estimationmilky-waynewtonian-gravitysolar systemtidal-effect
Tidal forces are experienced within a system moving in free fall around a bigger object because of the different strength of gravity over the system. The difference can be calculated between the far side and the near side.
Are there measurable tidal forces between the galactic centre and the solar system?
I've only skimmed the Wikipedia article you link to. From a quick look I'd say the paragraphs you quote are making points about what a theory of gravity needs to look like. For example you say "Curvature of spacetime in only required in order to explain tidal forces", but what that really means is that it's impossible to have a theory of gravity without curvature. That's because any theory of gravity inevitably has to describe tidal forces. You go on to say "as long as you ignore tidal forces, you can explain gravity without curvature", but you can't ignore tidal forces so you can't explain gravity without curvature.
To take your two specific questions:
Question 1. Gravity i.e. General Relativity isn't a theory of forces: it's a theory of curvature. By focussing on the "fictitious forces" you're getting the wrong idea of how GR works. When you solve the Einstein equation you get the geometry (curvature) of space. This predicts the path a freely falling object will take. We call this a geodesic and it's effectively a straight line in a curved spacetime. If you want the object to deviate away from a geodesic then you must apply a force - and there's nothing fictitious about it.
For example, GR predicts that spacetime is curved at the surface of the Earth, and if you and I were to follow geodesics we'd plummet to the core. That we don't do so is evidence that a force is pushing us away from the geodesic, and obviously that's the force between us and the Earth. But, and it's important to be clear about this, the force is not the force of gravity, it's the force between the atoms in us and the atoms in the Earth resisting the free falling motion along a geodesic.
Question 2. Again this is really just terminology. When you're free falling "gravity" is not eliminated. Remember that "gravity" is curvature, and in fact the curvature is the same for all observers regardless of their motion. That's because the curvature tensor is the same in all co-ordinate frames. The existance of tidal forces is proof that gravity/curvature is present.
When you're free falling you are moving along a geodesic. It is true to say that there are no forces acting, but this is always the case when you are moving along a geodesic. Remember a geodesic is a straight line and objects move in a straight line when no forces are acting. There would only be a force if you deviated from the geodesic e.g. by firing a rocket motor.
Response to fiftyeight's comment: this got a bit long to put in a comment so I thought I'd append it to my original answer.
I'm guessing your thinking that if you accelerate a spaceship it changes speed, so when you stop something has happened, but when the Earth accelerates you nothing seems to happen. The Earth can apply a force to your for as long as you want, and you never seem to go anywhere or change speed. Is that a fair interpretation of your comment?
If so, it's because of how you're looking at the situation. Suppose you and I start on the surface of the Earth, but you happen to be above a very deep mine shaft (and in a vacuum so there's no air resistance - hey, it's only a thought experiment :-). You feel no force because you're freely falling along a geodesic (into the Earth), while I feel a force between me and the Earth. From your point of view the force between me and the Earth is indeed accelerating me (at 9.81ms$^{-2}$). If you measure the distance between us you'll find I am accelerating away from you, which is exactly what you'd expect to see when a force is acting. If the force stopped, maybe because I stepping into mineshaft as well, then the acceleration between us would stop, though we'd now be moving at different velocities. This is exactly what you see when you stop accelerating the spaceship.
It's true that a third person standing alongside me doesn't think I'm accelerating anywhere, but that's because they are accelerating at the same rate. It's as though, to use my example of a spaceship, you attach a camera to the spaceship, then decide the rocket motor isn't doing anything because the spaceship doesn't accelerate away from the camera.
According to Fishbone, 1973, ApJ, 185, 43 the minimum density defining the Roche limit for something in the innermost stable, equatorial, circular orbit around a Kerr black hole is the same as that for an orbiting object in Newtonian physics$^1$.
Using Fig.2 in that paper, for a maximally spinning Kerr black hole, the limiting Roche density is $\sim 3 \times 10^{18} (M/M_\odot)^{-2}$ g/cm$^3$. The black hole that Miller's planet orbits has $M \sim 10^8M_\odot$ and so Miller's planet needs to have a density of $>300 $ g/cm$^3$ in order to survive.
Double-checking these figures, a simple use of the Newtonian Roche limit is that $$ \rho > (2.44)^3 \left(\frac{3}{4\pi}\right)\left(\frac{M}{r^3}\right)\ , $$ where $M$ is the black hole mass, $r \simeq 0.5 r_s$ is the orbital "radius" and the $(2.44)^3$ coefficient accounts for the planet being a deformable fluid ellipsoid (see here). For $M = 10^8M_\odot$, then $r = 1.48\times 10^{11}$ m, and this gives gives $\rho > 200$ g/cm$^3$, in fair agreement.
