Is there a singularity theorem in higher-dimensional Newtonian gravity? 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.
 A: This should follow from the following Virial-type 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.)
A: Following the lines  Willie followed,   but allowing for unequal  masses $m_i$, set $I(x) = \langle x, x \rangle $ where $\langle v, w \rangle = \Sigma m_i v_i \cdot w_i$ is the so-called mass metric built so that the kinetic energy is $K =\frac{1}{2}\langle v, v \rangle$ with $v = \dot x$. Here $x$ and $v$ lie in N copies of a d-dimensional Euclidean space.  Then Newton's equations read $\ddot x = - \nabla V (x)$ where the gradient $\nabla$ is with respect to the mass metric.  Suppose now that the potential $V$ is homogeneous of degree $-\alpha$. Using  Euler's identity $-\alpha V (x) = \langle x, \nabla V (x) \rangle$   a fun  elementary computation yields the  virial identity  $\ddot I = 4H +(2 \alpha - 4) V$, where $H = K +V$ is the total energy, which is conserved.  (This identity is called by mathematical celestial mechanicians the  Lagrange-Jacobi identity.)   Now for  $d$-dimensional ``Newtonian' gravity we have  $V < 0$ and $\alpha = d-2$ so that Lagrange-Jacobi reads $\ddot I = 4H +(8-4d)U$, with $U = -V > 0$.  Conclusion: for $d > 4$ and $H \le 0$ we have $\ddot I < 0$ and thus all such solutions either begin and end in total collision $I =0$,
(or have singularities before these total collision times),   as you conjectured.  (Asides.  If $d< 4$  then the virial identity implies  that $\ddot I > 0$ for $H \ge 0$ and consequently bounded solutions  have  $H < 0$. The case  $d =4$ is a wonderful bounding case between these  extremes. At the critical dimension $4$ solutions must lie on  spheres $I = const.$ and have energy  $H =0$  in order to be bounded.)
