That "simple" question is interesting, since it is pretty basic and can be answered at various different levels.
. Let me stay at a similar one as Grant used and just add a few further comments...
Firstly, the bodies with position vectors [tex]\vec{x_i}[/tex] and masses [tex]m_i[/tex] are supposed to satisfy
Newtons equations of motion. The first thing to do next, is to change variables: along with the
relative positions [tex]\vec{x_i} -\vec{x_j},\ i\ne j[/tex], let us introduce the
center-of-mass (=barycenter) position vector,
[tex]\vec{R} = \sum_i m_i\vec{x_i}/M[/tex],
where the
total mass [tex]M=\sum_i m_i[/tex]
of the system appears to be concentrated! In fact, one can readily derive from the original equations of the system that the center-of-mass vector satisfies the following Newton equation of motion:
[tex]\frac{d^2\, \vec{R}}{dt^2} =\frac{\vec{F}_{ext}}{M}[/tex]
an expression of Newton's second law, which states that the center-of-mass at [tex]\vec{R}(t)[/tex] moves as though it possessed the
total mass of the system and were acted upon by the total external force [tex]\vec{F}_{ext}[/tex]. Consequently, if [tex]\vec{F}_{ext}=0[/tex] then the center-off mass moves with constant velocity, as we all know.
Concerning the acceleration [tex]\frac{d^2\, \vec{R}}{dt^2}[/tex] of the center-of-mass, it's indeed the
sum of the masses, NOT their product that matters.
For a simple two-body gravitating system, for example, one encounters besides the total mass [tex]M=m_1+m_2[/tex] (that comes with the center-off-mass motion), the so-called
reduced mass,
[tex]\frac{1}{m_{red}} = \frac{1}{m_1} +\frac{1}{m_2},[/tex]
or equivalently
[tex]m_{red} = \frac{m_1\,m_2}{M},[/tex]
such that the product of masses as appearing in the force between the two bodies, factors once more,
into the product of the reduced mass and the total mass,
[tex]F =\frac{G m_{red}\, M}{r^2}[/tex]
Fridger
).
