What would a pair of Death Stars do to Earth?

Question

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?

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 ;-) — Cort Ammon, Nov 23 '14 at 19:32
Yes, let's ignore the politics -- that already caused some confusion around my older Death Star question. — Shokhet, Nov 23 '14 at 19:33
What mass do they have and what are their orbital parameters? — Philipp, Nov 23 '14 at 19:38
@Philipp I don't have exact numbers on the mass, but assuming a moon-sized piece of metal should be close enough. Orbital parameters are left open in the question -- they could be in geosynchronous orbit if necessary, but can also have a regular orbit. — Shokhet, Nov 23 '14 at 19:41
@Vincent I went looking for a source, but I was under the impression that it was designed to be around the same size as the (or a) moon. — Shokhet, Nov 24 '14 at 19:03
As big as a moon, but as Isaac Asimov pointed out at great length, the Earth has a huge moon by planetary standards. — DJClayworth, Nov 24 '14 at 23:22
My wife, reading just the title over my shoulder: "...Blow it up twice?" — Tim Pederick, Nov 25 '14 at 10:05

score 17 · Answer 1 · edited Sep 16 '15 at 19:48

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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.

edited Sep 16 '15 at 19:48

matts

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answered Nov 24 '14 at 05:37

Mark

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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)? – Nov 24 '14 at 06:32
@CreationEdge, roughly a degree across, or about twice the size of the Moon. – Mark Nov 24 '14 at 07:40
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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. – Loren Pechtel Nov 24 '14 at 21:26
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@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. – Mark Nov 24 '14 at 22:38
@Mark My point is that the energy of the tides is going to be small compared to the forces it's own drive exerts on it. Thus I'm not worried about it coming apart if it gets too close to a planet. – Loren Pechtel Nov 26 '14 at 00:29
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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. – Keith Thompson Nov 26 '14 at 03:57
@LorenPechtel The drives of Starwars are essentially magic. Since we do not know the rules of this magic drawing inferences from them to how the object would hold against physical phenomenae that we do understand is invalid. – Taemyr Sep 17 '15 at 09:13
The tidal force of the death star at 9Mm would be 22% larger than the tidal force of the moon. (10^18 kg/(9*10^6 m)^3)/((mass of moon)/(distance to moon)^3)=1.224 - At 175km it would be about 16000 times larger than the tidal force from the moon. – Taemyr Sep 17 '15 at 09:31
@KeithThompson You don't see large tides where the Death Star happens to be. Tidal bulges are rather small, big tides happens where those bulges hits land or get channeled through narrows. – Taemyr Sep 17 '15 at 09:34
@Taemyr, I think you're using the wrong distance. 9Mm and 175km are center-to-surface distances, and I think your formula needs center-to-center distances. – Mark Sep 17 '15 at 09:35
@Mark You are right. Is the 9Mm number from your calculation really center to surface? – Taemyr Sep 17 '15 at 09:42
@Taemyr, Looking back, I'm not sure. I just re-ran the calculation and got a center-to-surface distance of 10.7Mm. – Mark Sep 17 '15 at 09:51
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@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! – Bob Jarvis - Слава Україні Nov 16 '17 at 04:37
Wait...this ultimate galactic roving FTL superweapon has poor protection against fighters, an unshielded exhaust port, and is vulnerable to tides? I'm losing faith in the original design team... – user535733 Feb 18 '18 at 19:20

What would a pair of Death Stars do to Earth?

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