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A collision-free positive-mass configuration satisfies the displayed equation when every gravitational acceleration points proportionally toward a common centre. With $W=-{\sum }_{i<j}G{m}_i{m}_j/|{\mathbf{a}}_i-{\mathbf{a}}_j|$, homogeneity gives $\Lambda I=-W>0$, $I={\sum }_i{m}_i|{\mathbf{a}}_i{|}^2$. Pairwise force cancellation implies ${\sum }_i{m}_i{\mathbf{a}}_i=0$. Such a configuration generates a nonrotating homothetic gravitational motion by multiplying all position vectors by one positive scale factor.

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 ${\mathbf{L}}_{\mathbf{b}}={\mathbf{L}}_O-\mathbf{b}\times \mathbf{P}=\boxed{0}$.

For positive masses and distinct positions, Newtonian gravitational field givesInsert ${\mathbf{x}}_i=a(t){\mathbf{a}}_i$ with $a>0$. The separation numerator contributes a factor $a$ and the denominator $a^3$, soHence the time-dependent equations are satisfied wheneverThe fixed-shape condition is a Newtonian central configuration, and the resulting motion is a homothetic gravitational motion. It applies away from the collision scale $a=0$; a negative scale would require absolute values in the force scaling rather than the same positive-scale formula.

Multiplying the scalar equation by $\dot{a}$ givesThus its first integral is ${\dot{a}}^2/2-\Lambda /a=k/2$ for some real constant $k$.

In the pair potentialdifferentiating a pair with respect to ${\mathbf{a}}_i$ produces $G{m}_i{m}_j({\mathbf{a}}_i-{\mathbf{a}}_j)/|{\mathbf{a}}_i-{\mathbf{a}}_j{|}^3$. Every pair incident on $i$ contributes exactly once, so ${\nabla }_{i}W={\mathbf{G}}_i$. Moreover $W(\lambda {\mathbf{a}}_1,\dots ,\lambda {\mathbf{a}}_n)={\lambda }^{-1}W$ for $\lambda >0$. Differentiating with respect to $\lambda$ at one proves the relevant Euler homogeneous function theorem identity directly:Dot each fixed-shape equation with ${\mathbf{a}}_i$ and sum to obtainFor a nontrivial collision-free configuration with at least two positive masses, $W<0$ and $I>0$, and therefore $\Lambda >0$. Positivity requires an interacting pair; an isolated one-particle system is the degenerate zero-force exception rather than a positive-$\Lambda$ gravitational configuration. Pairwise cancellation also gives ${\sum }_i{\mathbf{G}}_i=0$, so the nondegenerate configuration has ${\sum }_i{m}_i{\mathbf{a}}_i=0$: its scaling origin is the centre of mass.

Finally, the system's kinetic energy is $I{\dot{a}}^2/2$ and its potential energy is $W/a$. ConsequentlyThis establishes the energy relation for every fixed configuration satisfying the algebraic condition, without assuming any particular particle arrangement.