Past exam of the mathematics course of the University of Cambridge / 2002 / ia / Paper 4 / 9E (Dynamics)

For distinct positions and positive masses, Newton's law of universal gravitation givesThe proposed homothetic gravitational motion has ${\ddot{\mathbf{x}}}_i=-(2/9)t^{-4/3}{\mathbf{a}}_i$ and pair distances $t^{2/3}|{\mathbf{a}}_i-{\mathbf{a}}_j|$, so cancellation of $t^{-4/3}$ gives the time-independent equationsThese define a Newtonian central configuration with this choice of scale. Fordifferentiation in ${\mathbf{a}}_i$ gives ${\nabla }_{{\mathbf{a}}_i}\Phi =(2/9)m_i{\mathbf{a}}_i-G{\sum }_{j\ne i}{m}_i{m}_j({\mathbf{a}}_i-{\mathbf{a}}_j)/|{\mathbf{a}}_i-{\mathbf{a}}_j{|}^3=0$, exactly the displayed equations. The double sum in the paper counts each pair twice, so it agrees with this expression.

Multiply the configuration equations by $m_i$ and sum. Pair contributions cancel, giving ${\sum }_i{m}_i{\mathbf{a}}_i=0$. Thus the total linear momentum is $\mathbf{P}=(2/3)t^{-1/3}{\sum }_i{m}_i{\mathbf{a}}_i=0$. Each position is parallel to its velocity, so every $m_i{\mathbf{x}}_i\times {\dot{\mathbf{x}}}_i=0$ and the total angular momentum is zero. About any other fixed point $\mathbf{b}$, it is ${\mathbf{L}}_{\mathbf{b}}={\mathbf{L}}_O-\mathbf{b}\times \mathbf{P}=\boxed{0}$.