# Roche Limit vs Death Star **URL:** \ **Category:** Factual Questions\ **Created:** [December 4, 2024, 2:36am UTC](https://boards.straightdope.com/t/roche-limit-vs-death-star/1011141 "2024-12-04T02:36:30Z")\ **Posts on this page:** 20\ **Page:** 1
**Author:** ![ArslanWilson](https://avatars.discourse-cdn.com/v4/letter/a/6de8d8/32.png) [@ArslanWilson](https://boards.straightdope.com/u/ArslanWilson)\ **Post date:** [December 4, 2024, 2:36am UTC](https://boards.straightdope.com/t/roche-limit-vs-death-star/1011141/1 "2024-12-04T02:36:31Z")
Pretty much what it sounds like. When orbiting the Forest Moon of Endor, assuming no space magic at play, would the physics behind the Roche limit not have ripped the DS to sparkly shreds? Far denser objects, like that Shoemaker comet, get broken into dozens of components by a world with only twice —or so— the gravity involved. Anyway, I ask as this is actually a plot point in an upcoming story I’m working on. LSS, dozens of 10 mile or so long starships, each massing about 90mio tons, must essentially repel down Sky Elevators to dismantled and use as material for colonizing a new world. But lately, I am wondering about how much of an issue that would actually be, starting from a geo-syn orbit. So naturally, Deathstar. ---
**Author:** ![Whack-a-Mole](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/whack-a-mole/32/141_2.png) [@Whack-a-Mole](https://boards.straightdope.com/u/Whack-a-Mole)\ **Post date:** [December 4, 2024, 2:43am UTC](https://boards.straightdope.com/t/roche-limit-vs-death-star/1011141/2 "2024-12-04T02:43:06Z")
The Roche Limit only applies to other celestial bodies that are held together by their own gravity. The Deathstar is an engineered thing and so much stronger and harder to pull apart than a comet would be. Also, the Deathstar is presumably made of Unobtainium™ (a sci-fi material which has properties no earthly material has). It’s been a while (so no link) but I read that something as long as the Executor class super star destroyer (Darth Vader’s command ship) would tear itself apart if made from steel. At 12 miles long pushing one end to turn it would cause stress at the other end. Greater stress than a modern material could handle so the ship would break. The Deathstar is considerably bigger. ---
**Author:** ![kenobi\_65](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/kenobi_65/32/13178_2.png) [@kenobi\_65](https://boards.straightdope.com/u/kenobi_65)\ **Post date:** [December 4, 2024, 2:44am UTC](https://boards.straightdope.com/t/roche-limit-vs-death-star/1011141/3 "2024-12-04T02:44:43Z")
> [@Whack-a-Mole](#): > > The Deathstar is considerably bigger. Indeed. Canonically, it’s 160 kilometers in diameter. > **[Death Star II](https://starwars.fandom.com/wiki/Death_Star_II)** > > The Death Star II was the second Death Star battle station to be constructed by the Galactic Empire. It was the same size as its predecessor, measuring 160 kilometers in diameter, and was built after the destruction of the first Death Star at the... ---
**Author:** ![Northern\_Piper](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/northern_piper/32/5304_2.png) [@Northern\_Piper](https://boards.straightdope.com/u/Northern_Piper)\ **Post date:** [December 4, 2024, 2:52am UTC](https://boards.straightdope.com/t/roche-limit-vs-death-star/1011141/4 "2024-12-04T02:52:30Z")
However, it is spherical, so wouldn’t the stress be more evenly distributed than Vader’s ship? ---
**Author:** ![md-2000](https://avatars.discourse-cdn.com/v4/letter/m/9d8465/32.png) [@md-2000](https://boards.straightdope.com/u/md-2000)\ **Post date:** [December 4, 2024, 3:44am UTC](https://boards.straightdope.com/t/roche-limit-vs-death-star/1011141/5 "2024-12-04T03:44:48Z")
Larry Niven’s _The Integral Trees_ posits a world (well, a toroidal cloud of breathable air held together by gravitational “magic”) where some people live on “trees” some 40km long. Due to differential gravitational force (tidal differences) they orient pointing toward the center of the graviational force. Tidal forces pull a body apart when the tidal difference exceeds the gravitational force holding it together. I.e. a body is 1,000km diameter. It orbits a much bigger body fairly close in. The first, closest surface is orbiting 500km closer than the center of the orbital body. The second, other surface is 500km further out than it should be for that speed. For the first surface, going too slow for it’s orbital location, it wants to fall inward. The second surface wants to fly away. When that difference is greater than the body;s gravity, the pieces start to fall apart. pieces of the first surface fall inward, dscribing a smaller, faster elliptical orbit. The second surface flies off, creating a elliptical orbit with a higher apogee. If the body is spinning, then this happens all along the equator as each aprt reaches there. (Not unlike a body spinning so fast stuff flies off its equator). As the body loses equator, presumably the whole sphere will rearrange smaller, etc. Some of the orbital energy tends to be lost to heating as the tides distort the body, thus losing even more monentum in the orbit over time… The Death Star etc being mechanically welded and riveted is less suceptible to 'stuff flying off" altough you would not want to be standing on the surface as it did a quick pass (or even inside the Star outside of the optimal orbital speed) or you get thrown about by the primary’s gravity, or in the worst case, see Nven’s _Neutron Star_ story. ---
