[Edit: Added the effect of the Moon's mass.]
Nugatory said:
Their natural trajectory under the influence of gravity will not keep them all aligned radially.
But gravity is not the only force on each section of the rope; there is also the tension in the rope. (And centrifugal force, if we work in the rotating frame in which the Moon is at rest, as I will below--but that's really included in the "natural trajectory".) On thinking this over, I think there *is* an equilibrium configuration, if the tension in the rope varies in the right way with altitude.
Consider a small piece of the rope, and ask what it takes to keep it on a circular trajectory around the Earth, but with the Moon's angular velocity (as opposed to the--larger--correct orbital angular velocity for its altitude). If we work in the rotating frame in which the Moon is at rest, then the condition for equilibrium is just that all the forces on the small piece of rope sum to zero. There are three such forces: gravity inward, centrifugal force outward, and tension in the rope, which can act both inward and outward. So we have (writing formulas for acceleration instead of force, since the mass of the small piece of rope is the same for all three forces and can be divided out)
$$
a_{total} = 0 = a_{grav} + a_{centrifugal} + a_{tension}
$$
Substituting ##a_{grav} = - G M_E / r^2 + G M_M / \left( R_M - r \right)^2## (where ##M_E## is the mass of the Earth and ##M_M## is the mass of the Moon) and ##a_{centrifugal} = \omega_M^2 r##, where ##\omega_M^2 = G M_E / R_M^3## is the angular velocity of the Moon in its orbit (##R_M## is the radius to the Moon's center), we have
$$
a_{tension} = \frac{G M_E}{r^2} - \frac{G M_M}{ \left(R_M - r \right)^2} - \frac{G M_E r}{R_M^3} = G M_E \left( \frac{1}{r^2} - \frac{1}{R_M^2} \frac{r}{R_M} \right) - \frac{G M_M}{ \left(R_M - r \right)^2}
$$
The rope will be in equilibrium if the proper acceleration due to tension in the rope satisfies this formula. Note that at the bottom of the rope, the RHS is clearly greater than zero, so ##a_{tension}## is directed outward; but there will be a "turning point" at which ##a_{tension}## goes to zero, and above that point it will be directed inward.
[Edit: Added the following to clarify the distinction between "proper acceleration due to tension" and tension.]
If we model the rope as having some constant mass per unit length ##\mu##, then the proper acceleration ##a_{tension}## is actually proportional to the *gradient* of the tension in the rope (i.e., it is due to the *net* force due to tension on each piece of the rope, which is the difference between the force from the piece above and the force from the piece below). The formula is (##T## is the tension in the rope as a function of radius ##r##):
$$
\frac{dT}{dr} = \mu a_{tension}
$$
Plugging in the formula above for ##a_{tension}##, we can easily integrate the result to obtain (##r_0## is the radius of the bottom end of the rope, and we have the boundary condition that the tension at that point is zero, i.e., ##T(r_0) = 0##)
$$
T = \mu G \left[ M_E \left( \frac{1}{r_0} - \frac{1}{r} - \frac{r^2 - r_0^2}{2 R_M^3} \right) - M_M \left( \frac{1}{R_M - r} - \frac{1}{R_M - r_0} \right) \right]
$$
This obviously satisfies the boundary condition.
, but I don't need to solve it to know that the solution is not going to be the rope hanging straight up and down - that follows just from considering the forces on the section at the free end of the rope.