Are orbiting masses in a uniform disc affected by masses outside its orbit? For a sphere of uniform density, we know an orbiting mass experiences a net attraction only by the mass inside its orbit, and that the mass outside its orbit exerts a net zero force (using Newton's Laws).
If we assume a uniform disc of mass such as an idealized galaxy, does the mass outside the orbit still NOT influence the orbit?
If interested, the motivation for this question comes from a recent publication on Modified Newtonian Dynamics (MOND). Unique prediction of 'modified gravity' challenges dark matter
"...MOND made a bold prediction: the internal motions of an object in the cosmos should not only depend on the mass of the object itself, but also the gravitational pull from all other masses in the universe--called "the external field effect" (EFE)"
Although MOND and Newtonian dynamics use different treatments depending on the acceleration scale, I am trying to understand how and if external masses affect galaxies in general when assuming Newtonian dynamics. ie. discriminating the differences between Newton and MOND.
Conceptual explanations are helpful along with math.
 A: 
Are orbiting masses in a uniform disc affected by masses outside its orbit?

Yes.
The gravitational potential inside a massive ring or annulus (in the plane of the ring or annulus) is not uniform. There is no “Ring Theorem” similar to the Shell Theorem. Since the potential is not uniform, there is a gravitational field from the outer mass.
The potential inside a ring can be calculated in the usual way by integrating $-G\,dM/r$, the potential due to each infinitesimal mass $dM$ along the ring. For a ring, you do a one-dimensional integral around the ring in terms of an angle $\theta$. The distance $r$ between the mass element and the position where you are calculating the potential is a varying function of $\theta$ (except at the center). The integral is messy; it can be expressed in terms of $K(k)$, the complete elliptic integral of the first kind. It doesn't seem important to give details; the point is that the potential turns out to be a function of the distance from the center rather than a constant.
An annulus can be treated as a collection of concentric rings.
A: There is in fact a generalization of Newton's shell theorem for ellipsoidal systems where the density is constant on homeoids (ellipsoidal shells). This has been known since the 19th Century, although it is often not mentioned in basic discussions of Newtonian gravity.
Quoting from Binney & Tremaine, Galactic Dynamics (2nd Ed.) (pages 60 and 87), we have:


*

*Newton’s first theorem: A body that is inside a spherical shell
of matter experiences no net gravitational force from that shell.


*Newton’s second theorem: The gravitational force on a body that lies outside a spherical shell of matter is the same as it would be if
all the shell’s matter were concentrated into a point at its center.

The first theorem turns out to be a generalization of



*Newton’s third theorem:  A mass that is inside a homoeoid experiences no net gravitational force from the homoeoid.


Note that "Newton's second theorem" does not apply in the case of a non-spherical homeoid, although the force due to a homeoidal shell of matter will approximate that from a point mass more and more closely as you move to larger and larger distances.
So, yes, you can in principle have an idealized disk galaxy where the shell theorem applies. Of course, the gravitational effects from non-symmetric distributions of matter outside the disk galaxy (satellite galaxies, the galaxy group or galaxy cluster a given galaxy is part of or near, etc.) will still have an effect on orbits within the galaxy. (And real disk galaxies will not actually have constant mass density on homeoids anyway.)
As for the connection to the Chae et al. 2020 paper you are referring to: their argument is that the effects of external, non-symmetric gravitational fields on orbits in a disk galaxy are different in the MOND case than they would be in a dark-matter scenario. E.g., from their Abstract: "Tidal effects from neighboring galaxies in the $\Lambda$ cold dark matter (CDM) context are not strong enough to explain these phenomena." So it doesn't actually have anything to do with whether the generalized shell theorem is valid in MOND or not.
A: There is apparently a result due to O. D. Kellogg that says the potential within a uniformly-charged ellipsoid is quadratic.  Wei Cai gives a closed form (eqn 80 & 83) for the potential within a uniformly-charged thin oblate ellipsoid (i.e. similar to a uniformly charged plate and perhaps equally relevant). This leads to the result that the field is given by $c\rho\pi^2r/a$.  Here, $\rho$ is the density, $2c$ is the thickness of the ellipsoid and $a$ is its outer radius.  If I'm interpreting this correctly, it says that the field varies linearly with radius, r, inside the ellipsoid, but inversely with the outer radius.  This result seems a little surprising. It does seem to say that the field at a given radius will drop as material is added outside that radius. So this differs from the result for a sphere.
A: An orbiting target mass in a uniform disc IS affected (a) by all masses within the disc AND (b) by masses outside the disc (which do not form spherical masses concentric with the target).
A conceptual idea that may be useful to you is to consider a target particle $P$ located somewhere on the x-axis, in a razor-thin quasi-2D disc of uniform density, at position $(r,0)$  at distance $r$ from the centre.

