[NOTES/CM-05007] Differential Equation of Orbit

Method-I: Using Euler Lagrange Equations

Here we obtain a second order differential equation for the orbit for motion in a spherically symmetric potential. For such a potential the orbits lie in a plane. This suggests use of plane polar coordinates \(r, \phi\). The angle \(\phi\) turns out to be a cyclic coordinate. This fact helps in converting the Euler Lagrange EOM into the differential equation of the orbit. The Lagrangian for a particle moving in a plane, in potential \(V(r)\), is given by \begin{eqnarray} \mathcal L&=& \frac{1}{2} \mu (\dot x^2 + \dot y^2) - V(r)\nonumber \\ &=& \frac{1}{2} \mu \dot{r}^{\,2} + \frac{1}{2} \mu r^{\,2} \dot{\phi}^{\,2} - V(r)\Label{EQ01} \qquad \qquad \mbox{\HighLight{Verify!}} \end{eqnarray}

The Euler Lagrangian equations are \begin{equation}\Label{EQ02} \frac{d}{dt} \left( \frac{\partial \mathcal L}{\partial \dot{r}}\right) - \frac{\partial \mathcal L}{\partial r} = 0 \end{equation} or \begin{equation}\Label{EQ03} \mu \ddot{r} - \mu r \dot{\phi}^{2} + \frac{\partial V}{ \partial r} = 0 \end{equation}

$\phi$ is a cyclic coordinate

Since \(\phi\) is a cyclic coordinate, we have \begin{equation} \Label{EQ04} \pp[L]{\dot{\phi}}=\text{constant, say} L, \Longrightarrow \mu r^{2} \dot{\phi} = L. \end{equation} \(L\) will turn out to be the angular momentum of the particle.  Let $f(r) = - \frac{\partial V}{ \partial r}$ be the force law, then \eqref{EQ03} and \eqref{EQ04} give  \begin{equation}\Label{EQ05} \mu\, \ddot{r} = f(r) + \frac{L^{2}}{\mu r^{3}} .\end{equation} Next use \eqref{EQ04} to convert $\ddot{r}$ into $ \frac{d^{2} r}{d \phi^{2}}$ \begin{eqnarray} \frac{dr}{dt} &=& \frac{dr}{d \phi} \left( \frac{d \phi}{dt} \right)\Label{EQ06}\\ &=& \frac{L}{\mu r^{2}} \frac{dr}{d \phi}\Label{EQ06A} \end{eqnarray} and \begin{equation}\Label{EQ07} \frac{d^{2} r}{dt^{2}} = \frac{L}{\mu r^{2}} \frac{d}{d \phi} \frac{L}{\mu r^{2}} \frac{dr}{d \phi}= \frac{L^2}{\mu^2r^2}\frac{d}{d \phi} \frac{1}{r^{2}} \frac{dr}{d \phi} \end{equation} The equation of the orbit takes the form \begin{equation} \mu \frac{L^2}{\mu^2r^2}\frac{d}{d \phi} \frac{1}{r^{2}} \frac{dr}{d \phi} = f(r) + \frac{L^2}{\mu r^3} \end{equation}

Method-II : Using Conservation Law

Here we will use conservation laws obtain a second order differential equation and solve it for the motion of the planets around the Sun.

The energy conservation for a particle in a spherically symmetric potential takes the form \begin{equation}\Label{eq21} E = \frac{1}{2}\mu \Big(\dd[r]{t}\Big)^2 + V(r) +\frac{L^2}{2\mu r^2} \end{equation} where \(\mu \) is the mass and \(L\) is the angular momentum of the particle. The sum of last two terms in \eqref{eq21} is just the effective potential for radial motion. Differentiating \eqref{eq21} w.r.t. time we get the EOM for radial coordinate \(r\):\begin{equation}\Label{eq22} \mu \DD[r]{t} = - \dd[V(r)]{r} + \frac{L^2}{\mu r^3}. \end{equation} The conservation of angular momentum in plane polar coordinate is of the form:\begin{equation} \Label{eq04} \mu r^{2} \dot{\phi} = \text{constant, say} L \end{equation} Let $f(r) = - \frac{\partial V}{ \partial r}$ be the force law, then \eqRef{eq22} and \eqref{eq04} give  \begin{equation}\Label{eq05} \mu\, \ddot{r} = f(r) + \frac{L^{2}}{\mu r^{3}} \end{equation}The steps now on wards are as in method-I and we arrive at the differential equation of the orbit already found by the first method.