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Here is my idea for a limitless* power. A gun on the moon fires moon rocks towards the Earth. The moon rocks fall in Earths gravity picking up speed and smash into Earth's atmosphere generating huge amounts of energy which is collected. Some of this energy is then fired by lasers back to the gun on the moon.

Is this in anyway a feasible way to generate energy? What are some of the hurdles?

Perhaps instead of rocks, the gun could fire ionised particles from moon dust.

(*For all intents and purposes. Say, for a few millennia.)

  • This is a good question, but it has been asked before. See this question. https://physics.stackexchange.com/q/178417/37364 – mmesser314 Nov 21 '18 at 03:27
  • @mmesser That's quite a different question. – PM 2Ring Nov 21 '18 at 06:18
  • Vaguely relevant xkcd What If. Also, https://xkcd.com/681/ – PM 2Ring Nov 21 '18 at 06:22
  • You are assuming that while mass is affected by gravity, the laser light would not be. This is not correct. Einstein talked about this. Otherwise similar simpler perpetual motion / infinite energy machine would have been possible. See this source: http://www.astro.umd.edu/~chris/Teaching/ASTR121_Spring_2010/class03.pdf – Steeven Nov 21 '18 at 12:06
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    @Steeven The laser light in the case asked about here is merely to provide the running power to get the rock mass out of the moon's gravity. It's not intended to provide all of the energy that is then harvested. – owjburnham Nov 21 '18 at 12:12
  • Eventually the moon is gone, so this isn't a renewable energy source of course. Either way, this better belongs into worldbuilding.SE or engineering.SE – DK2AX Nov 22 '18 at 09:41

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The Moon is finite, so this method can't supply limitless energy. But if you could do it with 100% efficiency, it would furnish the equivalent of current world energy consumption for over 7 billion years, although that BOTE calculation ignores the energy cost of cutting the Moon rock into convenient sized chunks.

This gravity well diagram from xkcd says that Earth's gravitational potential is equivalent to 6379 km in a constant $1g$ field, the figure for the Moon (also at $1g$) is 288 km. So the energy released by lifting mass $m$ from the Moon's surface and dropping it onto the Earth is $mg × 6091 km$.

Using $m=7.34767309 × 10^{22} kg$ for the mass of the Moon, and $g=9.81m/s^2$, we get $4.39043379×10^{30}$ joules.

That calculation ignores the fact that the Moon's gravity well gets shallower as we eat away at its mass. It also ignores the increase in the depth of Earth's gravity well as its mass increases due to the accumulated Moon rock.

However, there are enormous technical difficulties with this method, whether we catch the incoming Moon rock and somehow get it to turn a turbine, or allow it to burn up in the upper atmosphere and try to capture the heat it releases. Catching the heat would require the heat harvester to operate in the upper atmosphere. Catching the rocks themselves will put a lot of wear & tear on the equipment.

There's also the political dimension to consider. Plenty of people would not be happy with a railgun on the Moon that could be used as a weapon, or that could cause a disaster if anything went wrong with the aiming process.

PM 2Ring
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  • That's true. We'd have to consider the energy spent mining the moon rocks. And if this would cancel out the energy release when they fell to Earth. –  Nov 21 '18 at 13:34
  • @zooby At a guess, the energy necessary to mine the moon rocks is less than the energy required to lift them out of the lunar gravity well (the gravitational binding energy). It's basically the chemical binding energy. If the Moon were a big pile of loose pebbles, then we'd only need to worry about the gravitational binding energy. BTW, IMHO sending energy back to the Moon via laser just makes the system more inefficient. We could power the mining machines & railgun by solar power, OTOH, night lasts 2 weeks on the Moon... – PM 2Ring Nov 21 '18 at 13:49