Theis density looks 1-2 orders of magnitude too high for any kind of sensible planet. I'm quite unhappy with this answer, since Kip Thorne designed the world/black hole system so that it would survive tidal forces...??
If you make the planet a rigid, undeformable sphere, then the $(2.44)^3$ gets replaced by $(1.26)^3$, which reduces the threshold density to $\sim 30$ g/cm$^3$. This is still too high for a terrestrial-type rocky planet.
You really need the black hole to have had a mass $>2\times 10^8 M_\odot$ to give a sensibly dense planet a chance of surviving break-up. Apparently Kip Thorne didn't do that because he wanted to make the black hole mass a nice round number. In a footnote in Chapter 6 of his book The Science of Interstellar, where the black hole mass is discussed, Thorne says
A more reasonable value might be 200 million times the Sun’s mass, but I want to keep the numbers simple and there’s a lot of slop in this one, so I chose 100 million.
I guess that isn't the only (or even nearly the biggest) scientific inaccuracy in the film (see also Wouldn't Miller's planet be fried by blueshifted radiation?).
$^1$ Note that is by no means obvious that this should be the case. In general, the tidal force on an orbiting object is not given by the simple Newtonian formula. For example, the expression for the tidal forces in the radial direction for an object in a stable circular orbit around a Schwarzschild black hole are 1.5 times larger than the Newtonian value at the ISCO at $r=3r_s$ and would blow up to infinity at the unstable circular orbit at $r=3r_s/2$ (e.g. see chapter 9 of Exploring Black Holes by Taylor, Wheeler & Bertschinger).
Nevertheless if you consult chapter 19 of the same book, it gives general expressions for the tides for orbits around a Kerr black hole. If you work through these expressions, then when $a$ is maximal, and the ISCO is at $r = r_s/2$ then the tidal acceleration in the radial direction is given by $$ g_{\rm tidal} \simeq \frac{2GM}{r^3}\Delta r\ ,$$ as in Newtonian physics. I think this is because at this radius, the orbiting object is essentially co-rotating with the dragged spacetime.
Best Answer
The magnitude of the gravitational force, $F$, between two masses, $M$ and $m$, separated by a distance, $r$, is given by: $$ F = \frac{ G \ M \ m }{ r^{2} } \tag{0} $$ where $G$ is the gravitational constant.
Suppose I move $m$ by a distance $\Delta r$, then the force changes to: $$ F_{\Delta} = \frac{ G \ M \ m }{ \left( r + \Delta r \right)^{2} } = F \ \left[ 1 + \frac{ \Delta r }{ r } \right]^{-2} . \tag{1} $$ Note that the difference is entirely encompassed in the brackets.
One lightyear is roughly 9.461 trillion km or roughly 63,241 astronomical units (AU). Suppose we want to compare the force from the center of the Milky Way to one edge of Earth's surface to the opposite edge (i.e., line from Milky Way center through middle of Earth). Let's ignore the non-uniform and non-spherical mass distribution of the galaxy and just treat it like a point source. Note that 1 AU is roughly 23,455 times the Earth's mean equatorial radius (or ~215 solar radii).
The Solar System is located roughly 27,000 lightyears from the galactic center, or ~1.7 billion AU ~ $4 \times 10^{13} \ R_{E}$
So if our $\Delta r = 2 R_{E}$ and $r \sim 4 \times 10^{13} \ R_{E}$, then $\tfrac{ \Delta r }{ r } \sim 5 \times 10^{-15}$, i.e., we're getting into the rounding error level of small.
Even if we changed $\Delta r$ to correspond to Earth's orbit about the Sun (i.e., ~2 AU), $\tfrac{ \Delta r }{ r }$ only increases up to $\sim 10^{-9}$.
Are there tidal forces acting on massive objects due to the galactic center? Sure. Can we measure them? I am not aware of any instrument/detector accurate enough to do so (which may not mean much). This is further complicated by the fact that the galaxy is not a uniform, spherically symmetric, massive object. The geometry and distribution of mass outside the solar system's orbit about the galactic center is also not uniform or spherically symmetric. So all my estimates above would need to be altered accordingly. However, geometry is not going to increase the differences enough to make much of a difference.