# Newton's Third law of Motion, who can tell me how to deduct below?

The position vector of the centre of mass is defined as: $$\mathbf {r_{cm}}=\frac{\sum_{i=1}^{N}m_i \mathbf {r_i}}{M}$$ i.e., $$M \mathbf {r_{cm}}=m_1 \mathbf r_1+m_2 \mathbf r_2+ m_3 \mathbf r_3+...+m_N \mathbf r_N$$ $$\Rightarrow M \frac {d^2 \mathbf {r_{cm}}}{dt^2} = m_1 \frac {d^2 \mathbf {r_{1}}}{dt^2}+ m_2 \frac {d^2 \mathbf {r_{2}}}{dt^2}+...+m_N \frac {d^2 \mathbf {r_{N}}}{dt^2}$$ (Double Differentiating both sides) Here $$M=\sum_{i=1}^{N}m_i$$. Since single differentiation of the position vector gives velocity $$\mathbf v$$ and double differentiation gives acceleration $$\mathbf a$$. Therefore $$\frac {d^2 \mathbf {r_{cm}}}{dt^2} = \mathbf a$$ This means that $$\sum_{i=1}^{N} m_i \frac {d^2 \mathbf {r_i}}{dt^2} = M \frac {d^2 \mathbf {r_{cm}}}{dt^2} = M \mathbf a = \mathbf {F_{net}}$$

Now if the net force acting on the object is $$\mathbf 0$$ then $$M \mathbf a= M\frac {d^2 \mathbf {r_{cm}}}{dt^2}=\mathbf 0$$

Note that internal forces cannot cause acceleration as they always come in action-reaction pair.

Try this: take the definition

$$\vec {r_{cm}}=\frac{\sum_{i=1}^{N}m_i \vec {r_i}}{\sum_{i=1}^{N}m_i},$$

multiply both sides by $$M$$, and differentiate the equation twice with respect to time.