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.
Best Answer
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.