2
$\begingroup$

This question pertains to problem 3.10 in Lagrangian Dynamics by D.H. Wells (Schaum's Outlines).

The mass $m$ [figure omitted] is attracted to a stationary mass $M$ by the gravitational force $F=-G\frac{mM}{r^{2}}$. At an initial distance $r_{0}$, $m$ is given an initial velocity $v_{0}$in the $XY$ plane. Set up the equations of motion in $r,\theta$ coordinates. Show that the angular momentum $p_{\theta}=mr^{2}\dot{\theta}=$constant. With the aid of this, integrate the $r$ equation and show that the path is a conic.

As I understand the problem, I am to integrate

$F_{r}=\frac{d}{dt}\left[\frac{\partial T}{\partial\dot{r}}\right]-\frac{\partial T}{\partial r}=m\left(\ddot{r}+r\dot{\theta}^{2}\right)=-G\frac{mM}{r^{2}}$

with the aid of

$F_{\theta}=\frac{d}{dt}\left[\frac{\partial T}{\partial\dot{\theta}}\right]-\frac{\partial T}{\partial\theta}=\frac{d}{dt}\left[mr^{2}\dot{\theta}\right]=0=\dot{L}$.

As I show below, I can certainly obtain the result by integrating. I have the sense that Wells intends a more direct integration. He has not yet introduced Hamiltonian dynamics, nor action-angle variables, etc. All that should be needed are concepts of work, kinetic energy, linear and angular momentum, force and Lagrange's equation.

Does anybody care to share a different, more efficient way of integrating ``the $r$ equation'' than what I have used?

Is it necessary to introduce the conservation of energy?

Can $\ddot{r}+r\dot{\theta}^{2}=-G\frac{M}{r^{2}}$ be integrated directly, using $\mathscr{L}=r^{2}\dot{\theta}$?

I have a way of finding the path by integrating $\mathfrak{a}\times L\mathfrak{\hat{k}}$ which never discusses energy, but that doesn't use ``the $r$ equation'' either.

The following is my treatment of the traditional orbital motion problem. I have included more material than is needed to complete the exercise; simply because it is an essential component of modern Western European Civilization.

$\mathfrak{F}\equiv m\mathfrak{a}\equiv-G\frac{mM}{r^{2}}\hat{\mathfrak{r}}$

$T\equiv\frac{1}{2}m\mathfrak{v}^{2}$

$W\equiv\int_{\mathfrak{r}_{0}}^{\mathfrak{r}}dT$

$\mathfrak{r}\left[r,\theta\right]\equiv r\mathrm{e}^{\mathrm{i}\theta}$

$\mathfrak{v}\equiv\dot{\mathfrak{r}}\left[r,\theta\right]=\left(\dot{r}+\mathrm{i}r\dot{\theta}\right)\mathrm{e}^{\mathrm{i}\theta}$

$\mathfrak{a}\equiv\dot{\mathfrak{v}}$

Areal velocity.

$\mathfrak{v}=\dot{r}+\mathrm{i}r\dot{\theta}=v_{\parallel}+\mathrm{i}v_{\perp}$

$d\mathfrak{r}=\left(\dot{r}+\mathrm{i}r\dot{\theta}\right)dt=\left(\dot{r}\hat{\mathfrak{r}}+r\dot{\theta}\hat{\theta}\right)dt$

$\hat{\mathfrak{k}}d\mathscr{A}=\frac{1}{2}\mathfrak{r}\times\left(\mathfrak{r}+d\mathfrak{r}\right)=\frac{1}{2}\mathfrak{r}\times d\mathfrak{r}=\frac{1}{2}r^{2}\dot{\theta}dt\hat{\mathfrak{k}}$

$\mathfrak{A}\equiv\frac{d\mathscr{A}}{dt}\hat{\mathfrak{k}}=\frac{1}{2}\mathfrak{r}^{2}\dot{\theta}\hat{\mathfrak{k}}$

Area swept out by the radius vector over a change in $\theta$ is represented by$\mathscr{A}$.

Lagrangian expression of generalized force.

