In classical Newtonian gravity with 3 spatial dimensions, it's hard to get two particles to exactly collide, since at short distance the centrifugal force (~1/$r^3$) beats the gravitational attraction (~$1/r^2$). As a consequence, two particles can collide only if the angular momentum is exactly zero, which is measure zero.


In Newtonian gravity with $d$ spatial dimensions, the centrifugal force still goes like ~$1/r^3$ but the gravitational attraction goes like ~$1/r^{d-1}$. This means that for $d>4$, if two particles have negative total energy, they will collide regardless of their angular momentum. For two particles, having negative total energy is a sufficient condition to guarantee a collision after finite time. 


My question is whether this is still true with more than two particles? 


Conjecture: in $d>4$-dimensional Newtonian gravity, a collection of any number of point particles with negative total energy will experience a "singularity" (a distance=0 collision of at least two particles) within a finite time.



Is there a singularity theorem in higher-dimensional Newtonian gravity?
This should follow from the following [Virial-type](https://en.wikipedia.org/wiki/Virial_theorem) computation. 


For convenience we assume all particles have the same mass; this is not essential. 


Let $x\_i$ denote the position vector of the $i$th particle, then Newton's law of universal gravitation, suitably normalized, gives


$$ \ddot{x\_i} = (d-2) \sum\_{j \neq i} \frac{ x\_j - x\_i}{|x\_j - x\_i|^d } $$


This gives


$$ x\_i \cdot \ddot{x\_i} = (d-2) \sum\_{j \neq i} \frac{ x\_j \cdot x\_i - |x\_i|^2 }{|x\_j - x\_i|^d} $$


summing over all particles we get


$$ \sum\_{i} x\_i \cdot \ddot{x\_i} = (d-2) \sum\_{i,j, i\neq j} \frac{ x\_j \cdot x\_i - |x\_i|^2 }{|x\_j - x\_i|^d} = -\frac{d-2}{2} \sum\_{i,j, i\neq j} |x\_i - x\_j|^{2-d} \tag{1}$$


Conservation of energy, on the other hand, states that


$$ - \sum\_{i,j, i\neq j} |x\_i - x\_j|^{2-d} + \sum\_i |\dot{x\_i}|^2 = C < 0 \tag{2}$$


(by negative energy assumption). Summing (1) and (2) and using that $d \geq 4$ so that $(d-2)/2 \geq 1$ we get 


$$ \frac{d^2}{dt^2} \sum\_{i} |x\_i|^2 < 0 $$


Which shows that $\sum |x\_i|^2$ must impact $0$ at some finite time (either future or past). 


(Note: by "dropping particles at infinity" and using time-symmetric property of Newtonian mechanics we see that there exists trajectory where there is only one time of impact in the past, and none in the future.)


