General Solution for the Gravity Between Two 3D Triangles I would like to find the general solution for the gravity between two (flat) triangles in 3D, including the location $(x,y,z)$ where this force should be applied (in order to later account for rotational forces.)
Given two triangles $t_1$ and $t_2$, which are defined by the points $t_{1v_1}$, $t_{1v_2}$, $t_{1v_3}$, $t_{2v_1}$, $t_{2v_2}$, and $t_{2v_3}$.  Each of these points has $x$, $y$, and $z$ coordinates.  For example, $t_{1v_2}y$ would be triangle 1 $\to$ vertex 2 $\to$ y coordinate.
Note: This will ultimately be used in the physics engine for a 3D space game (used with JBullet).

My attempt at a solution:
The formula for gravity for point masses is:
$$F_{\text{gravity}}=G\frac{\text{mass}_1 \times \text{mass}_2}{\text{radius}^2}$$
Which is easily expanded out to:
$$F_{\text{gravity}}=G\frac{\text{mass}_1 \times \text{mass}_2}{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$$
With unit direction:  *which is not considered in the following equations, but ultimately needs to be
$$\text{dir} = \text{normal}(\text{point}_1-\text{point}_2)$$

We must consider the entirety of each triangle:
$$\int_0^1\int_0^1(t_{1v_3}x \times q+t_{1v_2}x \times (1-q)) \times r+t_{1v_1}x \times (1-r)\ d_q\ d_r$$



By combining the equation for gravity with the integral of a single variable over a triangle, we get:
$$\int_0^1\int_0^1\int_0^1\int_0^1 \left[ \frac{(\text{mass}_1 \times \text{mass}_2 \times G)}{(((((t_{1v_3}x \times q+t_{1v_2}x \times (1-q)) \times r+t_{1v_1}x \times (1-r))-((t_{2v_3}x \times s+t_{2v_2}x \times (1-s)) \times t+t_{2v_1}x \times (1-t)) ))^2+((((t_{1v_3}y \times q+t_{1v_2}y \times (1-q)) \times r+t_{1v_1}y \times (1-r))-((t_{2v_3}y \times s+t_{2v_2}y \times (1-s)) \times t+t_{2v_1}y \times (1-t)) ))^2+((((t_{1v_3}z \times q+t_{1v_2}z \times (1-q)) \times r+t_{1v_1}z \times (1-r))-((t_{2v_3}z \times s+t_{2v_2}z \times (1-s)) \times t+t_{2v_1}z \times (1-t)) ))^2 )} \right] \ d_q\ d_r\ d_s\ d_t $$

Is this correct?  What is the final solution? (including the position and/or angle of the force)  Alternatively... Is this problem unsolvable?
 A: This computation is essentially the same as computing the "Form Factor" in rendering (for computer graphics). It turns out that the integrals (once you include the direction of the force as you should) are not nice (i.e., involve dilogarithms). Here's a paper that describes the computation in some detail. 
A: I'm also interested in whether this integral has a nice closed form (I'm a bit skeptical). It will almost certainly involve choosing a clever set of coordinates (such as slicing one triangle intro strips parallel to the other) rather than brute-force barycentric interpolation. When searching for a solution to this integral, you may want to also search for the net Coulomb interaction between two triangular plates, and/or the integral of the gradient of the Green's function of the Laplacian over two triangles (which might have been addressed somewhere in the Boundary Element Method (BEM) literature).
For your purposes I'm going to suggest an alternative approach, though: instead of trying for an exact formula, approximate the integral using numerical quadrature. This PDF lists, towards the end, good quadrature weights and quadrature point locations for triangles. Numerical quadrature will work well provided that the two triangles are well-separated: if they touch or overlap you are going to need more sophisticated methods.
A: The coordinate transformation employed in this answer has a couple of references in: 

Example of a Problem Made Easier with Skew Coordinates

Suppose that the two triangles, with vertices numbered as $(0,1,2)$, are successively described by:
$$\begin{array}{lll}
(r) & : &\vec{r}(\xi,\eta) = \vec{r_0}+\xi\,(\vec{r_1}-\vec{r_0})+\eta\,(\vec{r_2}-\vec{r_0})\\
(\rho) & : &\vec{\rho}(\alpha,\beta) = \vec{\rho_0}+\alpha\,(\vec{\rho_1}-\vec{\rho_0})+\beta\,(\vec{\rho_2}-\vec{\rho_0})
\end{array}$$
with $\,0 \le \xi \le 1$ , $0 \le \eta \le 1-\xi\,$ and
$\,0 \le \alpha \le 1$ , $0 \le \beta \le 1-\alpha$ .