$T=\frac{1}{2}m\left(\dot{r}^{2}+\left(r\dot{\theta}\right){}^{2}\right)$

$F_{r}=\frac{d}{dt}\left[\frac{\partial T}{\partial\dot{r}}\right]-\frac{\partial T}{\partial r}=m\left(\ddot{r}+r\dot{\theta}^{2}\right)=-G\frac{mM}{r^{2}}$

$F_{\theta}=\frac{d}{dt}\left[\frac{\partial T}{\partial\dot{\theta}}\right]-\frac{\partial T}{\partial\theta}=\frac{d}{dt}\left[mr^{2}\dot{\theta}\right]=0=\dot{L}$

So equal areas are swept out in equal times (conservation of angular momentum):

$p_{\theta}=L\equiv mr^{2}\dot{\theta}=m2\left|\mathfrak{A}\right|$

$\mathscr{L}\equiv r^{2}\dot{\theta}$

Demonstrate conservation of energy.

$T-T_{0}=\int_{\mathfrak{r}_{0}}^{\mathfrak{r}}\mathfrak{F}d\mathfrak{r}=\int_{r_{0}}^{r}F_{r}dr=-GmM\left(-\frac{1}{r}\right)_{r_{0}}^{r}$

$=GmM\left(\frac{1}{r}-\frac{1}{r_{0}}\right)=\frac{1}{2}m\left(\mathfrak{v}^{2}-\mathfrak{v}_{0}^{2}\right)$

$E\equiv\frac{1}{2}m\mathfrak{v}^{2}-G\frac{mM}{r}=\frac{1}{2}m\mathfrak{v}_{0}^{2}-G\frac{mM}{r_{0}}$

Establish the differential relationship between $d\theta$ and $dr$, then integrate.

$\mathscr{E}\equiv\frac{E}{m}=\frac{1}{2}\mathfrak{v}^{2}-G\frac{M}{r}=\frac{1}{2}\mathfrak{v}_{0}^{2}-G\frac{M}{r_{0}}$

$\mathscr{E}=\frac{1}{2}\left(\dot{r}^{2}+\left(r\dot{\theta}\right){}^{2}\right)-G\frac{M}{r}$

$\frac{\mathscr{L}}{r^{2}}=\dot{\theta}$

$2\mathscr{E}+2G\frac{M}{r}-r^{2}\left(\frac{\mathscr{L}}{r^{2}}\right){}^{2}=\left(\frac{dr}{d\theta}\right)^{2}\left(\frac{\mathscr{L}}{r^{2}}\right)^{2}$

$\sqrt{2\mathscr{E}+2G\frac{M}{r}-\frac{\mathscr{L^{2}}}{r^{2}}}\left(\frac{\mathscr{L}}{r^{2}}\right)^{-1}=\frac{dr}{d\theta}$

$\frac{\mathscr{L}dr/r^{2}}{\sqrt{2\mathscr{E}+2G\frac{M}{r}-\frac{\mathscr{L^{2}}}{r^{2}}}}=d\theta$

$\mu=\frac{1}{r}$; $d\mu=-\frac{dr}{r^{2}}$

$-\frac{\mathscr{L}d\mu}{\sqrt{2\mathscr{E}+2GM\mu-\mathscr{L^{2}}\mu^{2}}}=d\theta$

$a=2\mathscr{E}$; $b=GM$; $h=\mathscr{L}^{2}$;

$-\frac{\sqrt{h}d\mu}{\sqrt{a+2b\mu-h\mu^{2}}}=d\arccos\left[\frac{b-h\mu}{\sqrt{b^{2}+ah}}\right]$

Using the integral form derived below:

$\int_{0}^{\theta}d\theta=\theta=\pm\arccos\left[\frac{b-h\mu}{\sqrt{b^{2}+ah}}\right]$

$\cos\left[\theta\right]=\frac{b-h\mu}{\sqrt{b^{2}+ah}}=\frac{GM-\mathscr{L}^{2}/r}{\sqrt{\left(GM\right)^{2}+2\mathscr{E}\mathscr{L}^{2}}}$

$\cos\left[\theta\right]=\frac{GM-\mathscr{L}^{2}/r}{\sqrt{\left(GM\right)^{2}+2\mathscr{E}\mathscr{L}^{2}}}$

$\alpha=\sqrt{\left(GM/\mathscr{L}\right)^{2}+2\mathscr{E}}$

$\cos\left[\theta\right]=\frac{GM-\mathscr{L}^{2}/r}{\mathscr{L}\alpha}$

$GM-\mathscr{L}\alpha\cos\left[\theta\right]=\mathscr{L}^{2}/r$

$r=\frac{\mathscr{L}^{2}/(GM)}{1-\frac{\mathscr{L}\alpha}{GM}\cos\left[\theta\right]}=\frac{p}{1-\mathscr{e}\cos\left[\theta\right]}$

Which is the equation of a conic section.


