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Y'all should know by now that I want to build a Death Star ;-)

Let's assume that we did that already -- and against the odds, we've built two of them, and they're currently in orbit around a planet; let's say planet Earth.

What would happen to our planet if two extra [metal] moons joined our orbit?

  • Tides would probably be affected; how?
  • Would sunlight patterns be affected?
  • Would there be differences if these gargantuan constructs stayed in place over one spot, or if they had a moon-like orbit?
  • Anything other major effects that I'm missing?
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    $\begingroup$ Are we ignoring the political effects of a pair of death stars orbiting the planet? I'm pretty sure the main-beam of the death star will have substantially more effect than changing the tides ;-) $\endgroup$ – Cort Ammon Nov 23 '14 at 19:32
  • $\begingroup$ Yes, let's ignore the politics -- that already caused some confusion around my older Death Star question. $\endgroup$ – Shokhet Nov 23 '14 at 19:33
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    $\begingroup$ The death star isn't solid $\endgroup$ – evandentremont Nov 23 '14 at 21:45
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    $\begingroup$ As big as a moon, but as Isaac Asimov pointed out at great length, the Earth has a huge moon by planetary standards. $\endgroup$ – DJClayworth Nov 24 '14 at 23:22
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    $\begingroup$ My wife, reading just the title over my shoulder: "...Blow it up twice?" $\endgroup$ – Tim Pederick Nov 25 '14 at 10:05
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What would happen? Not much.

Canonically, the Death Star is about 150 km in diameter. I can't find an official number for the mass, so I'll assume an average density about 10% that of steel, for an overall mass of 1.0×1018 kg. I'll also assume that the Death Star cannot orbit closer than its Roche limit, which, if I haven't botched the calculations, is about 9,000,000 m above the surface.

Feed the mass and distance into Newton's law of gravitation, and this gives a force on a 1 kg object on the Earth's surface of 8×10-7 N, or about 1/100,000,000th the force of Earth's gravity. Simple geometric calculations give an angular size of about 1 degree, twice that of the Moon.

Edit: Even in a much closer orbit (175 km, putting the lower edge just above the Karman line) will have little physical effect, producing a force of only 0.02% of Earth's gravity. The psychological effect of something with an angular diameter of 45 degrees passing overhead, trailing a plasma cloud from its lower edge, is a different matter.

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  • $\begingroup$ How large would the Death Star appear to be at 9,000 KM above the surface of Earth (at it's Roche limit you calculated)? $\endgroup$ – Web Head Nov 24 '14 at 6:32
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    $\begingroup$ I see no reason to assume the Roche limit is relevant. It's a big ball of metal meant to be boosted to extreme velocities. It's strong enough to hold together against planetary tides. $\endgroup$ – Loren Pechtel Nov 24 '14 at 21:26
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    $\begingroup$ @LorenPechtel, at the scale of the Death Star, you can't treat it as a rigid body. No, it won't fall apart under tidal forces, but it will flex in ways the designer would really rather it didn't. $\endgroup$ – Mark Nov 24 '14 at 22:38
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    $\begingroup$ If it's in close orbit, the force exerted by its gravity won't be huge, but the tides (which vary with the inverse cube of the distance rather than the inverse square) could be significant. We could see large ocean tides, localized to where the Death Star happens to be at the moment -- and perhaps ground tides and earthquakes as well. $\endgroup$ – Keith Thompson Nov 26 '14 at 3:57
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    $\begingroup$ @Mark - ah, but that "flexing in ways the designer would really rather it didn't" is why the framework of the Death Star is made from a handwavium-impossibilium alloy, which NOT ONLY absorbs all the tidal stresses and dissipates them harmlessly, it ALSO tones, stretches, slices, dices, shapes, increases, reduces, AND prevents hair loss! $\endgroup$ – Bob Jarvis Nov 16 '17 at 4:37

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