# Derivation of planetary orbit equation with Lagrangian

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TimeRip496
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]

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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##

vanhees71 and TimeRip496
TimeRip496
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##
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"

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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##.

TimeRip496
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##.
Thanks a lot for the reply! But i am stuck at this part
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##.
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.

Oh I see. But how did you obtain this $$\quad \delta =-\mathrm{sign} A_2 \arccos \left (\frac{A_1}{\sqrt{A_1^2+A_2^2}} \right).$$?