Why does the motion of the center of mass model the motion of the entire system of particles? A theorem in my physics textbook says:

The overall translational motion of a system of particles can be
analyzed using Newton’s laws as if all the mass were concentrated at
the center of mass and the total external force were applied at that
point.

Why does the translatory motion of the center of mass model the motion of the entire system? Is it something we simply observed or is there a deeper reason?
 A: Consider a system with $N$ point particles each has mass $m_{i}$ for $i = 1, 2, ..., n$. And each point particle is under an influence of a net force of $\vec{F}_{i}$ for $i = 1, 2, ..., n$.
Applying Newton's Second Law on each point particle, we have
\begin{align}
    \vec{F}_{i} \; = \; m_{i} \vec{a}_{i} \qquad i = 1, 2, ..., n \tag{1}
\end{align}
where $\vec{a}_{i}$ is the acceleration of the $i$-th point particle.
Hence, the net force for the whole system $\vec{F}_{sys}$ is simply summing up all the net forces acting on each point particle in the system
\begin{align}
    \vec{F}_{sys} \; &= \; \sum_{i=1}^{N} \vec{F}_{i} \\
&= \sum_{i=1}^{N} m_{i} \vec{a}_{i} \\
&= \sum_{i=1}^{N} m_{i}  \frac{d^{2}\vec{r}_{i}}{dt^{2}}  \qquad (\mathrm{definition \; of \; acceleration})\\
&= \frac{d^{2}}{dt^{2}} \Big( \sum_{i=1}^{N} m_{i}  \vec{r}_{i} \Big)\\
&= M \; \frac{d^{2}}{dt^{2}} \Big( \underbrace{\frac{1}{M} \sum_{i=1}^{N} m_{i}  \vec{r}_{i}}_{\vec{R}} \Big)\\
&= M \frac{d^{2}\vec{R}}{dt^{2}} \tag{2}
\end{align}
where $\vec{r}_{i}$ is the position vector of the $i$-th point particle. $M \equiv \sum_{i=1}^{N} m_{i}$ is the mass of the whole system, and $\vec{R}$ is known as the centre of mass of this system.
Look at Equation (2), it has the same mathematical form as Equation (1). Hence, the system translational motional, can be thought of as a point particle motion with total mass $M$ concentrated at the position $\vec{R}$ (centre of mass)
A: *Since I have joined recently I would really appreciate it if more experienced users edit my sloppy english.
First of all it does not model all the possible motions of the system for example the familiar case of rigid bodies in which you cannot know the behavior of the system simply by looking at the CM's position vector ,so it's important to emphasize the word translational .
Consider a system of particles (e.g. N particles) and we call the mass of the ith particle  mi .
Now assume that  we are looking at the system from an inertial frame , therefore for each particle we have $\vec{F}{_{i}}^{(tot)} = m{_i}{\vec{a}}{_i}$ in which $\vec{F}{_{i}}^{(tot)}$ is the total force acted on the ith particle , now this force can be separated into two terms :

*

*The force from all the particles inside our system which we will call
internal forces and we will show the force on the ith particle due to the jth particle by $\vec{f}{_i}{_j}$ and the
net internal force on the ith particle (just sum over j except for the
case i=j)  by  $\vec{f}{_i}$

*And the net of other forces which we call  external forces by $\vec{F}{_i}^{(e)}$ .

So now for each particle (any desired i) we have : $\vec{F}{_i}^{(e)}$ + $\vec{f}{_i}$ = $m{_i}\ddot{\vec{r}}{_i}$
in which $\vec{r}{_i}$  is the positon vector of the ith particle and the double dot presents two  time derivatives therefore acceleration of $m{_i}$ .
Now we sum over i  in the previous equation therefore the left side of our equality  will be sum of all forces in our system ,which is the mutual forces between all the particles and the net external force acting on our system ,if Newton's third law applies, i.e : $\vec{f}{_i}{_j}= - \vec{f}{_j}{_i}$.
The sum over internal forces will be zero and we are left with:  $\sum_{i=1}^{N}\vec{F}{_i}^{(e)} = m{_1}\ddot{\vec{r}}{_1}+m{_2}\ddot{\vec{r}}{_2}+...+m{_N}\ddot{\vec{r}}{_N}$
Now we have seen that in experience objects act like we have:
(the net force on them) =(mass)(acceleration)
*Note that Newton's laws were written for a point particle so this is not necessarily obvious.
now in our equation the left side is just like above and now if we sort of define the right side to be
$M\ddot{\vec{R}}{_c}{_m}$ in which M is the total mass of the system and then find $\vec{R}{_c}{_m}$ (to find that special point) , it can be explained(interpreted(?))in this way :
"The overall translational motion of a system of particles can be analyzed using Newton’s laws as if all the mass were concentrated at the center of mass and the total external force were applied at that point."
*For a better discussion of the above you can look at classical dynamics of particles and systems by Marion & Thornton - chapter 9 , the first 5 or 6 pages.
