Three asteroids, located at points \(P_{1}, P_{2},\) and \(P_{3}\), which are not in a line, and having known masses \(m_{1}, m_{2}\), and \(m_{3}\), interact with one another through their mutual gravitational forces only; they are isolated in space and do not interact with any other bodies. Let \(\sigma\) denote the axis going through the center of mass of the three asteroids, perpendicular to the triangle \(P_{1} P_{2} P_{3} .\) What conditions should the angular velocity \(\omega\) of the system (around the axis \(\sigma\) ) and the distances $$ P_{1} P_{2}=a_{12}, \quad P_{2} P_{3}=a_{23}, \quad P_{1} P_{3}=a_{13} $$ fulfill to allow the shape and size of the triangle \(P_{1} P_{2} P_{3}\) to remain unchanged during the motion of the system? That is, under what conditions does the system rotate around the axis \(\sigma\) as a rigid body?

When studying the fascinating cosmos, we frequently encounter the mysterious pull of gravitational forces. Imagine this force as an invisible thread connecting masses across the vacuum of space. For celestial bodies like asteroids, the gravitational attraction they exert on each other depends on their masses and the distances between them. It's Newton's law of universal gravitation that gives us the formula:

\[ \mathbf{F}_{ij} = G \frac{m_i m_j}{|\mathbf{r}_i - \mathbf{r}_j|^2} \hat{\mathbf{r}}_{ij}, \]
where \( G \) is the gravitational constant, \( m_i \) and \( m_j \) are the masses, and \( \mathbf{r}_i \) and \( \mathbf{r}_j \) are the position vectors. The vector nature of gravitational forces means that they have both magnitude and direction, determining how asteroids orbit or influence each other's paths, and play a fundamental role in maintaining the geometric stability in a system of rotating celestial bodies.

In the dance of rotating asteroids, the center of mass is like the choreographer, guiding the motion. It's a unique point where we could, in theory, balance the entire system on a needle's tip if we could indeed interact with it. Computationally, it is the weighted average position of all the individual masses within the system. For our trio of asteroids, the point is given by:

\[ \mathbf{R}_{\text{CM}} = \frac{m_1\mathbf{r}_1 + m_2\mathbf{r}_2 + m_3\mathbf{r}_3}{m_1+m_2+m_3}, \]
where \( \mathbf{R}_{\text{CM}} \) represents the system's balance point. Simplifying complex systems to this singular point allows us to analyze rotational dynamics and stability more efficiently. In our context, maintaining the triangle's integrity relies on the center of mass remaining static in its position relative to the rotating asteroids.

Circling asteroids maintain their pace around the center of mass thanks to angular velocity, denoted as \( \omega \). This scalar value describes how fast the asteroids whirl around an axis and is pivotal for ensuring rigid body rotation. Think of it as the rotational equivalent of linear speed, where instead of covering 'distance per time', a body covers 'angle per time'. For uniform motion, keeping the asteroid triangle intact, our \( \omega \) needs precise tuning. The formula looks like this:

\[ \omega^2 = G \left(\frac{m_{1} m_{2}}{a_{12}^{3}} + \frac{m_{2} m_{3}}{a_{23}^{3}} + \frac{m_{1} m_{3}}{a_{13}^{3}}\right) \left(\frac{1}{m_{1}+m_{2}+m_{3}}\right) \]
Big 'G' there is again the gravitational constant, and the values of \(a_{ij}\) represent the sides of our asteroid triangle. It's this angular velocity that determines whether the system can move as one solid, without changing its original form.

Torque, in this interstellar context, measures the effectiveness of gravitational forces to cause rotational motion. Simply put, it's the force that causes the asteroids to spin. To avoid them clashing into a cosmic ballet of chaos, the torques applied must cancel out, so the system spins uniformly without deforming. We use the cross product for calculating torque for an asteroid:

\[ \mathbf{\tau}_{\text{net}_i} = \mathbf{r}_i \times \mathbf{F}_{\text{net}_i}. \]
Here, \( \mathbf{r}_i \) is the position vector of a given asteroid, and \( \mathbf{F}_{\text{net}_i} \) is the net gravitational force on that asteroid. Zero net torque, as in \( \mathbf{\tau}_{\text{net}_1} + \mathbf{\tau}_{\text{net}_2} + \mathbf{\tau}_{\text{net}_3} = \mathbf{0} \), is our golden ticket to rigid body rotation, ensuring our asteroids pirouette around the axis \( \sigma \) as if locked in a celestial embrace, their formation unchanged.