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I Trying to solve the restricted two-body problem

  1. Aug 29, 2016 #1
    I'm just trying to solve the restricted two-body problem mentioned in passing in my textbook:

    ##\mathbf{\ddot{r}}+\frac{GM}{r^2}\left(\frac{\mathbf{r}}{r}\right)=\mathbf{0}##

    I understand the way to do it is to take the cross-product with the constant angular momentum ##\mathbf{h}## then integrate wrt time:

    ##\mathbf{\dot{r} \times h}= -\frac{GM}{r^2}\int \left(\frac{\mathbf{r}}{r}\right)\mathbf{\times h}\ dt##

    But I'm not sure what to do next. Can I use the fact that ##\mathbf{h}=\mathbf{r \times \dot{r}}## to make it:

    ##\mathbf{\dot{r} \times h}= -\frac{GM}{r^2}\int \left(\frac{\mathbf{r}}{r}\right)\mathbf{\times r \times \dot{r}}\ dt##
    Can anyone help?
    The answer I'm trying to get to is:
    ##\mu\left(\frac{\mathbf{r}}{r} + \mathbf{e}\right)##, where ##\mathbf{e}## is a constant vector of integration and presumably ##\mu## is some constant related to G.
     
  2. jcsd
  3. Aug 29, 2016 #2
    Hi,

    After you manage to use the angular momentum conservation, you need to do variable change from r to u = 1/r.
    Also there are many online texts which solve this problem.

    BR,
    Ohad
     
  4. Aug 30, 2016 #3
    Thanks. The problem I'm having is that the cross product of a vector with itself is zero, so I seem to get 0 in the integrand. Actually, I think the ##r^{-2}## term I factored out should also be in there, but that would still give me an integrand of:
    ##\int \frac{\mathbf{r \times r\times \dot{r}}}{r^3} dt##
    Isn't the inside of the integrand then zero?
     
  5. Aug 30, 2016 #4

    vanhees71

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    It's simpler to work with differential equations. Starting from your EoM
    $$\ddot{\vec{r}}=-\frac{G M}{r^3} \vec{r},$$
    you can very easily derive the conservation of angular momentum (here everything is divided by ##m##, the mass of the "test particle"):
    $$\vec{r} \times \ddot{\vec{r}}=0.$$
    But on the other hand
    $$\frac{\mathrm{d}}{\mathrm{d} t} (\vec{r} \times \dot{\vec{r}})=\dot{\vec{r}} \times \dot{\vec{r}} + \vec{r} \times \ddot{\vec{r}}=\vec{r} \times \ddot{\vec{r}}.$$
    Thus you have with ##\vec{h}=\vec{r} \times \dot{\vec{r}}##
    $$\dot{\vec{h}}=0 \; \Rightarrow \; \vec{h}=\text{const}.$$
    For ##\vec{h} \neq 0## it's clear that the motion is in a plane perpendicular to ##\vec{h}##. The standard way to proceed further is to use the conservation of energy as a 2nd first integral and solve for the orbit ##r=r(\phi)##, where ##r## and ##\phi## are polar coordinates in the plane.

    Obviously your textbook follows another very elegant way, using the huge symmetry of the Kepler problem. The point is that for the ##1/r## potential there is an extra symmetry, which leads to another conserved vector, the Runge-Lenz vector in the plane of motion. To derive it we use
    $$\frac{\mathrm{d}}{\mathrm{d} t} (\dot{\vec{r}} \times \vec{h})=\ddot{\vec{r}} \times \vec{h}=-\frac{GM}{r^3} \vec{r} \times \vec{h} \qquad (*).$$
    On the other hand we have
    $$\frac{\mathrm{d}}{\mathrm{d} t}=\frac{\vec{r}}{r} = \frac{r \dot{\vec{r}}-\dot{r} \vec{r}}{r^2}.$$
    Now we have (check it!)
    $$\dot{\vec{r}} = \frac{\mathrm{d}}{\mathrm{d} t} \sqrt{\vec{r} \cdot \vec{r}}=\frac{\vec{r} \cdot \dot{\vec{r}}}{r}=-\frac{\vec{r} \times \vec{h}}{r^3}.$$
    Multiplying with ##GM## and subtracting from (*) you get
    $$\frac{\mathrm{d}}{\mathrm{d} t} \left [\dot{\vec{r}} \times \vec{h} - G M \frac{\vec{r}}{r} \right]=0,$$
    i.e., the Runge-Lenz vector
    $$\vec{e}= \dot{\vec{r}} \times \vec{h} - G M \frac{\vec{r}}{r}=\text{const} \qquad(**).$$
    From this it's very easy to find the shape of the orbit by introducing polar coordinates in the orbital plane with ##\vec{e}## as the polar axis. Multiplying (**) with ##\vec{r}## you get
    $$\vec{r} \cdot \vec{e} = r e \cos \varphi.$$
    On the other hand from the definition of the Runge-Lenz vector) you have
    $$\vec{r} \cdot \vec{e} = \vec{r} \cdot (\dot{\vec{r}} \times \vec{h})-GM r = (\vec{r} \times \dot{\vec{r}}) \cdot \vec{h} - G M r =h^2-G M r.$$
    Together with the previous equation you find
    $$r=\frac{h^2}{GM+e \cos \varphi}=\frac{h^2/(GM)}{1+e/(GM) \cos \varphi}.$$
    This is a conic section with the excentricity ##\epsilon=e/(GM)##, i.e., an ellipse for ##0<\epsilon<1##, a circle for ##\epsilon=0##, a parabola for ##\epsilon=1##, and a hyperbola for ##\epsilon>1##.
     
  6. Aug 30, 2016 #5
    Thanks, that's an awesome answer. I'm going to come back to it to work through properly, and make sure I understand it fully.
    Thanks for taking the time out.
     
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