Let's take a closed system of point particles with only pair-particle interactions at a distance, i.e., the paradigmatic Newtonian model. Then you have
$$m_i \ddot{\vec{x}_i}=\sum_{j \neq i} \vec{F}_{ij}(\vec{x}_i-\vec{x}_j).$$
Thanks to Newton's Lex III we have
$$\vec{F}_{ij}=-\vec{F}_{ji}, \quad F_{ii}=0.$$
This implies
$$\sum_i m_i \ddot{\vec{x}_i} = \sum_j \sum_{i \neq j} \vec{F}_{ij} = \frac{1}{2} \sum_j \sum_{i \neq j} [\vec{F}_{ij} +\vec{F}_{ji}]=0,$$
i.e., the center of mass,
$$\vec{R}=\frac{1}{M} \sum m_i \vec{x}_i, \quad M=\sum_i m_i$$
moves like a free particle,
$$M \ddot{\vec{R}}=0.$$
If you have in addition a homogeneous gravitational field like close to the Earth, then you add
$$m_i \ddot{\vec{x}}_i=\sum_{j \neq i} \vec{F}_{ij}+m_i \vec{g},$$
and then for the center of mass you have
$$M \ddot{\vec{R}}=\vec{g} \sum_i m_i = M \vec{g},$$
i.e., the center of mass is freely falling.
That's what can be said with really simple math about a Newtonian system of point particles that's closed or one in an external homogeneous gravitational field.
For the special case of a closed two-particle system, you can reduce the problem to the free motion of the center of mass and an effective one-body problem moving in an external force field by simply introducing ##\vec{R}## and the relative coordinates ##\vec{r}=\vec{r}_1-\vec{r}_2##. Then you get
$$M \ddot{\vec{R}}=0, \quad \mu \ddot{\vec{r}}=\vec{F}_{12}(\vec{r}),$$
where ##\mu=m_1 m_2/M## is the reduced mass.
The case of a rigid body is treated in my above quoted manuscript:
https://itp.uni-frankfurt.de/~hees/pf-faq/spinning-top.pdf
using, however, the Lagrangian form of the Hamilton principle.
