My $.02, again:
I think you are insisting on taking the hard apporach to the problem, after I"ve attempted to point out that there is an easier way.
But as long as you get the same answers in the end, it doesn't necessarily matter.
Under this assumption, after all other effects are accounted for, the Moon should be receding from the Earth at a speed of 2.75 cm/yr as calculated by the OP. However, I don't think you can measure this recession for the following reason. Consider an experment carried out in a remote region of space where forces are nil. Place two neutrons at rest with respect to each other and at a distance from each other equal to the radius of the Moon's orbit. Come back a year later and see where they've gotten to. In my opinion, they will not have moved
You are basically close to being on the right track here, except that there is something that you have not considered.
a(t) is not a truly linear function of time. If a(t) were linear, you'd be correct.
The next level of approximation is to see what happens when a(t) is modeled with a linear term and a square law term. If assuming that a(t) is linear gives no effect, you need to carry out the analysis to the next order to find out what the actual effect will be.
Exercise: calculate what happens between the neutrons if a(t) is not linear. You should find that they accelerate towards each other if the second derivative of the scale factor d^2a/dt^2 < 0, away from each other if the second derivative is positive
Exercise: a(t) satisfies the Friedman equation. A convenient form of this is, using currently accepted values for the various constants
http://www.astro.ucla.edu/~wright/Distances_details.gif
http://en.wikipedia.org/wiki/Friedmann_equations
<br />
da(t)/dt = H0 \sqrt{\Omega_m/a + \Omega_{vac}a^2}<br />
also written as
<br />
[da(t)/dt] / a(t) = H0 \sqrt{\Omega_m/a^3 + \Omega_{vac}}<br />
You can see that these are equivalent
here a(t) is the changing scale factor of the universe
H0 is the current value of Hubble's constant.
H(t) is, by defintion, (da/dt) / a
H0 is the value of H(t) now.
This gives you numbers that you can actually plug into find the second derivatve of a(t). The simplest case to analyze in the one that Cooperstock also analyzes. This is the case with no cosmological constant. You can then assume
\Omega_{vac}=0 and \Omega_m=1
Compare with
http://xxx.lanl.gov/PS_cache/astro-ph/pdf/9803/9803097.pdf
equation 3.1
You will note that in this model, with no cosmological constant, the neutrons accelerate
towards each other as previously noted.
Exercise: compare this acceleration with the gravitational acceleration due to the matter density rho in the sphere between the two neutrons using Newton's law. (Matter in the hollow sphere outside the two neutrons won't contribute). You'll need to find the matter density rho via the equation
rho = 3H^2 / 8 Pi G
(This is the assumed "critical" value of matter density equivalent to \Omega_m=1, the one that makes space flat, i.e. k=0.)
As a result of this calculation, you should then see that this provides exactly the same answer as the previoius methods did.
If you wish to go beyond this "no-cosmological constant" model to the consensus Lambda-CDM model would be
\Omega_{m}, the contribution due to matter is .27
\Omega_{vac}, the contribution of the vacuum is .73
I've decided in retrospect to snip discussion of this - better to get one point through than to lose it in a sea of information.