Stable equilibrium in a gravitational field with 2 masses Suppose you have two point masses $M_1, M_2$ in space and you locate the point $A$ on the line between them such that an object placed at $A$ is in equilibrium (the only forces acting on the obect are the gravitational forces from each point mass). Apparently $A$ is an unstable equilibrium for points along the line joining $M_1,M_2$, but a stable equilibrium for points along  the line passing through $A$ and perpindicular to the line joining $M_1,M_2$.
How can check this (1) intuitively and (2) mathematically?
 A: Intuitively, your point $A$ is the first Lagrange point, L1.  The other unstable points L2, L3 are infinitely far away in the absence of angular momentum.
$A$ is an unstable equilibrium because it is a saddle point in the overall potential
$$
\frac {U(\vec x)}{-G} = \frac{M_1}{|\vec x_1 -\vec x|} + \frac{M_2}{|\vec x_2 -\vec x|}
$$
In the special case where $M_1=M_2$, there is a plane midway between the two masses where the force $\vec F = -\nabla U$ points towards neither mass; however trajectories along this plane aren't stable, either, and small perturbations will send a mass towards one point or another.
If the masses $M_1\neq M_2$ are unequal there is not such a plane. 
Instead there is a paraboloid where the potential energy due to $M_1$ and $M_2$ is the same, as Sofia has shown, whose vertex is your point $A$.
As the ratio between the two masses gets farther from unity, the volume of space where you can reasonably talk about "trajectories that head towards $A$" gets smaller and smaller.  This is probably related to the fact that the stability of the Lagrangian points L4 and L5 requires the ratio $M_1/M_2 \gtrsim 25$.
A: The equilibrium line perpendicular on the line $M_1M_2$ is the given by the equation
$G\frac {M_1 M_A cos(\theta_1)}{|\vec r_{M_1,A}|^2} = G\frac {M_2 M_A cos(\theta_2)}{|\vec r_{M_2,A}|^2}$
where $\theta_1$ is the angle between the line $M_1A$ and the line $M_1M_2$, while $\theta_2$ is the angle between the line $M_2A$ and the line $M_1M_2$.
Thus, the desired equilibrium line is given by the relation
(1) $\frac {M_1 cos(\theta_1)}{|\vec r_{M_1,A}|^2} = \frac {M_2 cos(\theta_2)}{|\vec r_{M_2,A}|^2}$.
For the moment I don't know if it is a straight line, so let me introduce new variables
$cos(\theta_1) = x, \ \ \ cos(\theta_1) = L - x$,
where $L$ is the distance $M_1M_2$.
$|\vec r_{M_1,A}|^2 = x^2 + y^2, \ \ \ |\vec r_{M_2,A}|^2 = (L- x)^2 + y^2$,
where $y$ is the distance on vertical between the point $A$ and the line $M_1M_2$.
Introducing all this in (1)
(2) $\frac {M_1 x}{x^2 + y^2} = \frac {M_2 (L - x)}{(L- x)^2 + y^2}$.
Simplifying a bit,
(3) $M_1 x [(L - x)^2 + y^2] = M_2 (L - x)[x^2 + y^2]$.
Moving everything to the left-hand-side,
(4) $M_1(x^3 -2Lx^2 + L^2x + xy^2) + M_2(x^3 - Lx^2 + xy^2 - Ly^2) = 0$,
from which we get the curve
$x^3 - (1 + \frac{M_1}{M_1 + M_2})Lx^2 + \frac{M_1}{M_1 + M_2}L^2x + (x - \frac {M_2}{M_1 + M_2} L)y^2 = 0$.
It's not a straight line. One would be tempted to ask whether the vertical $x = \frac {M_2}{M_1 + M_2}L$ could be the sought vertical. The answer is no, unless $M_1 = M_2$.
