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.