Now construct a line $L$ that is parallel to the y-axis and passes through point $P$. The segment of the disk that lies to the $x+$ side of $P$ may be called area $A1$.  Now reflect area $A1$ across line $L$ to produce area $A2$.  Let the combined area be called $A$ such that $A = A1+A2$.
Now, from symmetry, it should be evident that the gravitational pull from any source element inside area $A$ can be added together (vectorially) with the pulls from equivalent source elements lying at three positions obtained by mirroring (i) in the $L$ line, (ii) the x-axis and (iii) both.  The vector sum of the 4 pulls will be zero.  The same applies to any element within area $A$.
If you look at the area of the disc outside area $A$ (lets call it area $B$) you will see the remaining elements which together exert pulls on particle $P$.  It is evident that, due to symmetry the net pull will be towards the centre.
Next you can imagine/sketch/model how the size and shape and location of areas $A$ and $B$ will change as the test particle moves along the x-axis from the centre to the edge.  With $P$ at the centre, area $A$ occupies the full disc and the net force is zero. With $P$ at the disc edge, area $A$ is zero and the whole of the disc acts to exert a net pull towards the centre.
It follows, intuitively, that as $P$ moves from the centre to the edge of the disc, so the net gravitation pull, towards the disc centre, will increase in magnitude from zero to some maximum value.
The following diagrams (with semi-log and log:log axes) (based on a simple numerical model of multiple concentric rings weighted by circumference)  illustrate how specfic forces ($F/m$) vary across and beyond a uniform circular disc, both for a Newtonian Force (1/D^2) and for an inverse distance force.  Also shown are the specific forces that would apply if all the disc mass was situated at the disc centre point.  Also shown are the circular velocities $V_{circ}$ resulting from the forces (for the disc mass, not the point mass) according to $F/m = V_{circ}^2/r$.

Interesting from a MOND perspective: the $V_{circ}$ data plotted with linear axes:-

A: To answer I would use gauss theorem:
$$
\iint_{S} \vec{g} \cdot d \vec{S}=-4 \pi G \iiint_{V} \rho_{m} \mathrm{~d} V=-4 \pi G M_{\mathrm{int}}
$$
Assuming the disk is a very thin cylinder we could write $S$ as:
$$ S = S_{top} \cup S_{bottom} \cup S_{side}$$
Splitting the first integral you'll obtain terms not equal to zero:
$$LHS = 2\pi R e g + \pi R^2 g + \pi R^2g $$
If $e$ is the thickness of the cylinder:
$$2\pi Rg(e+R) = -4\pi G (\pi R^2e)$$
This gives us:
$$ g = -\frac{2\pi GRe}{e+R} $$
So I would say the outside doesn't affect the inside but I'm very skeptical on my own answer:

*

*I remember applying Gauss theorem in non spherical case can be tricky

*I hadn't use this theorem for a while so I'm not sure about the computation

However I'm very interested by your question because I'm currently having an interest for MOND theory, I'll follow this.
