IPhO 1989 · Theory problem 2

Three gravitating masses rotating as a rigid body

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Three non-collinear points \(P_1\), \(P_2\) and \(P_3\), with known masses \(m_1\), \(m_2\) and \(m_3\), interact with one another through their mutual gravitational forces only; they are isolated in free space and do not interact with any other bodies. Let \(\sigma\) denote the axis going through the center-of-mass of the three masses, and perpendicular to the triangle \(P_1P_2P_3\). What conditions should the angular velocities \(\omega\) of the system (around the axis \(\sigma\)) and the distances:

\[ P_1P_2=a_{12},\qquad P_2P_3=a_{23},\qquad P_1P_3=a_{13}, \]

fulfill to allow the shape and size of the triangle \(P_1P_2P_3\) unchanged during the motion of the system, i.e. under what conditions does the system rotate around the axis \(\sigma\) as a rigid body?

Official English text of the problem.

Answers

  • Problem 2: The triangle must be equilateral: \[ \boxed{a_{12}=a_{23}=a_{13}=a>0.} \] All three masses must rotate with the same constant angular velocity, whose magnitude satisfies \[ \boxed{\omega^2=\frac{G(m_1+m_2+m_3)}{a^3}.} \] Either sense of rotation is possible. These conditions are necessary and sufficient for any positive masses; the masses need not be equal.

Solution

Problem 2

Use Newton’s laws of motion and Newtonian gravitation for point masses. Define \[ M=m_1+m_2+m_3. \] There is no external force, so the center of mass moves uniformly. Work in an inertial frame in which it is at rest, and let \(\mathbf r_i\) be the position of \(P_i\) relative to it. Then \[ \sum_{i=1}^{3}m_i\mathbf r_i=\mathbf 0. \]

For rigid rotation about the perpendicular axis \(\sigma\), all three masses have a common angular velocity \(\boldsymbol\omega\). Their angular momentum about the center of mass is \[ \mathbf L=I\boldsymbol\omega, \qquad I=\sum_{i=1}^{3}m_i|\mathbf r_i|^2. \] The gravitational forces are internal and central, so their total torque vanishes and \(\mathbf L\) is conserved. Because a rigid triangle has constant \(I\), \(\boldsymbol\omega\) is constant. Consequently, \[ \ddot{\mathbf r}_i=-\omega^2\mathbf r_i. \]

Newton’s second law, after dividing by \(m_i\), therefore requires \[ -\omega^2\mathbf r_i = G\sum_{j\ne i} \frac{m_j}{a_{ij}^3}(\mathbf r_j-\mathbf r_i). \tag{1} \] The center-of-mass relation also gives \[ \sum_{j\ne i}m_j(\mathbf r_j-\mathbf r_i) = -M\mathbf r_i. \tag{2} \] Substituting (2) into the left-hand side of (1) and rearranging, \[ \sum_{j\ne i}m_j \left(\frac{G}{a_{ij}^3}-\frac{\omega^2}{M}\right) (\mathbf r_j-\mathbf r_i)=\mathbf 0. \tag{3} \]

At each vertex, the two vectors pointing toward the other vertices are linearly independent, since the three points are non-collinear. Thus both coefficients in (3) must vanish. With positive masses, this implies, for every pair, \[ \frac{G}{a_{ij}^3}=\frac{\omega^2}{M}. \] Hence \[ a_{12}=a_{23}=a_{13}=a, \qquad \omega^2=\frac{GM}{a^3}. \] This proves necessity.

To establish sufficiency, take an equilateral configuration of side \(a\). The gravitational acceleration of any mass is \[ \ddot{\mathbf r}_i = \frac{G}{a^3}\sum_{j\ne i}m_j(\mathbf r_j-\mathbf r_i) = -\frac{GM}{a^3}\mathbf r_i. \] Thus its resultant gravitational force points toward the center of mass and provides exactly the centripetal acceleration required when \[ \omega^2=\frac{GM}{a^3}. \]

Choose the initial velocities in the center-of-mass frame as \[ \dot{\mathbf r}_i(0) = \boldsymbol\omega\times\mathbf r_i(0), \qquad \boldsymbol\omega\parallel\sigma. \] The corresponding uniform rotation preserves the equilateral triangle and satisfies Newton’s equations at every time. The conditions are therefore sufficient as well. An arbitrary uniform motion of the center of mass may be added.

The solution was checked against the official solution.