Invert “Gravitational” Force Function or Solve an Intersection

Recall "gravitational"-type force functions, by which I mean anything of the form:

$f(x,y,z) = \frac{k}{((x-x_0)^2+(y-y_0)^2+(z-z_0)^2)^p}, p\in\Re_{>0}, k\in\Re, (x,y,z) \neq(x_0,y_0,z_0)$
(e.g., for gravity, $p=1,k=G m_1 m_2$)

Define a function $g(x,y,z) = f_1(x,y,z) + f_2(x,y,z) + f_3(x,y,z) +\cdots+f_n(x,y,z)$ (a sum of several force functions, presumably each with different $k$ (though probably not $p$).

I have two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ such that $g(x_1,y_1,z_1)<a$ and $g(x_2,y_2,z_2)>a$. My problem is to find the location $(x_3,y_3,z_3)$ such that $g(x_3,y_3,z_3)=a$ where $(x_3,y_3,z_3)$ lies on the straight line between $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$. If it helps, I know that exactly two of the xs, ys, and zs are equal.

I'm currently solving this numerically, but it will eventually be solved on an OpenCL kernel where I would very much like a closed form solution.

However, I am having trouble getting such a solution. I'm not even sure it's possible. I tried inverting $g$, but got basically nowhere. Ideas? Thanks.

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The usual gravitational potential is actually $p = \frac12$ in your formulation, because your denominator represents the square of the distance between the two points. – Rahul Jun 16 '12 at 5:13
In any case, I doubt there will be a closed-form solution for general $p$. Nonlinear equations usually can't be solved analytically. – Rahul Jun 16 '12 at 5:18
Nope should be 1 :-) It's over r^2, but in calulating r I'm using the Pythagorean theorem, which has a square root, so the 1/2 and 2 powers cancel. Anyway, I figured it would be unsolvable, but I wanted to check just in case. It should be easy to solve if you have only one term, but of course I have several . . . – imallett Jun 16 '12 at 5:35
The magnitude of the gravitational force is proportional to $1/r^2$. The corresponding potential, which the force is a derivative of, is proportional to $1/r$. – Rahul Jun 16 '12 at 5:46
Ackkk that's what I meant (you probably mean the derivative the other way). I've edited the question. Thanks! – imallett Jun 16 '12 at 6:14

Let us consider an extremely simplified version of the problem: $p=1$ and just two terms centered at $(0,\bar y_1,0)$ and $(\bar x_2,\bar y_2,0)$, with the line on which the unknown point lies being $y_1 = y_2 = 0$ and $z_1 = z_2 = 0$. Then you want to solve the equation $$\frac{k_1}{x^2+\bar y_1^2} + \frac{k_2}{(x-\bar x_2)^2+\bar y_2^2} = a,$$ which expands to the fourth-degree polynomial equation in $x$, $$k_1\big((x-\bar x_2)^2+\bar y_2^2\big) + k_2\big(x^2+\bar y_1^2\big) = a\big(x^2+\bar y_1^2\big)\big((x-\bar x_2)^2+\bar y_2^2\big).$$ Mathematica finds an analytical solution that I cannot reproduce here because its $\LaTeX$ code has 11887 characters in it. If you have more than two terms, of course, you'll get a polynomial equation of degree greater than five, and those are known to have no analytical solutions in general.