What exactly is the microgravity field in orbit? The ISS and other objects in orbit still experience small acceleration outside from the perfect line of orbit (of the system CM).  For instance, two objects in the ISS that are let to be at rest will pass by each other twice as the station makes one orbit because the two items are in separate orbits, and all orbits pass over the same point because they are great circles over the Earth.
My question is how would you actually quantify this?  Regarding the ISS, the entire craft could rotate just so it is effectively tidally locked with the Earth.  If you assume it does that, then you have one axis along which it is totally acceleration-less relative to the craft.  Looking forward along that line, moving to the left or right would create an acceleration back to the line.  Moving up or down would also create an acceleration toward the line since they would assume more elliptical orbits.  But I'm really curious as to whether there would also be a acceleration parallel to the line of orbit, and I'm also really curious if it would be unstable or stable.
So if the line of orbit is x, down toward the Earth is the negative z, and right of the line of motion is y, then my intuition is that the system is stable on the y-axis, it is stable up-and-down for the z-axis, but movement up the z-axis would create acceleration in the negative x-axis.
One major consequence would be if you left a hammer just outisde the airlock, the stability of these fields would dictate whether it sticks around or leaves.  Could we find a simple $(x,y,z)$ equation for the acceleration?  I can't find anything that quite answers this, and I my attempts result in more questions than answers.
 A: I can't answer the question in full, but I did get some result that kind of helps.  Start with the Wikipedia article on this.  We can employ the equation specific to an orbiting reference frame.
$$ \frac {d^2 \mathbf{x}_{A}}{dt^2}  =\mathbf{ \Omega \ \times } \left( \mathbf{ \Omega \times} (\mathbf{ X}_{AB}+\mathbf{x}_B) \right) + \mathbf{a}_B + 2\ \boldsymbol{\Omega} \times\mathbf{v}_B\ \ $$
This should give the right answer if applied correctly.  My understanding is that the acceleration in the A reference frame is gravity.  Then rearrange to have an expression for $\mathbf{a}_B$, and that is the general nature of the thing being sought here.  Here are some more notes about the notation.


*

*x - axis along the direction of motion

*y - axis perpendicular to Earth's surface and the direction of motion

*z - axis' positive part points directly away from the center of Earth

*B if the rotating and orbiting reference frame

*A is the stationary reference frame centered about center of Earth

*$\mathbf{X}_{AB} = R < \cos{\omega t}, 0, \sin{\omega t}>$ is the (dynamic) vector between the center of Earth and the origin of the B reference frame

*$\mathbf{x}_{B}$ is the vector between the B origin and the object.  This uses the axises of the A reference frame, which I found confusing, but was able to implement. 

*$\mathbf{\Omega} = <0,\omega,0>$ is the angular speed vector for the reference orbit

*$\omega^2 = G M / R^3$ is the scalar angular speed of the orbit

*$M $ mass of Earth

*$R$ scalar radius of the orbit


Using these mathematics, I was able to obtain a partial answer.  This is a partial simplified version of the acceleration of an object reported in the B reference frame.
$$ a \approx <\frac{d^2 x }{dt^2} , \frac{d^2 y}{dt^2}, \frac{d^2 z}{dt^2}> \approx < -2 \omega v_z, - \omega^2 y, 2 \omega v_x > $$
I'm almost positive that these terms are at least part of the answer.  Consider, if you're traveling in orbit and you hold out an object to your right, it will "accelerate" to you according to the term in the above expression.
The above answer is incomplete because one thing I employed was the assuming that the radius of the object from the center of the Earth (denote $R'$) was about equal to the radius of the B reference frame origin from the center of the Earth ($R$).  A better approximation I found was:
$$ R' \approx \sqrt{ R^2 + (x^2+y^2) + 2 R z } \approx R + \frac{(x^2+y^2) + 2 R z}{2R} $$
One could also neglect the contribution from the $x$ and $y$ displacement in many cases since it's obvious that the vertical displacement is much greater.  Once correctly accounted for, breaking the $R'=R$ assumption I used I think should get a revision that looks something like this:
$$ a_x = -\omega^2 \frac{3 z }{ R } -2 \omega v_z$$
$$ a_z = \omega^2 \frac{3 z}{R} + 2 \omega v_x $$
Within the xz-plane, this would allow an object to "orbit" the ISS or something.  Imagine it is higher than the ISS with zero velocity.  Being at a higher altitude, it's speed is no longer sufficient for the altitude of orbit, so it begins to both trail behind the ISS linearly in the direction of motion, as well as fall.  As it begins to move in the back-down direction, the $\omega v$ terms kick in and accelerate it forward-down, accelerating it toward the ISS.  A similar argument could be made for all points in the circle, demonstrating that it orbits.
A: For small velocities and displacements around a circular Kepler orbit, the equations of motion are the Hill-Clohessy-Wiltshire equations. They can be exactly solved to give

(The above switches your convention for y and z.)
Your approach of finding the forces in a rotating frame works; there is a derivation here.
The motion is in general a sort of looping that is unstable. A pure-z initial velocity or a pure-y initial velocity or displacement leads to periodic solutions (using your notation), but any other initial displacement or velocity leads to run-away. (However, the run-away is linear in time, not exponential.)
I made some plots of the trajectories here.
A: MTW (eqn 32.24b) lays out the tidal accelerations (for that is what these are) for the separations $\boldsymbol{\xi}$ of two test masses in a Schwarzschild geometry:  
$$ \begin{align*}
\boldsymbol{\xi} & = \xi^{\hat{j}} \boldsymbol{e_{\hat{j}}}\\
\frac{D^2 \xi^{\hat{\rho}}}{d \tau^2} & = + \left(2M/r^3 \right) \xi^{\hat{\rho}} \\
\frac{D^2 \xi^{\hat{\theta}}}{d \tau^2} & = - \left(M/r^3 \right) \xi^{\hat{\theta}} \\
\frac{D^2 \xi^{\hat{\phi}}}{d \tau^2} & = - \left(M/r^3 \right) \xi^{\hat{\phi}}
\end{align*} $$
Replace $M$ by $GM$ to "unitize".
