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In another article we reduced the two-body problem to a one dimensional problem. Now the potential energy only depends on the distance of some point $r$. The resulting force of this potential is called a central force \begin{align} F= -\frac{\partial U}{\partial \mathbf{r}}= -\frac{\partial U}{\partial r} \frac{\mathbf{r}}{r} \end{align} We have already shown, that the angular momentum for such a system is conserved. \begin{align} \mathbf{L}=\mathbf{r} \times \mathbf{p} \end{align} The vector $\mathbf{L}$ is always orthogonal to $\mathbf{r}$. Since $\mathbf{L}$ is constant $\mathbf{r}$ moves in a plane orthogonal to $\mathbf{L}$. In polar coordinates we can characterize movement in the plane by $r$ and $\varphi$ \begin{align} \mathcal{L}=\frac{m}{2} ( \dot{r}^2 + r^2 \dot{\varphi}^2)- U(r) \end{align} The Lagrangian does not explicitly depend on the coordinate $\varphi$, such coordinates are called cyclic. The derivative of the Lagrangian with respect to cyclic coordinates is zero. From the Euler-Lagrange Equation we know \begin{align} \frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{q_i}}=\frac{\partial \mathcal{L}}{\partial q_i}=0 \end{align} this means the canonical momentum $p_i$ \begin{align} \frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{q_i}}=\frac{d}{dt}{p_i}=0 \end{align} is conserved \begin{align} \frac{\partial \mathcal{L}}{\partial \dot{q_i}}={p_i}=const. \end{align} For our problem this means \begin{align} \frac{\partial \mathcal{L}}{\partial \dot{q_i}}={p_{\varphi}}=m \cdot r^2 \cdot \dot{\varphi}=const. \end{align} We define the area $df=\frac{1}{2} r \cdot r \cdot d\varphi$

This equation shows, that the area-velocity $\dot{f}$ is also constant. \begin{align} {p_{\varphi}}=m \cdot r^2 \cdot \dot{\varphi}=2\cdot m \cdot \dot{f}=const. \end{align} This is also the central statement of Kepler's second law. To finally solve the problem, the easiest way is to make use of energy conservation. \begin{align} E=\frac{m}{2} ( \dot{r}^2 + r^2 \dot{\varphi}^2)+ U(r) \end{align} To this end we express $\dot{\varphi}$ in terms of $\mathbf{L}$ \begin{align} L=m \cdot r^2 \cdot \dot{\varphi}\\ \dot{\varphi}=\frac{L}{m \cdot r^2} \end{align} Now kinetic energy can be written as \begin{align} E=\frac{m \cdot \dot{r}^2}{2} + \frac{L^2}{2m r^2} + U(r)=\frac{m \cdot \dot{r}^2}{2}+U_{eff} \tag{1} \end{align} with an additional term that has a dependence on $r$. We define a new function \begin{align} U_{eff}=\frac{L^2}{2m r^2} + U(r) \end{align} and call it the effective potential $U_{eff}$. To find the equations of motion we rearrange (1) \begin{align} \frac{dr}{dt}=\sqrt{\frac{2}{m}[E-U(r)]-\frac{L^2}{m^2r^2}} \tag{2} \end{align} and integrate \begin{align} t=\int \frac{dr}{\sqrt{\frac{2}{m}[E-U(r)]-\frac{L^2}{m^2r^2}}}+const. \end{align} with the eqution $d\varphi = \frac{L}{mr^2} dt$ we find by plugging in $dt$ from (2) \begin{align} \varphi=\frac{M}{mr^2} \int \frac{dr}{\sqrt{\frac{2}{m}[E-U(r)]-\frac{M^2}{m^2r^2}}}+const. \end{align} \begin{align} \varphi= \int \frac{\frac{L}{r^2}dr}{\sqrt{2m[E-U(r)]-\frac{M^2}{r^2}}}+const. \end{align}