Satellites around flat Earths Apparently the latest great idea of flat earth crackpots is launching a satellite to "prove" that earth is actually flat. Now, without actually commenting on this "plan" (let's not feed the trolls), I started wondering what an orbit around a perfectly flat uniform density disk would like look like. 
Far enough away, I'd expect we'd get the usual conic sections, but nearby, I'm not sure what orbits would look like - do stable or somewhat stable orbits even exist? What would they look like if they did? 
 A: Axisymmetric potentials: great subject! The treatment is surprisingly similar to the central force potentials actually, with the difference that orbits are not necessarily planar. Here is a terse introduction but don't hold your breadth for orbits as I haven't worked out the computations to completion.
Of course, one chooses a Cartesian frame $Oxyz$ where $Oz$ is the axis of symmetry (where $O$ would be the centre of the disk in your example, and where $Oz$ would be perpendicular to it), and then one immediately moves to cylindrical coordinates $(r,\theta,z)$:
$$\begin{aligned}
\vec{x} &= r\hat{u}_r+z\hat{u}_z,\\
\vec{v} &= \dot{r}\hat{u}_r + r\dot{\theta}\hat{u}_\theta+\dot{z}\hat{u}_z,\\
\vec{a} &= (\ddot{r}-r\dot{\theta}^2)\hat{u}_r+(2\dot{r}\dot{\theta}+r\ddot{\theta})\hat{u}_\theta+\ddot{z}\hat{u}_z.
\end{aligned}$$
The gravitational potential $\Phi$, by symmetry, depends only on $r$ and $z$ and not on $\theta$. So $m\vec{a}=-m\nabla\Phi$ reads
$$\begin{align}
\ddot{r}-r\dot{\theta}^2&=-\Phi'_r,\\
2\dot{r}\dot{\theta}+r\ddot{\theta} &= 0,\\
\ddot{z}&=-\Phi'_z. \tag{1}
\end{align}$$
The second equation expresses the conservation of
$$L=r^2\dot{\theta},\tag{2}$$
which is the component of the angular momentum about the $z$-axis. This is exactly the same conserved quantity as for a central force, and we can therefore use the same tricks, by removing $\dot{\theta}$ from the first equation,
$$\ddot{r} = -\Psi'_r,\tag{3}$$
where $\Psi$ is the effective potential
$$\Psi(r,z) = \Phi(r,z)+\frac{L^2}{2r^2}. $$
The only difference with the central force is that we have a mouvement in the $(r,z)$ plane. That plane is itself rotating at a varying angular speed since $\dot{\theta}$ depends on $r$ through eqn (2). But luckily, the equations for $(r,z)$ are decoupled: eqn (1) and (3).
The crucial point, as in the case of a central force, is the term $\frac{L^2}{2r^2}$ in the effective potential: the famous centrifugal barrier which prevents the orbiting mass to fall back on the planet. More precisely, it enables the existence of a minimum of the effective potential $\Psi$. Since this term depends only on $r$, this enables a minimum at $r_g>0$ but only at $z=0$, i.e. in the plane of the disk. However, it is perfectly possible to have $r_g > a$: then there is a circular orbit $r=r_g,\ z=0$, and then there are slightly perturbated orbits about that circular orbit, which can be computed by expanding the potential as a Taylor series about $(r,z)=(r_g, 0)$. There will be motions in the $(r,z)$ plane, which will not hit the disk thanks to $r_g>a$ and small enough a perturbation, combined with a rotation of that plane, as already stated above.
The potential for a flat disk is quite a beast, involving elliptic integrals [1, eqn (14)]. As a service for those who don't have access to paid journals, and also to give you a flavour of the difficulty, I have reproduced the formulae below. Here is a contour plot of the effective potential for $a=1$ and a well-chosen value of $L$ (the lighter the colour, the higher the potential): there is clearly a shallow minimum for $(r,z)\approx(2.5\,a, a)$.

Formula of the potential
For a disk of radius $a$ and uniform density $\sigma$,
$$\Phi(r,z)=2G\sigma\left(\pi|z|-d(r,z)E(k)-\frac{a^2-r^2}{d(r,z)}K(k)-\frac{a-r}{a+r}\frac{z^2}{d(r,z)}\Pi(n^2,k)\right)$$
where
$$\begin{align}
d(r,z)&=\sqrt{z^2+(a+r)^2}\\
k^2 &= \frac{4ar}{d(r,z)^2}\\
n^2 &=\frac{4ar}{(a+r)^2},
\end{align}$$
and where $K$, $E$ and $\Pi$ are the integrals
$$\begin{align}
K(k)&=\int_0^{\frac{\pi}{2}}\frac{d\varphi}{\sqrt{1-k^2\sin^2\varphi}},\\
E(k)&=\int_0^{\frac{\pi}{2}}\sqrt{1-k^2\sin^2\varphi},\\
\Pi(n^2,k)&=\int_0^{\frac{\pi}{2}}\frac{d\varphi}{(1-n^2\sin^2\varphi)\sqrt{1-k^2\sin^2\varphi}}.\\
\end{align}$$
1 Harry Lass and Leon Blitzer. The gravitational potential due to uniform disks and rings. Celestial Mechanics, 30:225–228, 1983.
