the planet moves along an ellipse with the sun in one of its foci. That's Kepler's first law.
     
  9. Jan 27, 2013 #8
    1.Switch to polar coordinates
    2.Eliminate the angular coordinate by the conservation of angular momentum
    3.Use above mentioned trick u=1/r to get a r(θ) solution.

    If you try to solve for r(t) you'll get a non elementary integral, feed it into mathematica if you're still interested.
     
  10. Jan 27, 2013 #9

    VVS

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    Hi,

    I realized what I needed to do this morning. I have attached the derivation again for completeness till we get the final differential equation. Now I would like to solve this differential equation with initial conditions: position and speed. I will try to figure it out myself and if I can't do it I will ask for help.

    I have also derived a differential equation at the end using the time derivative expressions for energy and angular momentum. This DE is in d2/dt2(r) and r. I wonder whether this DE is useful for anything.

    Thanks a lot for the support so far, this forum rocks!
     

    Attached Files:

  11. Jan 27, 2013 #10

    VVS

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    Hi,

    I've done it! Please see the attached pdf. I have a formula for the trajectory as a function of theta and the initial conditions. I rechecked it with my father, there shouldn't be any errors inside.

    Thanks for everyone's help!
     

    Attached Files:

  12. Jan 27, 2013 #11

    VVS

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    Hello,

    I started plugging in numbers and I started to get weird graphs as you can see at the end of my attached file.

    What is wrong in my derivation of the Coefficients for the solution of the differential equation?
     

    Attached Files:

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