Then the force upon an infinitesimal part $(d\alpha\,d\beta)$ of triangle $(\rho)$
being attracted by an infinitesimal part $(d\xi\,d\eta)$ of triangle $(r)$ is:
$$
d\vec{F} = G\cdot m_r\cdot m_\rho \frac{\vec{r}(\xi,\eta)-\vec{\rho}(\alpha,\beta)}
{\left|\vec{r}(\xi,\eta)-\vec{\rho}(\alpha,\beta)\right|^{3/2}}
2\,d\xi\,d\eta\cdot 2\,d\alpha\,d\beta
$$
Where the factors $2$ comes from the fact that the parent triangles have area $1/2$ instead of $1$ :
$$
2\int_{\xi=0}^{\xi=1} d\xi \int_{\eta=0}^{\eta=1-\xi} d\eta =
2\int_{\alpha=0}^{\alpha=1} d\alpha \int_{\beta=0}^{\beta=1-\alpha} d\beta = 1
$$
The force as a whole with $\vec{r} = (x,y,z)$ , $\vec{\rho} = (a,b,c)$ is:
$$\left[ \begin{array}{c} F_x\\F_y\\F_z \end{array} \right] = 
\left[\begin{array}{c}
\int_{\xi=0}^{\xi=1}\int_{\eta=0}^{\eta=1-\xi}\int_{\alpha=0}^{\alpha=1}\int_{\beta=0}^{\beta=1-\alpha}
\frac{4\,G\cdot m_r\cdot m_\rho\cdot (x-a)\cdot d\xi\,d\eta\cdot d\alpha\,d\beta}
{\left|(x-a)^2+(y-b)^2+(z-c)^2\right|^{3/2}} \\
\int_{\xi=0}^{\xi=1}\int_{\eta=0}^{\eta=1-\xi}\int_{\alpha=0}^{\alpha=1}\int_{\beta=0}^{\beta=1-\alpha}
\frac{4\,G\cdot m_r\cdot m_\rho\cdot (y-b)\cdot  d\xi\,d\eta\cdot d\alpha\,d\beta}
{\left|(x-a)^2+(y-b)^2+(z-c)^2\right|^{3/2}} \\
\int_{\xi=0}^{\xi=1}\int_{\eta=0}^{\eta=1-\xi}\int_{\alpha=0}^{\alpha=1}\int_{\beta=0}^{\beta=1-\alpha}
\frac{4\,G\cdot m_r\cdot m_\rho\cdot (z-c)\cdot  d\xi\,d\eta\cdot d\alpha\,d\beta}
{\left|(x-a)^2+(y-b)^2+(z-c)^2\right|^{3/2}} \end{array} \right]
$$
Where:
$$
\left\{\begin{array}{l}
x = x_0+\xi\,(x_1-x_0)+\eta\,(x_2-x_0)\\
y = y_0+\xi\,(y_1-y_0)+\eta\,(y_2-y_0)\\
z = z_0+\xi\,(z_1-z_0)+\eta\,(z_2-z_0)\end{array}\right.
\qquad \qquad \left\{\begin{array}{l}
a = a_0+\alpha\,(a_1-a_0)+\beta\,(a_2-a_0)\\
b = b_0+\alpha\,(b_1-b_0)+\beta\,(b_2-b_0)\\
c = c_0+\alpha\,(c_1-c_0)+\beta\,(c_2-c_0)\end{array}\right.
$$
A short excercise with MAPLE learns that even calculating the innermost integrals is a disaster.
And it doesn't finish at all with calculating the innermost double integrals. Therefore a "closed form"
seems to be out of reach anyway, unless someone can devise some clever trick.
It has been suggested that it would be easier to calculate the potential energy of a triangle
$(r)$ with respect to a triangle $(\rho)$ :
$$V = 
\int_{\xi=0}^{\xi=1}\int_{\eta=0}^{\eta=1-\xi}\int_{\alpha=0}^{\alpha=1}\int_{\beta=0}^{\beta=1-\alpha}
\frac{4\,G\cdot m_r\cdot m_\rho\cdot d\xi\,d\eta\cdot d\alpha\,d\beta}
{\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}}
$$
But, apart from calculating the integrals, I have no idea how to proceed from there to obtain the force.

Numerically
As suggested in a comment by Mr. G , The easiest thing to do would be to just consider both
of the triangles as point masses located at their centers of mass for the purposes of gravity
(this is an approximation, however) :
$$
\vec{F} = G\cdot m_r\cdot m_\rho \frac{\vec{r}(\xi,\eta)-\vec{\rho}(\alpha,\beta)}
{\left|\vec{r}(\xi,\eta)-\vec{\rho}(\alpha,\beta)\right|^{3/2}}
\qquad \mbox{with} \qquad \xi = \eta = \alpha = \beta = \frac{1}{3}
$$
With a subdivision of both triangles, like in the above picture on the right,
the approximation can be made more accurate. Define $N^2$ midpoints in one triangle and $M^2$ in the other:
$$\begin{array}{l}
\vec{r}(i,j) = \vec{r_0}+\frac{i+(1/3\, \vee\, 2/3)}{N}\,(\vec{r_1}-\vec{r_0})+\frac{j+(1/3\,\vee\, 2/3)}{N}\,(\vec{r_2}-\vec{r_0})\\
\vec{\rho}(k,l) = \vec{\rho_0}+\frac{k+(1/3\,\vee\, 2/3)}{M}\,(\vec{\rho_1}-\vec{\rho_0})+\frac{l+(1/3\,\vee\, 2/3)}{M}\,(\vec{\rho_2}-\vec{\rho_0})
\end{array}$$
For suitable integers $\,0 \le i+j < N(-1) \; ; \; 0 \le k+l < M(-1)$ , namely such that all midpoints are inside the main triangles.
But I don't think that implementing such a subdivision is intended by the OP,
because the gravitational force between the two main triangles is meant as simple case in itself. But anyway, then the net force would become:
$$\vec{F} = \frac{G\cdot m_r\cdot m_\rho}{N^2 M^2} \sum_i \sum_j \sum_k \sum_l 
\frac{\vec{r}(i,j)-\vec{\rho}(k,l)}
{\left|\vec{r}(i,j)-\vec{\rho}(k,l)\right|^{3/2}}
$$
