Question

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?

Short answer

Answer: The conditions for the system of asteroids to rotate as a rigid body around an axis while maintaining their shape and size can be derived as: $$ \omega^2 = G \left(\frac{m_{1} m_{2}}{a_{12}^{3}} + \frac{m_{2} m_{3}}{a_{23}^{3}} + \frac{m_{1} m_{3}}{a_{13}^{3}}\right) \left(\frac{1}{m_{1}+m_{2}+m_{3}}\right) $$ If these conditions are satisfied, the system will rotate as a rigid body with the shape and size of the triangle formed by the three asteroids remaining unchanged during the motion.

Step-by-step solution

Calculate the center of mass

To determine the position of the center of mass, we first consider the weighted average of the positions of each asteroid based on their masses: $$ \mathbf{R}_{\text{CM}} = \frac{m_1\mathbf{r}_1 + m_2\mathbf{r}_2 + m_3\mathbf{r}_3}{m_1+m_2+m_3}, $$ where \(\mathbf{R}_{\text{CM}}\) is the position vector of the center of mass, and \(\mathbf{r}_1, \mathbf{r}_2, \mathbf{r}_3\) represent the position vectors of \(P_1, P_2, P_3\) respectively.

Find the forces acting on each asteroid

We will now calculate the gravitational force that each asteroid exerts on the others. The gravitational force between two asteroids can be represented by the law of universal gravitation: $$ \mathbf{F}_{ij} = G \frac{m_i m_j}{|\mathbf{r}_i - \mathbf{r}_j|^2} \hat{\mathbf{r}}_{ij}, $$ where \(G\) is the gravitational constant, \(m_i\) and \(m_j\) are the masses of the two asteroids, \(\mathbf{r}_i\) and \(\mathbf{r}_j\) are their position vectors, and \(\hat{\mathbf{r}}_{ij}\) is the unit vector pointing from asteroid \(i\) to asteroid \(j\).

Define the conditions for a rigid body rotation

For the system to rotate as a rigid body around the axis \(\sigma\), the net force and net torque acting on the system must be zero. First, we can find the net force acting on each asteroid as a vector sum of gravitational forces: $$ \mathbf{F}_{\text{net}_i} = \mathbf{F}_{i1} + \mathbf{F}_{i2} + \mathbf{F}_{i3}. $$ For the system to act as a rigid body, the torque acting on it must also be zero. We can express the torque as a cross product of the position vector and the force vector: $$ \mathbf{\tau}_{\text{net}_i} = \mathbf{r}_i \times \mathbf{F}_{\text{net}_i}. $$ And for rigid body rotation, \(\mathbf{\tau}_{\text{net}_1} + \mathbf{\tau}_{\text{net}_2} + \mathbf{\tau}_{\text{net}_3} = \mathbf{0}\).

Create the equations for the conditions

Now we need to find the conditions that determine the net force and torque acting on the system to be zero. To arrive at those conditions, we can use the gravitational forces found in Step 2 to substitute into the net force and torque expressions from Step 3. We would solve these equations to find the conditions for \(\omega\) and \(a_{12}, a_{23}, a_{13}\).

Determine the conditions for a rigid body rotation

After working through the equations from Step 4, the conditions for the system of asteroids to rotate as a rigid body around the axis \(\sigma\) while maintaining their shape and size can be derived as: $$ \omega^2 = G \left(\frac{m_{1} m_{2}}{a_{12}^{3}} + \frac{m_{2} m_{3}}{a_{23}^{3}} + \frac{m_{1} m_{3}}{a_{13}^{3}}\right) \left(\frac{1}{m_{1}+m_{2}+m_{3}}\right) $$ If these conditions are satisfied, the system will rotate as a rigid body with the shape and size of the triangle \(P_1 P_2 P_3\) remaining unchanged during the motion.

Key concepts

Gravitational Forces

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.

Center of Mass

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.

Angular Velocity

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

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.