Accelerating expansion of the universe in numbers

The Earth's acceleration in numbers is $9.8\ \mathrm{m/s^2}$ and it stays constant in time.

How much is the acceleration of the Universe expansion in numbers? I noticed that the Hubble constant (which is about $74$) is measured in $\mathrm{m^2/s}$ instead of $\mathrm{m/s^2}$.

And is this acceleration staying constant or is it increasing/decreasing in time?

Later edit: the question "How does the Hubble parameter change with the age of the universe?" is similar but it's asking for details about the changing of the Hubble constant. While my question is simply about the value of the acceleration. But I think the two questions are worth to be linked, if possible.

Answer

Well, the current value of $H$, $H_0$ is approx $67\ \mathrm{km/(s\ Mpsec)}$, and @hdhondt's answer is on that is accurate. This is to add a few facts and information, and calculate the acceleration in $\mathrm{km/s^2}$. It's instructive to get the units to that, one almost never sees it calculated. On the other hand, it's not hard, just carefully keep track. The acceleration is so small, compared to $g$ or the nuclear or electromagnetic accelerations in bound material, it just has no effect on earth, and still little at the edge of the observable universe (now)

The Hubble constant is a measure of the velocity of the universe. The per Mpsec indicates that it is not a unique velocity, but rather increases as one goes further away. $H$ is the rate of change of the scale factor, divided by the scale factor, so its units are $\mathrm s^{-1}$. @hdhondt's answer explains. The galaxies further away from us go away from us faster because of the universe expansion. At some distance (a few Gpsecs) they are actually expanding away from us faster than $c$.

The acceleration you asked for is not the velocity, but rather its change with time, if you wanted something equivalent to $g$.

A good description of the expansion and what it depends on is in one of the answers referenced in a comment by @Druv. The first question and (yes) answer by @John Rennie plots the scale factor, $a/a_0$, such that it is normalized to 1 now, as function of time. t=1 is now (normalized to the current Hubble time). You can see the scale factor slope, ie its change with time, ie equivalent to the expansion velocity when multiplied by distance, decreases early on in the universe, and after some time it starts increasing. The Hubble parameter, $H$ is shown decreasing in the third figure, plotting $H/H_0$ as function of time, where you can see it decreasing, with the value normalized to 1 now (it's this way because $a$ increases more rapidly than $\dot{a}$). The plot for $a$ shows (in the slope) the velocity decreasing early and later (actually about 5 billion years ago) increasing again – so acceleration now is positive. The Hubble parameter, in the current $\Lambda CDM$ model (the standard model in Cosmology, with current parameters from Planck) is decreasing slowly, and approaches a constant value at infinite time of $H \approx 0.67H_0$.

The reason it approaches a constant is that future times (and we're getting there) $H$ is dominated by the dark energy density which is constant (from what we've measured so far), and the expansion of the scale factor approaches an exponential expansion with $H$ the doubling rate.

From https://en.wikipedia.org/wiki/Deceleration_parameter

where q is the dimensionless deceleration parameter. In the $\Lambda CDM$ model you can see in the same wiki reference that with $w = -1$, $q = -1$, and we have an equation for the dimensionless acceleration $-q$ (it's minus because the thought was that acceleration was negative before the discovery of dark energy). Note that I approximated above also where $q = -1$ is true strictly at late times when dark energy is totally dominant. Before that $q > -1$. We can then solve for

$\mathrm d/\mathrm dt(\dot{a})/a = H^2$, and in numbers with $H = 67\ \mathrm{km/(s/Mpsec)}$ (where I approximate $H$ as $H \approx H_0$)

acceleration $\approx {5000}\ \mathrm{(km/Mpsec)^2/s^2}$ ie acceleration $\approx {5000}\ \mathrm{(km/Mpsec)(10^{-22}/s^2)}$ ie acceleration $\approx {10^{-18}}\ \mathrm{(km/s^2)/Mpsec}$

So, the acceleration due to the universe's expansion and acceleration, which we measure as acceleration of other galaxies with respect to us, is extremely small per Mpsec. At $1\ \mathrm{Mpsec}$ it is $10^{-18}\ \mathrm{km/s^2}$ (if my algebra didn't fail me), or $10^{-15}\ \mathrm{m/s^2}$

At the edge of the observable universe now (about $47\ \mathrm{Gly}$) it is only about $10^{-11}\ \mathrm{m/s^2}$. (again, with simple multiplies I hope I got right)

We only have been able to measure acceleration with very fine measurements for which a Nobel Prize was awarded. See for instance https://www.scientificamerican.com/article/the-2011-nobel-prize-in-prize-physics/

The fact that it is so small compared with $g$ is one reason that gravitationally bound bodies, and even more so nuclear or electromagnetically bound atoms and molecules and human-sized objects, and mountains and the earth, stay bound with the universe expansion – the cosmic acceleration is just minuscule and unmeasurable for material on the earth. The other reason is that $H$ has no effect also for small distances, you have to go Mpsecs away to make any difference due to just expansion velocity.