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Although it seems intuitive, I haven't found proof of this fact in my textbooks from a purely mathematical standpoint. I was wondering if anybody knows of a particularly elegant proof of this fact, and if so would they be kind enough to share a link?

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marked as duplicate by Kyle Kanos, Aaron Stevens, Jon Custer, Qmechanic Feb 21 at 17:26

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You may generalize to an arbitrary number of masses $m_j$ at positions $\vec{r}_j$. The center of mass is defined as $$ \vec{R} = \frac{\sum_j \vec{r}_j m_j}{M}, $$ where $$ M = \sum_j m_j.$$ Its time derivative is $$ \vec{V} = \frac{d\vec{R}}{dt} = \frac{\sum_j \vec{v}_jm_j}{M} = \frac{\sum_j \vec{p}_j}{M} \Rightarrow \vec{P} = M\vec{V} = \sum_j\vec{p}_j,$$ where $\vec{p}_j$ is the momentum of mass $j$, and $\vec{P}$ is the momentum of the center of mass. The time derivative of the last equation gives $$ \frac{d\vec{P}}{dt} = \sum_j\frac{d\vec{p}_j}{dt}.$$ According to Newton's second law we find $$ \frac{d\vec{P}}{dt} = \sum_j\vec{F}_j,$$ where $\vec{F}_j$ is the net force acting on mass $j$. This net force can be written as $$ \vec{F}_j = \sum_{i}\vec{F}_{ij}^{\mathrm{(int)}} + \vec{F}_j^{\mathrm{(ext)}},$$ where $ \vec{F}_{ij}^{(int)}$ is the force exerted by mass $i$ on mass $j$ (internal force), whereas $\vec{F}_j^{\mathrm{(ext)}}$ is the net force on mass $j$ by forces that are external to the system of masses. Inserting this gives $$ \frac{d\vec{P}}{dt} = \sum_{ij}\vec{F}_{ij}^{\mathrm{(int)}} + \sum_j\vec{F}_j^{\mathrm{(ext)}}.$$ According to Newton's third law (actio = reactio) we have $$ \vec{F}_{ij}^{\mathrm{(int)}} = -\vec{F}_{ji}^{\mathrm{(int)}},$$ and therefore $$ \sum_{ij}\vec{F}_{ij}^{\mathrm{(int)}} = 0.$$ On the other hand, $$ \sum_j\vec{F}_j^{\mathrm{(ext)}} = \vec{F}^{\mathrm{(ext)}}$$ is the total external force acting on the system of masses. As a result $$ \frac{d\vec{P}}{dt} = \vec{F}^{\mathrm{(ext)}},$$ i.e., the change in time of the momentum of the center of mass is equal to the total external force.

If the masses are isolated, as you ask in your question, $\vec{F}^{\mathrm{(ext)}}=0$, and therefore the center of mass moves with constant momentum (i.e., constant velocity). If you choose the center of mass to be the origin of your frame of reference in which you describe the motion of the masses, then this point remains fixed, and the masses may orbit around this point under the influence of the internal forces, if appropriate initial conditions are given.

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I don't think there can be a simple proof. The reason is that by "orbit the center of mass" we mean that each orbit is an ellipse with the center of mass at one focus. This is nontrivial, and requires working out the full Kepler problem. For example, two masses joined by a spring also describe ellipses, but with the center of mass at the center.

In short, the problem is one of definitions: to prove that the masses orbit their center of mass you need to define mathematically what that means, and we already have two situations where it means slightly different things.

The closest you can get is to observe that in a frame where the center of mass is at rest, the two masses are always at opposite sides of it, and maintaining a constant proportion between their distances to it.

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