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I Derivation of planetary orbit equation with Lagrangian

  1. May 7, 2016 #1
    I am stuck at this part page 1,
    $$\frac{\partial{L}}{\partial{\dot{φ}}}=\mu{r^2}\dot{φ}=const=l------->\dot{φ}=\frac{l}{\mu{r^2}}...........................(8)$$
    Why is this a constant? Isn't r and dφ/dt variables of time?

    Source: http://www.pha.jhu.edu/~kknizhni/Mechanics/Derivation_of_Planetary_Orbit_Equation.pdf [Broken]
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. May 7, 2016 #2
    it is because ##\frac{\partial L}{\partial \varphi}=0## and due to the Lagrange equation: ##\frac{d}{dt}\frac{\partial L}{\partial \dot\varphi}=\frac{\partial L}{\partial \varphi}## one gets ##\frac{d}{dt}\frac{\partial L}{\partial \dot\varphi}=0##
     
  4. May 8, 2016 #3
    ok Thanks a lot!
    I have one more question as to how did the source get this w(φ) = Acos(φ+d)? The equation is below eqn(14) which is stated in here " which has the well known solution w(φ) = Acos(φ+d), where both A and are constants. We can always choose d= 0 by a convenient choice of φ"
     
    Last edited: May 8, 2016
  5. May 9, 2016 #4

    vanhees71

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    The most general solution of the differential equation
    $$\frac{\mathrm{d}^2 w}{\mathrm{d} \varphi^2}=-w$$
    is given by
    $$w(\varphi)=A_1 \cos \varphi+A_2 \sin \varphi,$$
    where ##A_1## and ##A_2## are constants of integration. This follows, because for a homogeneous 2nd-order linear differential equation the solution is the linear combination of two arbitrary linearly independent solutions, and cos and sin fulfill these conditions.

    Now you can write this general solution also in another form,
    $$w(\varphi)=A \cos(\varphi+\delta),$$
    where ##A## and ##\delta## are constants. To see this just use the addition theorem for cos
    $$w(\varphi)=A \cos \delta \cos \varphi - A \sin \delta \sin \varphi.$$
    Comparing with the general solution above you have
    $$A_1=A \cos \delta, \quad A_2=-A \sin \delta,$$
    and you always find ##A>0## and ##\delta \in (-\pi,\pi]## to fulfill these equations. First we have
    $$A=\sqrt{A_1^2+A_2^2}, \quad \delta =-\mathrm{sign} A_2 \arccos \left (\frac{A_1}{\sqrt{A_1^2+A_2^2}} \right).$$
    So indeed the given solution is the most general one for the differential equation for ##w##.
     
  6. May 11, 2016 #5
    Thanks a lot for the reply! But i am stuck at this part
    Isn't $$\quad \delta=arccos(\frac{A_1}{A})=-arcsin(\frac{A_2}{A})$$?
    Besides do you mean that $$w(\varphi)=A \cos(\varphi+\delta),$$ is more general than $$w(\varphi)=A_1 \cos \varphi+A_2 \sin \varphi,$$? If so, I can't see the reason behind it.
     
  7. May 11, 2016 #6

    vanhees71

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    No, both equations are equivalent, as demonstrated in the previous posting. You can always map from the pair ##(A_1,A_2)## of integration constants to the pair ##(A,\delta)##. It just depends on what you consider more convenient. In this case, it's easier to see that the orbit is a conic section (ellipse, parabola, hyperbola), using the ##(A,\delta)## notation.
     
  8. May 11, 2016 #7
    Oh I see. But how did you obtain this $$\quad \delta =-\mathrm{sign} A_2 \arccos \left (\frac{A_1}{\sqrt{A_1^2+A_2^2}} \right).$$?
     
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