# Motion of the center of mass of rigid bodies in space

For the classic two body problem, I know that the motion of the center of mass is a straight line (with respect to an inertial frame), provided that the bodies are considered as point particles.

Now suppose you have two rigid bodies, each consisting of $$N_1$$ and $$N_2$$ point particles respectively. Again let them interact gravitationally. Is the motion of the center of mass of the two bodies still an exact straight line or not? My simulation shows that it is not a straight line, but I doubt...

Note that during the gravitational interaction, I am assuming that the center of mass of the body $$1$$ is attracted by $$N_2$$ particles, while the center of mass of the body $$2$$ is attracted by $$N_1$$ particles.

• I think the last paragraph does a double counting. Once a center of mass is defined with the total mass and momentum of each ensemble, one ignores the distribution within each. The motion of the center of mass has to be a straight line rom conservation of momentum. Even in the non inertial frame of the earth moon system, once the center of mass is defined one ignores tides etc as far as the system sun earth.It is the center of mass that is plotted in the eliptical orbit. Commented Dec 13, 2018 at 19:59

In the absence of external forces, the motion of the centre of mass of the whole system will be a straight line. Total momentum must be conserved, and this may be written as the total mass times the centre-of-mass velocity vector; or, equivalently, as the sum of the centre-of-mass momenta of the two composite rigid bodies: $$M \dot{\mathbf{R}} = \sum_j m_j \dot{\mathbf{r}_j} = \sum_{j\in 1} m_j \dot{\mathbf{r}_j} + \sum_{j\in 2} m_j \dot{\mathbf{r}_j} = M_1\dot{\mathbf{R}}_1 + M_2\dot{\mathbf{R}}_2 = \text{constant}$$ where $$M=\sum_j m_j$$, $$M_1=\sum_{j\in 1} m_j$$, $$M_2=\sum_{j\in 2} m_j$$, and we have defined the centre of mass positions $$\mathbf{R} = \frac{1}{M}\sum_j m_j \mathbf{r}_j , \quad \mathbf{R}_1 = \frac{1}{M_1}\sum_{j\in 1} m_j \mathbf{r}_j , \quad \mathbf{R}_2 = \frac{1}{M_2}\sum_{j\in 2} m_j \mathbf{r}_j .$$
• Sorry for the delay. Indeed since there are no external forces, the COM shall follow a straight line. Anyway I found that the problem arose from a logical error in my program. As far as what you said about the gravitational interaction between each pair of point particles, this is true, but you can skip calculations (the double sum), by assuming that the motion of the whole rigid body is determined by the sum of forces acted upon its COM only. So if the body 2 consists of $N_2$ particles, then $N_2$ forces will pull the COM of body 1. The same applies for the body 1.