First off, I know this wouldn't work due to it breaking thermodynamics and that there would also be heaps of energy dissipated to the surroundings, and fusion doesn't work yet, but I'm more interested in why, even if no energy was dissipated, it wouldn't work.

  1. Start with some nuclei that can be fused (like hydrogen), a set distance off the ground, on a platform that powers a generator when it is lowered
  2. Allow the platform that they are on to drop, powering the generator, putting the energy (from g.p.e. of the nuclei) into battery A
  3. At the bottom, fuse them together, and put the energy from that into battery B
  4. Use the energy from battery A to raise it again. This will use slightly less energy than was used to raise it due to the new larger nuclei being lighter than the initial nuclei (because of the mass defect), so some energy will remain in the battery
  5. Use the energy from battery B to split that nucleus into the lighter nuclei that we started with. If my understanding is correct, this should lead to having exactly 0 energy in battery B
  6. We are now left with the same nuclei at the same height, with more energy in the batteries than we started with

Because there is no net energy transfer due to splitting and fusing the nuclei, the small amount of energy gained from lifting less than was dropped.

To reiterate, I just want to know why it wouldn't work, not just that it wouldn't. Also please keep answers simple as I am only GCSE level physics

  • $\begingroup$ Gravity couples to energy not mass. $\endgroup$ – Qmechanic Dec 18 '18 at 19:51

The problem is that you're not applying $E = mc^2$ to everything. In particular, you put energy into battery B at the bottom, but take it out at the top. This corresponds to raising a mass up, which precisely cancels the claimed extra energy you get.

To make this perfectly clear, let's substitute the word "energy" for "mass" everywhere in your setup.

  1. Start with a pile of rocks on a platform
  2. Allow the platform to drop, powering battery A
  3. At the bottom, take one rock out and put it into battery B
  4. Raise the platform with battery A. This will use slightly less energy than was used to raise it, so some energy will remain in the battery.
  5. Take the rock out of battery B and put it back on the platform

It should be clear that the extra energy needed to do step (5) is exactly equal to the energy you stored in battery A.

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  • $\begingroup$ Thanks, that seems to make some sense, but I'm not sure what is meant by "This corresponds to raising a mass up". Does this mean because of the mass of the battery? Or is it some weird relativistic effect of raising energy to a higher level? (I don't know much about such things, being GCSE level) If the former is true, could you simply bypass that by transferring the energy using a wire? $\endgroup$ – Joe Robinson Dec 18 '18 at 20:04
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    $\begingroup$ @JoeRobinson It simply costs energy to raise energy to a higher level, no matter what form it takes. For example, if you transferred the energy by shining light up, it would lose energy due to gravitational redshift. A similar thing will happen with a wire, though it's more complicated. $\endgroup$ – knzhou Dec 18 '18 at 20:08

The energy you use to perpetually drive this system, or any one like it, did not come from nowhere. It was contained in the raw materials you started with. As such, the helium feedstock is fuel and your process is intended to extract useful work from that fuel, just like a gasoline engine burns the gas to perform work.

By the way, your process- since it relies on nuclear fusion to power it- could in principle be replaced by a fusion reactor driving a generator.

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