There is an error in thinking that the high-altitude helium will be at a lower temperature than the low altitude helium. If the helium is not being continually stirred (which would take energy input), then the equilibrium state has the density decreasing with height, but the temperature is uniform. The high-altitude atoms are just as energetic as the low-altitude atoms. This is a basic fact about thermal equilibria: each of the degrees of freedom has the same time-averaged energy. Temperature is just energy per degree of freedom.
If the initial setup is not in that equilibrium state, then you can extract work from it, but only a finite amount as it approaches equilibrium.
The continual stirring of the Earth’s atmosphere is a substantial contributor to the decrease of temperature with height.
You raise a point about thermal equilibrium, and you’re absolutely right that in a static equilibrium, temperature would be uniform throughout the gravitational field.
However, the system I’m describing isn’t in thermal equilibrium as we “start” the setup (e.g. insert the atmosphere). It’s in a state with continuous evaporative cooling. When the fastest atoms escape the gravity well, they remove more than the average kinetic energy from the atmosphere (that’s why they can escape).
This is exactly analogous to evaporative cooling of water, just with gravity instead of intermolecular forces.
The key is that the CMB at 5K provides continuous heat input to compensate for this cooling. So we have:
1. Fast atoms escape → top of atmosphere cools below 5K
2. CMB radiation heats the atmosphere back toward 5K
3. This maintains a steady temperature gradient
4. If we waited enough, eventually an equal amount of atoms would be falling back to the asteroid as is evaporating
5. Eventually we would reach an equilibrium
6. However, we do not let that happen, since as soon as density accumulates near the shell, we enclose the atoms (metaphrically like just screwing a lid on a jar “trapping” some air), and send them down in a way where we can extract some of the potential energy before releasing them.
7. This requires no knowledge of individual atoms. One can calculate statistically when there will be a higher density near the shell (or have a measuring device).
Short answer is no. Here on Earth space “starts” at around 100 km. Above this we kind of have a vacuum. The same would be true for the asteroid (above a certain point the pressure is very low). As long as we release the lowered atoms above this point there would be no atmospheric pressure to fight against.
You’re sorta talking here about extracting work from an initial pressure differential by converting it to a temperature differential, just like in the Planet X example.
That would be fine, but it contradicts your post, where you specifically state that everything starts in thermal equilibrium at 5K. The CMB is still irrelevant and unneeded, and does not provide the kind of T gradient you’re claiming. (6) does not work, and (7) is not true.
There is an error in thinking that the high-altitude helium will be at a lower temperature than the low altitude helium. If the helium is not being continually stirred (which would take energy input), then the equilibrium state has the density decreasing with height, but the temperature is uniform. The high-altitude atoms are just as energetic as the low-altitude atoms. This is a basic fact about thermal equilibria: each of the degrees of freedom has the same time-averaged energy. Temperature is just energy per degree of freedom.
If the initial setup is not in that equilibrium state, then you can extract work from it, but only a finite amount as it approaches equilibrium.
The continual stirring of the Earth’s atmosphere is a substantial contributor to the decrease of temperature with height.
You raise a point about thermal equilibrium, and you’re absolutely right that in a static equilibrium, temperature would be uniform throughout the gravitational field.
However, the system I’m describing isn’t in thermal equilibrium as we “start” the setup (e.g. insert the atmosphere). It’s in a state with continuous evaporative cooling. When the fastest atoms escape the gravity well, they remove more than the average kinetic energy from the atmosphere (that’s why they can escape).
This is exactly analogous to evaporative cooling of water, just with gravity instead of intermolecular forces. The key is that the CMB at 5K provides continuous heat input to compensate for this cooling. So we have:
1. Fast atoms escape → top of atmosphere cools below 5K
2. CMB radiation heats the atmosphere back toward 5K
3. This maintains a steady temperature gradient
4. If we waited enough, eventually an equal amount of atoms would be falling back to the asteroid as is evaporating
5. Eventually we would reach an equilibrium
6. However, we do not let that happen, since as soon as density accumulates near the shell, we enclose the atoms (metaphrically like just screwing a lid on a jar “trapping” some air), and send them down in a way where we can extract some of the potential energy before releasing them.
7. This requires no knowledge of individual atoms. One can calculate statistically when there will be a higher density near the shell (or have a measuring device).
When you release the lowered atoms back at the surface, do you have to fight against the atmospheric pressure?
Good question!
Short answer is no. Here on Earth space “starts” at around 100 km. Above this we kind of have a vacuum. The same would be true for the asteroid (above a certain point the pressure is very low). As long as we release the lowered atoms above this point there would be no atmospheric pressure to fight against.
You’re sorta talking here about extracting work from an initial pressure differential by converting it to a temperature differential, just like in the Planet X example.
That would be fine, but it contradicts your post, where you specifically state that everything starts in thermal equilibrium at 5K. The CMB is still irrelevant and unneeded, and does not provide the kind of T gradient you’re claiming. (6) does not work, and (7) is not true.