The expansion of space does not affect gravitationally bound systems such as a galaxy or our solar system.

Perhaps $0$????

@AdrianHoward I’ve heard that said many times, but I have a hard time believing that the effect is theoretically exactly zero. I can easily believe that it is so small that it is unmeasureable.

@AdrianHoward Actually, I want to make a stronger statement. I think it is obvious based on the geodesic equation that the geodesics for two gravitating bodies in an expanding universe are not the geodesics of those two gravitating bodies in a non-expanding universe.

@AdrianHoward As a concrete example, consider the geodesics for test particles around a de Sitter-Schwarzschild black hole. I don’t see how they could be the same for zero and positive cosmological constant.

@G.Smith You're right - the effect is not exactly zero. However, you can look at two gravitating test masses in the non-relativistic limit and you find that $a\cdot R$ - the product of the scale factor and comoving orbital radius - is constant up to ludicrously small correction terms, on the scale of the time it takes light to traverse the system divided by the Hubble time (and even that might be squared). The effect of the expanding universe is to very slightly alter the effective gravitational attraction and to add a very, very small drag-like term.

@G.Smith See e.g. here for the calculation

@J.Murray Thanks for that reference.

There is absolutely no local effect whatsoever from the expansion of the universe. Local accelerations come only from locally present matter, moving however it's locally moving. See this answer.

@benrg Can you provide a reference to a published paper rather than a PSE answer?

@G.Smith Can you provide a reference … The original Einstein–Straus, 1945 paper (aka Swiss cheese model) would suffice. It is possible to find OA pdf of it.

@AdrianHoward - As I mentioned in the question, any experiment you cite takes place in curved space. If your experiment tells you that space is flat, then your experiment is giving you the wrong answer. If your experiment shows a small acceleration, then you need to filter out the effects of the Earth, the moon, the sun, the planets and the galactic core. You seem very confident of your answer, so please show me, even a thought experiment, that removes these influences.

While gravity propagates forever, different galaxies are not gravitationally bound to each other, while the stars in an individual galaxy are. i myself have often wondered if this currently accepted theory is true. But as G. Smith said he could easily believe, I also believe it to be to small to be measured, so with no evidence otherwise I, for now, follow the accepted theory.

@AdrianHoward - No evidence? How can you look at the Tully Fisher relationship and say there's no evidence? If spacetime were flat, then when you looked at galaxies, you'd see a relation between the luminosity (proxy for mass) and the product of the radius and the square of the tangential velocity. You don't. Explain to me how the failure of this prediction isn't overwhelming evidence that your model is wrong?

@GluonSoup When did I say spacetime was flat? If you are referring to the asymptotic rotation of a disc galaxy, dark matter theories are most commonly cited for discrepancies.

@AdrianHoward - You said "The expansion of space does not affect gravitationally bound systems such as a galaxy or our solar system". The only way this statement can be valid is if you take the four-force equation:$$\Sigma F^{\mu}= m\left(\frac{d}{d\tau}U^{\mu} + \Gamma^{\mu}{}_{\nu\eta}U^{\nu}U^{\eta}\right)$$and assume that the Christoffel symbols vanish (i.e. Flat Space), leaving you with $$\Sigma F = ma$$ You're the one who implied that the local geometry of spacetime was flat.

Wow, good luck with that.

This should really be three different questions. One about local effects of the expansion of the universe, one about the meaning of local flatness in GR, and one about the smallest measurable acceleration.