**Author:** ![Princhester](https://avatars.discourse-cdn.com/v4/letter/p/3e96dc/32.png) [@Princhester](https://boards.straightdope.com/u/Princhester)\ **Post date:** [December 4, 2024, 3:46am UTC](https://boards.straightdope.com/t/roche-limit-vs-death-star/1011141/6 "2024-12-04T03:46:08Z")
The critical part of Whack-a-Mole’s post is the stress created by: > [@Whack-a-Mole](#): > > pushing one end to turn it… combined (I assume) with measurements gleaned from the footage within the well known Star Wars documentaries, to determine rates of turn. I’m not an expert on those documentaries but don’t recall there being any explanation of the Death Stars’ means of propulsion. If it is some sort of field (perhaps related to the “tractor beam” technology demonstrated in the first-distributed episode) maybe it acts on the whole mass of the Death Star at once, inducing no internal strain? ---
**Author:** ![Princhester](https://avatars.discourse-cdn.com/v4/letter/p/3e96dc/32.png) [@Princhester](https://boards.straightdope.com/u/Princhester)\ **Post date:** [December 4, 2024, 3:50am UTC](https://boards.straightdope.com/t/roche-limit-vs-death-star/1011141/7 "2024-12-04T03:50:20Z")
> [@md-2000](#): > > The Death Star etc being mechanically welded and riveted is less suceptible to 'stuff flying off" altough you would not want to be standing on the surface as it did a quick pass (or even inside the Star outside of the optimal orbital speed) or you get thrown about by the primary’s gravity, or in the worst case, see Nven’s _Neutron Star_ story. Given that the Star Wars documentaries clearly show that personnel within the Death Stars can walk upright at all times, and the Death Stars would not be sufficiently heavy to generate their own gravity, and they don’t spin, there is clearly some form of artificial gravity operating which overcomes these difficulties. ---
**Author:** ![kenobi\_65](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/kenobi_65/32/13178_2.png) [@kenobi\_65](https://boards.straightdope.com/u/kenobi_65)\ **Post date:** [December 4, 2024, 3:51am UTC](https://boards.straightdope.com/t/roche-limit-vs-death-star/1011141/8 "2024-12-04T03:51:52Z")
> [@Princhester](#): > > I’m not an expert on those documentaries but don’t recall there being any explanation of the Death Stars’ means of propulsion. If it is some sort of field (perhaps related to the “tractor beam” technology demonstrated in the first-distributed episode) maybe it acts on the whole mass of the Death Star at once, inducing no internal strain? Any attempts to apply real-world physics to the Star Wars universe will just make you cry. And, no, there was never any “explanation” of how it moves in the movies (though there’s all manner of semi-canon or non-canon details like that that have undoubtedly been written, with varying levels of technobabble). Obviously, I’m a big Star Wars fan, but in no way, shape, or form is it anything even vaguely approaching hard science fiction; it’s a set of fantasy movies that happen to take place in space. ---
**Author:** ![Horatius](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/horatius/32/3572_2.png) [@Horatius](https://boards.straightdope.com/u/Horatius)\ **Post date:** [December 4, 2024, 4:02am UTC](https://boards.straightdope.com/t/roche-limit-vs-death-star/1011141/9 "2024-12-04T04:02:25Z")
> [@Princhester](#): > > there is clearly some form of artificial gravity operating which overcomes these difficulties In addition to this, we know the have several kinds of force fields and tractor beams. Those could provide greater structural integrity that just normal physical materials. ---
**Author:** ![Whack-a-Mole](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/whack-a-mole/32/141_2.png) [@Whack-a-Mole](https://boards.straightdope.com/u/Whack-a-Mole)\ **Post date:** [December 4, 2024, 4:05am UTC](https://boards.straightdope.com/t/roche-limit-vs-death-star/1011141/10 "2024-12-04T04:05:24Z")
> [@Horatius](#): > > In addition to this, we know the have several kinds of force fields and tractor beams. Those could provide greater structural integrity that just normal physical materials. The same notion is used a lot in Star Trek too. ---
