It is entirely legitimate to ask the following question: Where is the centre of gravity of a group of bodies? If many people are on a raft, its stability will depend on the location of their (together with the raft’s) centre of gravity.

The meaning of this concept remains the same. The centre of gravity is the point of application of the sum of the gravitational forces of all the bodies in the group under consideration.

We know the result of the computation for two bodies. If two bodies of weights \(F_{1}\) and \(F_{2}\) are located at a distance \(x\) from each other, their centre of gravity is situated at distances \(x_{1}\) from the first body and \(x_{2}\) from the second, where \[x_{1} + x_{2} = x \quad \textrm{and} \quad \dfrac{F_{1}}{F_{2}} = \dfrac{x_{2}}{x_{1}}\] Since weight may be represented as a product \(mg\), the centre of gravity of the pair of bodies satisfies the condition \[m_{1}x_{1} = m_{2}x_{2}\] i.e. lies at the point which divides the distance between the masses into segments inversely proportional to the masses.

Let us now recall the firing of a gun attached to a platform. The momenta of the gun and the shell are equal in magnitude and opposite in direction. The following equalities hold: \[m_{1} v_{1} = m_{2} v_{2} \quad \textrm{or} \quad \dfrac{v_{2}}{v_{1}} = \dfrac{m_{1}}{m_{2}} \label{ellipse}\] where the ratio of the speeds retains this value during the entire interaction. In the course of the motion arising as a result of the recoil, the gun and the shell are displaced with respect to their initial positions by distances \(x_{1}\) and \(x_{2}\) in opposite directions. The distances \(x_{1}\) and \(x_{2}\)—the paths covered by the two bodies—increase, but for a constant ratio of speeds, they will also be in the same ratio to each other all the time: \[\dfrac{x_{2}}{x_{1}} = \dfrac{m_{1}}{m_{2}} \quad \textrm{or} \quad x_{1} m_{1} = x_{2} m_{2}\] Here \(x_{1}\) and \(x_{2}\) are the distances of the gun and the shell from their original positions. Comparing this formula with the formula determining the position of the centre of gravity, we observe their complete identity. It immediately follows from this that the centre of gravity of the gun and the shell remains at its original position all the time after the firing.

In other words, we have arrived at the very interesting result—the centre of gravity of the gun and the shell remains stationary after the firing.

Such a conclusion is always true: if the centre of gravity of two bodies was initially stationary, their interaction, regardless of its nature, cannot change the position of the centre of gravity. This is precisely why it is impossible to pick oneself up by the hair or pull oneself up to the Moon by the method of the French writer Cyrano de Bergerac, who proposed (jokingly, of course) to this end that one threw a magnet upwards while holding a piece of iron which would be attracted by the magnet.

A stationary centre of gravity is moving uniformly from the point of view of a different inertial frame of reference. Hence, a centre of gravity is either stationary or moving uniformly and rectilinearly.

What we have said about the centre of gravity of two bodies is also true for a group of many bodies. Of course, for an isolated group of bodies; this is always stipulated when we are applying the law of conservation of momentum.

Consequently, every group of interacting bodies has a point which is stationary or is moving uniformly, and this point is their centre of gravity.

To emphasize the new property of this point, we give it an additional name: the centre of mass. As a matter of fact, the question of, say, the weight of the solar system (and hence its centre of gravity) can have only a hypothetical meaning.

No matter how the bodies forming a closed group move, the centre of mass (gravity) will be stationary or, in another frame of reference, will move by inertia.