Further manipulations lead to additional enlightening results. Dimensions of an ellipse Ellipse in polar coordinates $\mathscr{e}=\frac{\mathscr{L}\alpha}{GM}$; $p=\frac{\mathscr{L}^{2}}{GM}$;

$\mathscr{e}=\frac{\mathscr{L}}{GM}\sqrt{\left(GM/\mathscr{L}\right)^{2}+2\mathscr{E}}$

$=\sqrt{1+2\left(\frac{\mathscr{L}}{GM}\right)^{2}\mathscr{E}}$

$=\sqrt{1+2\left(\frac{\mathscr{L}}{GM}\right)^{2}\left(\frac{1}{2}\mathfrak{v}_{0}^{2}-G\frac{M}{r_{0}}\right)}$

Planets move in an ellipse with the sun at one focus.

$\mathscr{e}<1\implies\frac{1}{2}m\mathfrak{v}_{0}^{2}<G\frac{mM}{r_{0}}$

Let $a,b,c$ be the semimajor axis, the semiminor axis, and the center-focal distance respectively. Recall that $c=a\mathrm{e}$, and $a^{2}=b^{2}+c^{2}$.

$r\left[\frac{\pi}{2}\right]=\frac{p}{1-\mathscr{e}\cos\left[\frac{\pi}{2}\right]}=p$

$p^{2}+4c^{2}=\left(2a-p\right)^{2}=4a^{2}-4ap+p^{2}$

$a^{2}-c^{2}=ap$

$p=\frac{b^{2}}{a}=a\left(1-\mathscr{e}^{2}\right)=\frac{\mathscr{L}^{2}}{GM}$

$pa^{3}=b^{2}a^{2}=\frac{\mathscr{L}^{2}}{GM}a^{3}$

The area of an ellipse.

$4\int_{0}^{a}y\,dx=4\int_{\frac{\pi}{2}}^{a}b\sin\left[\theta\right]\frac{d}{d\theta}\left[a\cos\left[\theta\right]\right]d\theta$

$=-4ab\int_{\frac{\pi}{2}}^{0}\sin^{2}\left[\theta\right]d\theta$

$=-4ab\int_{\frac{\pi}{2}}^{0}\frac{1}{2}\left(1-\cos\left[2\theta\right]\right)d\theta$

$=-4ab\int_{\pi}^{0}\frac{1}{4}\left(1-\cos\left[u\right]\right)du$

$=\pi ab$

Let$\mathscr{T}$ represent the period of orbit.

$\mathscr{L}=r^{2}\dot{\theta}=2\left|\mathfrak{A}\right|$

$\mathscr{T}\frac{1}{2}r^{2}\dot{\theta}=\frac{1}{2}\mathscr{T}\mathscr{L}=\pi\sqrt{\frac{\mathscr{L}^{2}}{GM}a^{3}}$

$\mathscr{T}^{2}=\frac{4\pi^{2}}{GM}a^{3}$

The period of orbit is constantly proportional to the semi-major axis to the$\frac{3}{2}$ power.

Combining

$\mathscr{e}=\frac{\mathscr{L}}{GM}\sqrt{\left(GM/\mathscr{L}\right)^{2}+2\mathscr{E}}$

and

$p=a\left(1-\mathscr{e}^{2}\right)=\frac{\mathscr{L}^{2}}{GM}$

$=a\left(1-\left(\frac{\mathscr{L}}{GM}\right)^{2}\left(\left(GM/\mathscr{L}\right)^{2}+2\mathscr{E}\right)\right)$

$=a\left(1-1-2\mathscr{E}\left(\frac{\mathscr{L}}{GM}\right)^{2}\right)$

$=-2a\mathscr{E}\left(\frac{\mathscr{L}}{GM}\right)^{2}$

$-\mathscr{E}=\frac{GM}{2a}$

$m\mathscr{E}=-\frac{GmM}{2a}=E=\frac{1}{2}m\mathfrak{v}^{2}-G\frac{mM}{r}=\frac{1}{2}m\mathfrak{v}_{0}^{2}-G\frac{mM}{r_{0}}$

