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Considering the fact that you have the entire moon as available surface, you'd rather lift a giant rock with that power and store all the potential gravitationa

• deleted 4y ago
  [deleted]
• sbierwagen 4y ago
  >A 10 ton rock, lifted up to 50m high, will store up to 810M Joules at moon gravity Wrong prefix? I think you meant 810K Joules. For Earth gravity: 10,000 kg times 9.8 m/s^2 times 50 m = 4.9 megajoules, 1.36 kilowatthours. (Anker sells a 2.04 KWh battery pack for US$1999: https://www.anker.com/products/a1780 https://www.anker.com/products/a1780 The manual says it masses 30.5 kg, so that's a cool 491.88x difference in mass power density.) Again, that's for Earth gravity. Lunar gravity: 10,000 kg times 1.62 m/s^2 times 50 m = 0.81 megajoules, 0.225 kilowatthours. (2,976x difference in mass power density vs the lithium battery.) Gravity storage isn't economical on Earth, and it's really not economical on the Moon, where gravity is lighter.
• sacred_numbers 4y ago
  Unfortunately I think you're off by an order of magnitude. I think it would be 810 Kilojoules, which is approximately equivalent to a 1kg lithium-ion battery. Of course, you could move thousands of rocks up and down a big crater, rather than just one, but it would still be a lot of infrastructure for a fairly small amount of energy storage.
  • sbierwagen 4y ago
    Three orders of magnitude*