How much energy one would get by tethering an imaginary rope to a galaxy and let it unwind until reaching the Hubble horizon? Based on the tethered galaxies gedankenexperiment 1, let's imagine we attach an imaginary cosmologically-long string to a galaxy that is at a sufficiently long distance to be drawn away by the Hubble flow.
In an accelerating universe (like our own), there would be always tension to the string and that tension is used to make energy (as it is indicated in 1). We let the string unwind as the galaxy recedes until reaching the Hubble horizon, where the string would break.
My question is: How much energy would we harness with this? Would it be equal to the energy corresponding to the rest mass of the galaxy (given by the famous equation E=mc²)? Or would it be less or greater than that?
 A: A simple argument in a de Sitter background gives an extracted energy of $mc^2$.
The usual static coordinates for de Sitter space are
$$ds^2 = \left(1-\frac{r^2}{R^2}\right)dt^2 - \left(1-\frac{r^2}{R^2}\right)^{-1}dr^2 - r^2 dΩ^2$$
The event horizon is at $r=R$. The angular coordinates are irrelevant to this problem so I'll drop them.
In a manifold with metric $ds^2 = f(r)^2 dt^2 - f(r)^{-2} dr^2$, the proper acceleration of a worldline at constant $r$ is $f'(r)$ inward. The work done on a slow-moving test object of small mass moving from $r$ to $r+dr$ is the mass times the acceleration times the metric distance moved, i.e., $-m\,f'(r)\,f(r)^{-1}\,dr$. However, if you transmit this energy to $r_0$, due to gravitational redshift it is reduced by a factor of $f(r)/f(r_0)$ when it arrives. The total extractable energy is therefore
$$-\frac{m}{f(r_0)} \int_{r_i}^{r_f} f'(r)\,dr = m\frac{f(r_i)-f(r_f)}{f(r_0)}$$
If $r_i=r_0$ (lowering from your location) and $f(r_f)=0$ (lowering to an event horizon), then the extracted energy is $m(c^2)$.
This argument is independent of $f$, and works with other metrics that have this form, such as Schwarzschild, Reissner-Nordström, and Schwarzschild-de Sitter black holes.
A: Let's assume that galaxy is in distance L from us. The average velocity of this Galaxy can be explained by the initial universe expanding and is equal:
$$v\sim HL\Rightarrow W_{kin}= M_{galaxy}v^2/2\approx M_{galaxy}H^2L^2/2$$
Considering the probable infinite accelerating of expanding (we cannot be sure in it), we can calculate the inertial forces that can experience object in the distance L from us. Let's set the easiest relation $H=H_0+kt$ (really $\dot{H}=-H^2(1+q)$ but we don't know q), because the times of major change in function are very big. So a=kL and we'll get energy (from relativistic 2nd newton law):
$$F=M_{galaxy}(a\gamma +(v^2/c^2)\gamma^{3}a)=kLM_{galaxy}\gamma(1 +(v^2/c^2)\gamma^2)=_{(v=HL)}=kM_{galaxy}L\dfrac{1}{\sqrt{1+H^2L^2/c^2}}(1 +\dfrac{H^2L^2}{c^2+H^2L^2})$$
This is not conservative force so the energy will depend on how are you pulling your rope. But if we will just integrate this force we will get:
$$W_{in_1}\approx \int_0^{F(v=c)}FdL=\dfrac{kM_{galaxy}c^2}{H^2} (-3 + 2.5\sqrt{2})$$
In some models Hubble constant decreases with time (k<0) but each galaxy will still accelerating because will also experience inertial force, associated with changing inertial system of body due to moving from us (just like coriolis force):
$$a_{in_2}=Hv_{radial}\approx H^2L\Rightarrow F=H^2LM_{galaxy}\dfrac{1}{\sqrt{1+H^2L^2/c^2}}(1 +\dfrac{H^2L^2}{c^2+H^2L^2})$$
This is also not conservative force so the energy will depend on how are you pulling your rope. But if we will just integrate this force we will get:
$$W_{in_2}\approx \int_0^{F(v=c)}FdL=M_{galaxy}c^2 (-3 + 2.5\sqrt{2})$$
Too simplify calculations I assumed that initial L is much smaller than $L_{HubHor}$. Also we can see factors like $mc^2$.
Let's also consider gravitational forces that in our model assumed weaker than influence of dark matter. Average density of universe is approximately $10^{-26}kg/m^3$, so the gravitational force will depend linearly from distance:
$$F=\dfrac{M_{galaxy}G\rho L^3}{L^2}\Rightarrow W_{grav}=-M_{galaxy}G\rho(L_{HubHor}^2-L^2)/2$$
That's the simpliest calculations that we can make for you question. For any clarifications you need to refer to specific models like Friedmann universe and manipulating with maximum of non-conservative force's work.