$-\frac{GmM}{2a}=E=T+U$


The following integral form was used to show the path is a conic section:

$\pm\int\frac{d\mu}{\sqrt{a+2b\mu-h\mu^{2}}}=\frac{1}{\sqrt{h}}\arccos\left[\frac{b-h\mu}{\sqrt{b^{2}+ah}}\right]$

It is derived as follows:

$a+2b\mu-h\mu^{2}=-h\left(\mu^{2}-2\frac{b}{h}\mu-\frac{a}{h}\right)$

$=-h\left(\mu^{2}-2\frac{b}{h}\mu+\left(\frac{b}{h}\right)^{2}-\left(\frac{b}{h}\right)^{2}-\frac{a}{h}\right)$

$=h\left(\frac{b^{2}+ah}{h^{2}}-\left(\frac{b-h\mu}{h}\right)^{2}\right)$

$=\frac{b^{2}+ah}{h}\left(1-\frac{\left(b-h\mu\right)^{2}}{b^{2}+ah}\right)$

$=\frac{b^{2}+ah}{h}\left(1-u^{2}\right)$,

where the substitution

$u=\frac{b-h\mu}{\sqrt{b^{2}+ah}}$

was used.

Using $du=-\frac{h}{\sqrt{b^{2}+ah}}d\mu$ results in

$\sqrt{a+2b\mu-h\mu^{2}}=\sqrt{h}\sqrt{\frac{b^{2}+ah}{h^{2}}}\sqrt{1-u^{2}}$

$=-\sqrt{h}\sqrt{1-u^{2}}\frac{d\mu}{du}$.

So

$\frac{\sqrt{h}d\mu}{\sqrt{a+2b\mu-h\mu^{2}}}=-\frac{du}{\sqrt{1-u^{2}}}$.

$1=\frac{d\arccos\left[u\right]}{d\theta}=\frac{d\arccos\left[u\right]}{du}\left(-\sin\left[\theta\right]\right)$

$d\arccos\left[u\right]=-\frac{du}{\sqrt{1-u^{2}}}=\frac{\sqrt{h}d\mu}{\sqrt{a+2b\mu-h\mu^{2}}}$

$\pm\int\frac{\sqrt{h}d\mu}{\sqrt{a+2b\mu-h\mu^{2}}}=\arccos\left[\frac{b-h\mu}{\sqrt{b^{2}+ah}}\right]$

The $\pm$ follows from $\cos\left[\theta\right]=\cos\left[-\theta\right].$

$\endgroup$

closed as off-topic by sammy gerbil, Jon Custer, Kyle Kanos, honeste_vivere, Wolpertinger Sep 4 '17 at 13:19

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – sammy gerbil, Jon Custer, Kyle Kanos, honeste_vivere, Wolpertinger
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ The integration using the conservation of energy is standard when the force field is integrable, however you can use another constants of the motion to find the trajectory, you can plug the angular momentum and solve the differential equation (as you stated) or use the the conservation of the Laplace-Runge-Lenz vector. $\endgroup$ – David Leonardo Ramos Aug 26 '17 at 1:37
  • $\begingroup$ $\frac{d}{dt}\left[m\mathfrak{v}\times L\mathfrak{\hat{k}}+\vec{\mathrm{e}}\right]=m\mathfrak{a}\times L\mathfrak{\hat{k}}$ is what I use in the other derivation. Apparently that is the derivative of the Laplace-Runge-Lenz vector. But I never knew it by that name. I just cribbed some notes from David Goldstein's The Mechanical Universe lecture videos. Thanks for the pointer. $\endgroup$ – Steven Thomas Hatton Aug 26 '17 at 2:22
  • 1
    $\begingroup$ You really need to make a digest of your question or format it so it's less fearsome to read. $\endgroup$ – ZeroTheHero Aug 27 '17 at 15:42
  • 1
    $\begingroup$ I'll be brutally honest with you, no one is going to read that. $\endgroup$ – Javier Aug 27 '17 at 15:54
  • $\begingroup$ @Javier That's kind of my point. The essential part lies above the first horizontal rule. I admit that it is lengthy. But that is the most concise solution I have come up with. Years ago, I thought I had a much more direct means of solving the one-body Kepler problem. Unfortunately, when I found my notes, they were so cryptic I couldn't reproduce what I had done. Or thought I had done. $\endgroup$ – Steven Thomas Hatton Aug 30 '17 at 16:13