 Solution

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.