Ok, here is an updated mathematical treatment. I'm going to treat a scenario that somewhat combines scenarios proposed by
@Jorrie and
@PAllen. The analysis that follows, btw, will show that some of my previous posts were incorrect in some respects.
First, let's verify that the weak field approximation is ok. The simplest way to check is to verify that the maximum possible values of all terms in the metric that would make it different from the flat Minkowski metric are small (absolute value much less than ##1##). The two terms of interest are ##2M / r##, where ##M## is the mass of the "marbles" we are going to use as our experimental objects, and ##K r^2##, where ##K## is an appropriate constant related to dark energy. (In the notations of previous posts, this constant would be ##\Lambda / 3##, where ##\Lambda## is the cosmological constant, or ##8 \pi \rho / 3##, where ##\rho## is the dark energy density. The small numerical factors here are not important for the order of magnitude analysis we are about to do.)
The maximum value of ##2M / r##, if we assume ##M## is 1 kg, will be the one corresponding to the minimum value of ##r##, i.e., the minimum distance apart that our experimental objects will be. Let's assume that is 1 meter; i.e., we are only going to test separations of 1 meter or larger. Then we have ##2 M / r = 2 \times 7.42 \times 10^{-28} / 1##, which is obviously small. (In fact we could make ##r## as small as the size of an atomic nucleus, about ##10^{-15}## meters, and still have ##2M / r## be small. So we have no issue here.)
The maximum value of ##K r^2## will be the one corresponding to the
maximum distance apart that our experimental objects will be. Let's assume that is 1 light year; i.e., we are only going to test separations of 1 light year or smaller. Then we have ##K r^2 = (8 \pi / 3) \times 4.92 \times 10^{-54} \times ( 9.30 \times 10^{15} )^2##, which is also obviously small. So we are quite justified in using the weak field approximation.
Before writing down the weak field approximation to the metric, we want to first make a few adjustments to our coordinates. First, we will only consider motion along a single line, so we can eliminate angular coordinates and we can use ##x## instead of ##r## for our coordinate along the line. Second, we will put a "lab" at the point ##x = 0##, which is defined to be equidistant between our two experimental objects; the "lab" is always in free fall and serves as a "comoving" reference point. Third, we start our two experimental objects, which are each "marbles" with mass ##M##, at coordinates ##x = D## and ##x = -D##, and with zero coordinate velocity. We then seek the conditions under which the objects, in free fall, will remain at their starting coordinates.
With these stipulations, the weak field approximation to the metric then is:
$$
ds^2 = - \left( 1 - \frac{2M}{D - x} - \frac{2M}{D + x} - K x^2 \right) dt^2 + \frac{1}{1 - \frac{2M}{D - x} - \frac{2M}{D + x} - K x^2} dx^2
$$
Note that the only use we are actually making of the weak field approximation is to allow us to add both of the ##2M / r## terms together without any nonlinear effects.
The relevant equation for evaluating the motion of our experimental objects is the geodesic equation for the ##x## component of 4-velocity. That equation is:
$$
\frac{d U^x}{d \tau} + \Gamma^x{}_{\mu \nu} U^\mu U^\nu = 0
$$
We are looking for the conditions that will give us ##U^x = 0## for all ##\tau## for the worldlines of our experimental objects, which means we want ##d U^x / d\tau = 0## at the values of ##x## where our experimental objects are. That means the second term above must vanish. Since the only nonzero 4-velocity component under these conditions is ##U^t##, for the second term to vanish means we must have ##\Gamma^x{}_{tt} = 0##. This gives (including only nonzero terms, of which there is only one since our metric is diagonal and its components are only functions of ##x##):
$$
\Gamma^x{}_{tt} = 0 = - \frac{1}{2} g^{xx} g_{tt,x}
$$
which in turn gives, formally,
$$
\left( 1 - \frac{2M}{D - x} - \frac{2M}{D + x} - K x^2 \right) \left( \frac{M}{\left( D + x \right)^2} - \frac{M}{\left( D - x \right)^2} - K x \right) = 0
$$
This will be zero if the factor inside the second parentheses is zero (the factor inside the first parentheses will always be close to ##1## when the weak field approximation is valid, so we don't need to consider that one).
I say "formally" because, as noted above, we want this to hold at the values of ##x## where our objects are, i.e., ##x = D## and ##x = - D##, but our expression is obviously singular there. The hand-wavy non-rigorous way we will handle this is by simply ignoring the singular terms (this amounts to assuming that each experimental object does not respond to its own self-field but only to the dark energy and the field of the other experimental object). Doing this, the evaluations at ##x = D## and ##x = - D## both give the same condition:
$$
\frac{M}{4 D^2} = K D
$$
or, using ##K = 8 \pi \rho / 3##,
$$
D = \left( \frac{M}{4 K} \right)^{\frac{1}{3}} = \left( \frac{3 M}{32 \pi \rho} \right)^{\frac{1}{3}}
$$
Plugging in numbers gives ##D = 2.62 \times 10^8## meters.
Now, looking at the condition above, it certainly looks like we are just equating the gravitational force (strictly speaking, the acceleration) between the two balls at distance ##2D## with the "dark energy force" (or acceleration). (Although note that the "distance" in the dark energy force term is just ##D##, not ##2 D##. Why isn't it ##2 D##? I'll leave that possible issue for another discussion.)
(Note, however, that there is a subtlety here. What is ##D##? Well, it's the ##x## coordinate of the balls. But the ##x## coordinate is
not the same as proper distance. The ##g_{rr}## term in the metric is not ##1##. So the proper distance between the balls will not be exactly ##2 D##, and the Newtonian gravitational acceleration between the balls will not be exactly the same as ##M / (2D)^2##. I might have more to say about this in a future post.)
In a follow-up post I'lll compare the above with the case where we use idealized test objects and the only nonzero stress-energy is dark energy.
