Orbits in an universal expansion?

Question

Has anyone worked out the math for the motion of objects in distant orbits accounting for both gravity and universal expansion?

Objects in very distant orbits in expanding space-time should orbit more slowly than when accounting for gravity alone. Eventually at some great distance an orbit will not even be possible. Using NASA's latest value of H, my own attempts indicate that the sun should not be able to hold anything beyond the mid Kuiper belt, so I must be doing something wrong. Any help?

related/relevant: How does the inclusion of vacuum energy modify Newtonian gravity equation of motion? — AccidentalFourierTransform, Dec 16 '16 at 16:40
Solar system effects in Schwarzschild–de Sitter space–time — John Rennie, Dec 16 '16 at 16:46
I get the distance where the CC balances out the Sun's gravity to be about 300 light years. Somewhat greater than the orbital radius of the Kuiper belt. — John Rennie, Dec 16 '16 at 17:17
@JohnRennie I get a little farther away (see comments: http://physics.stackexchange.com/a/253557/8521) — DilithiumMatrix, Dec 16 '16 at 17:22
Possible duplicates: http://physics.stackexchange.com/q/2110/2451 , http://physics.stackexchange.com/q/71649/2451 and links therein. — Qmechanic, Dec 16 '16 at 17:40

score 2 · Answer 1 · edited Apr 13 '17 at 12:40

Using the equation conveniently proved in @AccidentalFourierTransform's answer on a related question (note: strange units): $$g=-\partial_r\Phi=-\frac{GM}{r^2}+\frac{1}{3}\Lambda r\tag{3}$$ Then we can calculate the radius at which the cosmological constants (CC; $\Lambda \approx 10^{-29} \textrm{ g cm}^{-3}$ [1]) counteracts an objects gravity as: $$r = \left[\frac{3M}{\Lambda}\right]^{1/3} $$

For the Sun, this is roughly, $r\approx 273 \textrm{ pc}$. This is likely completely unobservable, because within that same radius there are roughly 1 million other stars, and their effect on dynamics will be much stronger than the cosmological constant. If we look at smaller, scales, for example the Earth's orbit at $1 \textrm{ AU} \approx 10^{13} \textrm{ cm}$, the relative effect of the CC is only about $10^{-22}$---again, completely unobservable. The issue is that typical densities within galaxies are larger than the (energy)density of the cosmological constant. Only on very large scales, when matter becomes very sparse, does the cosmological constant become important.

If we use the entire Milky Way mass, for example, we find that the distance for CC to cancel out gravitational acceleration is roughly $r \approx 3 \textrm{ Mpc}$. So it's really only on large scales that CC has a noticeable affect [2].

Notes:
[1]: The value $\Lambda \approx 10^{-29} \textrm{ g cm}^{-3}$ can be dug out of Wikipedia: Cosmological Constant. Known the correct expression to use is non-trivial... but note that including traditional units the expression should be $(8\pi G / c^2 )\Lambda \approx 10^{-29} \textrm{ g cm}^{-3}$.

[2]: Even on $3 \textrm{ Mpc}$ scales, we still have to worry about dwarf galaxies, Andromeda, and the circum-galactic medium... so really it's not until $10$s--$100$s of Mpc that the effects of CC really become clear.

You see the CC quoted in all sorts of units. Wikipedia quotes it as $1.19 \times 10^{-52} \text{m}^{-2}$, and to get that you have to multiply DilithiumMatrix's value by $8\pi G/c^2$. This is the trouble with causally setting constants to unity then forgetting you've done it :-) — John Rennie, Dec 16 '16 at 17:43

Orbits in an universal expansion?

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