I'm self-studying Lagrangian Mechanics using Goldstein's Classical Mechanics, supplemented with Lemos's Analytical Mechanics, a modern version of the same, and Landau and Lifshitz's Mechanics. Currently I'm studying Kepler's two-body problem. I've tried to come up with an approximate solution for the same by considering $M\gg m$. That is, we ignore the movements of the more massive body and consider it fixed at the origin.

The lagrangian for the system will be $$\mathcal L=\frac12m\dot r^2+\frac12mr^2\dot\phi^2+\frac{GMm}r.$$ $\phi$ is a cyclic coordinate, thus we have $$mr^2\dot\phi=l=\text{constant.}$$ And using $$\frac{\mathrm d}{\mathrm dt}\frac{\partial\mathcal L}{\partial\dot r}=\frac{\partial\mathcal L}{\partial r}$$ we have $$m\ddot r=mr\dot\phi^2-\frac{GMm}{r^2}.$$ Now I used chain rule $$\frac{\mathrm d}{\mathrm dt}=\dot\phi\frac{\mathrm d}{\mathrm d\phi}$$ to get $$\ddot r=-\frac{2l^2}{m^2r^5}\Big(\frac{\mathrm dr}{\mathrm d\phi}\Big)^2+\frac{l^2}{m^2r^4}\frac{\mathrm d^2r}{\mathrm d\phi^2}.$$ The final DE simplifies a lot with the substitution $r=\dfrac1u$. $$\frac{\mathrm d^2u}{\mathrm d\phi^2}+u=\frac{GMm^2}{l^2}.$$ Now, I assumed that the mass $m$ starts with $\mathbf r_0=r_0\mathbf i$ and $\mathbf v_0=v_0\mathbf j$. And the angular momentum $l$ is the same everytime, we can write $l=mr_0v_0$, giving us $$\frac{\mathrm d^2u}{\mathrm d\phi^2}+u=\frac{GM}{r_0^2v_0^2}.$$ This can be trivially solved $$u(\phi)=\frac1{r(\phi)}=\frac{GM}{r_0^2v_0^2}+A\cos\phi+B\sin\phi.$$ The initial conditions $r(0)=r_0$ and $v(0)=v_0$ give $$A=\frac1{r_0}-\frac{GM}{r_0^2v_0^2},\quad B=-\frac1{r_0}.$$ where to find $B$, I used the previously mentioned chain rule $$\frac{\mathrm dr}{\mathrm d\phi}=\frac{mr^2}l\frac{\mathrm dr}{\mathrm dt}.$$ This gives us the final trajectory of $m$ $$r(\phi)=\frac{r_0^2v_0^2}{GM+(r_0v_0^2-GM)\cos\phi-r_0v_0^2\sin\phi}.$$ Now, here is the problem I'm facing. I had used the book Calculus: Early Transcendentals by Anton, Bivens, Davis for studying calculus as approximately a fresher. I had used the 10th edition, where (don't know about the other editions) the author had provided a derivation for the approximate trajectory I want to achieve (same assumptions as mine), purely using the two vector products and some common vector function theorems he had discussed before.

There, he derived $$r(\phi)=\frac{r_0^2v_0^2}{GM+(r_0v_0^2-GM)\cos\phi}.$$ So somehow, $B$ has to be zero. But I don't see how that's possible, given the initial conditions.

Kindly help me fix this. Tell me if there are any mistakes in my solution or arguments.


1 Answer 1


The initial conditions imply $B=0$. This is so, because by your initial conditions. In polar coordinates \begin{equation} \vec{r} = r\vec{e_r} \end{equation} \begin{equation} \dot{\vec{r}} = \dot{r}\vec{e_r} + r\dot{\phi}\vec{e_{\phi}} \end{equation} Thus $\vec{r} \cdot \dot{\vec{r}}\; \Big|_{t=0} = 0$ means that since $\vec{e_r} \perp \vec{e_{\phi}}$: \begin{equation} 0=\frac{d}{dt}r = \frac{d\phi}{dt}\frac{dr}{d\phi} = -\frac{\dot{\phi}}{u^2}\frac{du}{d\phi}, \end{equation} since $r_0 \neq 0$. Since $\dot{\phi}\neq 0 $, it must follow that $\frac{du}{d\phi}\Big|_{t=0}=0$. Thus $B=0$ as desired.

  • $\begingroup$ Thanks a lot. I was thinking if the manipulation I did is even valid or not. Your method is quite good, formal and correct. Thanks a lot again. :) $\endgroup$ Commented Jul 17, 2022 at 23:12
  • $\begingroup$ You’re welcome ^^ $\endgroup$ Commented Jul 17, 2022 at 23:31
  • $\begingroup$ Any ideas why my chain rule didn't work? $\endgroup$ Commented Jul 18, 2022 at 0:59
  • $\begingroup$ Also, doesn't this imply that $v(0)=0$. Why would that be the case as the body has initial speed $v_0$? $\endgroup$ Commented Jul 18, 2022 at 1:04
  • $\begingroup$ You have to differentiate between the coordinate vector $\vec{r(t)}$ and $r(t)$ as the radial coordinate in polar coordinates. Thus you set erroneously $v_0 =\dot{r}$ instead of $\vec{v_0} = \dot{\vec{r(0)}} = r_0 \dot{\phi} \vec{e_{\phi}} $. $\endgroup$ Commented Jul 18, 2022 at 1:56

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