March 30, 2022
A central force is the one in which the force between two particles is solely a function of distance between them, thus
$$ \vec F_{Central}=f(r)\hat r\tag{6.1} $$
Consider a system of two masses $m_1$ and $m_2$ at points $\vec r_1$ and $\vec r_2$ from the origin, the only force between them being a central one, thus
$$ \vec r=\vec r_1-\vec r_2\\ \\m_1\ddot{\vec r}_1=f(r)\hat r\\ \\m_2\ddot{\vec r}_2=-f(r)\hat r\tag{6.2,\quad6.3,\quad6.4} $$
Where a minus sign comes in equation (6.4) due to conservation of momentum.
The center of mass of the system is
$$ \vec R=\frac{m_1\vec r_1+m_2\vec r_2}{m_1+m_2}\tag{6.5} $$
Since there are no external forces, $\ddot{\vec R}=0$ which on integration gives
$$ \vec R=\vec R_0+\vec Vt\tag{6.6} $$
If we let the origin at the center of mass and make it stationary then the above equation becomes $0$.
From equations (6.2-4) we get
$$ \ddot{\vec r}_1-\ddot{\vec r}_2=\Big(\frac 1 m_1+\frac 1 m_2\Big)f(r)\hat r \\ \\ \mu\ddot{\vec r}=f(r)\hat r\tag{6.7,\quad 6.8} $$
Where $\mu$ is the reduced mass.
Thus the problem of 2 bodies has been reduced to a one-body problem of an object with a mass of $\mu$.
Since the force is radial, there are no torques and hence angular momentum is conserved. Therefore, $\vec L=\vec r\times\mu\dot{\vec r}=$ a constant. Now since $\vec r\perp\vec L$, and $\vec L$ cant change the direction, it follows that the motion of the object lies in a plane.