Centre of mass moves with constant velocity

Question

The centre of mass of the Newton $n$-body problem is given by $$S=\frac{1}{M} \sum m_ix_i$$ with $M=\sum m_i$. Show that it moves with contant speed and hence has no acceleration.

I don't understand as if I differentiate, I'll surely just get $$S'=\frac{1}{M} \sum m_ix'_i$$ which is not constant...is it?

Answer 1

This system is closed, that is $\sum\overrightarrow{F}=\vec{0}=\sum m_i\ddot{x}_i=M\ddot{S}$ which implies $\dot{S}=constant$

Answer 2

You need to differentiate twice to get the acceleration $\ddot{S}$(by doing it only once, you get the velocity $\dot{S}$, which will be constant — not necessarily $0$); once this is done, you should also use the expression of the forces (which for an isolated $M$-body system are only the pairwise gravitation forces between the $M$ bodies) to show that $\dot{S}=0$.

Edit: metacompactness' solution above is more straightforward and elegant.