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It is proved that the two body problem with central potential \(V(|\vec{r}_1-\vec{r}_2|)\) can be reduced to one body problem. In the reduced problem the body has reduced mass and moves in spherically symmetric potential \(V(r)\). In this case the center of mass moves like a free particle.
Reduction of Two Body Problem
Consider two particles interacting through a potential which depends only on the distance between the two. The Lagrangian can be written as
\begin{equation}\label{EQ01}
\mathcal L = \frac{1}{2} m_{1} \dot{\vec{r}}^{\,2}_{1} + \frac{1}{2} m_{2}
\dot{\vec{r}}^{\,2}_{2} - V(| \vec{r_2} - \vec{r_1}|).
\end{equation}
We now introduce the position vector of the centre of mass $\vec{R}$, and the relative coordinate $\vec{r}$ by means of
\begin{equation}\label{EQ02}
\vec{R} = \frac{m_{1} \vec{r}_{1} + m_{2}\vec{r}_{2}}{(m_{1} + m_{2})}, \qquad
\vec{r} = (\vec{r_{1}} - \vec{r_{2}}),
\end{equation}
and express the Lagrangian in terms of the new variables takes the form \begin{equation}\label{EQ03}
{\mathcal L }= \frac{1}{2} M \dot{\vec{R}}^{\,2} + \frac{1}{2} \mu \dot{\vec{r}}^{2} -
V(r).
\end{equation}
where \(M\) is the total mass, and $\mu$ is reduced mass.
\begin{equation}\label{EQ04}
M = m_{1} + m_{2}, \qquad \mu = \frac{m_{1}m_{2}}{m_{1} +
m_{2}}.
\end{equation}
Since components of \(\vec{R}\) are cyclic coordinates, the equations of motion for $\vec{R}$ are simple and just state that the center of mass moves with constant velocity. The rest of the terms in the Lagrangian describe motion of a particle of mass $\mu$ in a potential \(V(r)\). The potential \(V(r)\) depends only on $r = |\vec{r}|$ and is independent of $\theta$, $\phi$. Such potentials are called spherically symmetric potentials.
