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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?

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    $\begingroup$ I’d say a better way to put it is, How an Orbit about a Super-Dense Galaxy would look like. The orbit that flat Earthers expect would be a circular orbit above land. Funny enough, don’t believe that they can put a satellite into stable orbit without using correct orbital mechanics which obviously assumes a round Earth. As to your question, I believe that a stable orbit isn’t possible due to the fact that orbits are about “moving forward as fast as you are falling down”. If you consider the gravity explanation of flat Earthers then no normal orbit. Otherwise, a elongated elliptical orbit works $\endgroup$ – Alpha Mineron Oct 3 '17 at 18:08
  • $\begingroup$ -1. No research effort. What is your attempt to find out or calculate what the neaby orbits look like? $\endgroup$ – sammy gerbil Oct 4 '17 at 12:56
  • $\begingroup$ @sammy gerbil - I haven't done orbital mechanics for a couple of years, so I'm not going to post an attempt. I don't think it's a trivial problem that could be given as homework, so I'm really not sure what the issue is. $\endgroup$ – nbubis Oct 4 '17 at 13:06
  • $\begingroup$ @nbubis The issue is that this is not a problem-solving website. The purpose of the site is to explain the concepts of physics, not to provide solutions to challenging problems. $\endgroup$ – sammy gerbil Oct 4 '17 at 13:13
  • $\begingroup$ Hi nbubis. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$ – Qmechanic Oct 4 '17 at 20:57
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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 $(r,z)\approx(2.5\,a, a)$.

enter image description here

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.

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  • $\begingroup$ Thank you! I think it's worth adding that the answer to the original question is a resounding yes: first, a single circular orbit in the $z=0$ plane, and second, additional orbits that solve the equations you posted, and that are limited to bounded areas in the $r-z$ plane. $\endgroup$ – nbubis Oct 5 '17 at 19:04
  • $\begingroup$ I did add some more information to make it clearer what kind of orbits we can get. $\endgroup$ – user154997 Oct 7 '17 at 17:01

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