**Author:** ![Exapno\_Mapcase](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/exapno_mapcase/32/1051_2.png) [@Exapno\_Mapcase](https://boards.straightdope.com/u/Exapno_Mapcase)\ **Post date:** [December 4, 2024, 4:17am UTC](https://boards.straightdope.com/t/roche-limit-vs-death-star/1011141/11 "2024-12-04T04:17:38Z")
> [@Whack-a-Mole](#): > > Also, the Deathstar is presumably made of Unobtainium™ Most stuff in media sf is made from Plotium. Whatever the plot demands, the stuff is capable of. That’s just a special case of the more general Hollywoodium. _Whatever_ the plot demands happens. Don’t ask questions. ---
**Author:** ![Dr.Strangelove](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/dr.strangelove/32/6613_2.png) [@Dr.Strangelove](https://boards.straightdope.com/u/Dr.Strangelove)\ **Post date:** [December 4, 2024, 4:56am UTC](https://boards.straightdope.com/t/roche-limit-vs-death-star/1011141/12 "2024-12-04T04:56:09Z")
A wizard space wizard did it. ---
**Author:** ![Der\_Trihs](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/der_trihs/32/233_2.png) [@Der\_Trihs](https://boards.straightdope.com/u/Der_Trihs)\ **Post date:** [December 4, 2024, 5:05am UTC](https://boards.straightdope.com/t/roche-limit-vs-death-star/1011141/13 "2024-12-04T05:05:41Z")
Given how large it is, the fact that it hasn’t collapsed on itself demonstrates that it’s either very structurally strong or held in shape by “space magic” such as the obvious artificial gravity. That makes applying the Roche Limit problematic because the Limit applies to objects held together only by their own gravity - that’s why near- Earth satellites and spacecraft aren’t just immediately torn apart. How close it can approach a planet is going to depend on how strong whatever is holding it together is, which isn’t specified as far as I know. ---
**Author:** ![Whack-a-Mole](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/whack-a-mole/32/141_2.png) [@Whack-a-Mole](https://boards.straightdope.com/u/Whack-a-Mole)\ **Post date:** [December 4, 2024, 5:11am UTC](https://boards.straightdope.com/t/roche-limit-vs-death-star/1011141/14 "2024-12-04T05:11:51Z")
> [@Northern\_Piper](#): > > However, it is spherical, so wouldn’t the stress be more evenly distributed than Vader’s ship? Wouldn’t rotational stress cause the same problem? If the Death Star rotates to aim its gun the outer shell will be moving faster than the bits at the center which puts stress on the structure. I’d think that would be a problem for something that size (if made from non-magic, earthly materials). ---
**Author:** ![Der\_Trihs](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/der_trihs/32/233_2.png) [@Der\_Trihs](https://boards.straightdope.com/u/Der_Trihs)\ **Post date:** [December 4, 2024, 5:31am UTC](https://boards.straightdope.com/t/roche-limit-vs-death-star/1011141/15 "2024-12-04T05:31:26Z")
> [@Whack-a-Mole](#): > > Wouldn’t rotational stress cause the same problem? That depends on the method; if it’s using some kind of gravity manipulation then possibly not, since the force being applied can be kept evenly across the structure. ---
**Author:** ![Princhester](https://avatars.discourse-cdn.com/v4/letter/p/3e96dc/32.png) [@Princhester](https://boards.straightdope.com/u/Princhester)\ **Post date:** [December 4, 2024, 6:26am UTC](https://boards.straightdope.com/t/roche-limit-vs-death-star/1011141/16 "2024-12-04T06:26:12Z")
It just depends on acceleration. Let’s say Darth needs to shoot up a planet and the DS is facing exactly the wrong way so he wants to turn the DS 180 degrees. If you assume that the DS has a diameter of 160km (given above) and can cope with accelerating at 1/3G at the circumference, it could make the turn in about ten minutes (starting and ending with no spin). This seems like it should be quick enough. It would accelerate at 1/3G for above five minutes, reach max angular velocity halfway through the turn, then decelerate at the same 1/3G. At the halfway point there would be about 1G of centripetal acceleration at the surface. [According to this table](https://en.wikipedia.org/wiki/Specific_strength) carbon fibre can support its own weight at 1G at up to 250km in length. So it’s not immediately apparent that the limited factors I’m considering here would make the exercise impossible. Always assuming (as is unlikely) I haven’t borked the calculations and/or dropped a zero or two somewhere along the line. ---
**Author:** ![Whack-a-Mole](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/whack-a-mole/32/141_2.png) [@Whack-a-Mole](https://boards.straightdope.com/u/Whack-a-Mole)\ **Post date:** [December 4, 2024, 6:42am UTC](https://boards.straightdope.com/t/roche-limit-vs-death-star/1011141/17 "2024-12-04T06:42:25Z")
My only question about this (and I really do not know): Does this assume a solid sphere as opposed to a hollow sphere with some structure? Also, the wiggle each time adds up. Same as if you take a metal spoon and bend it back and forth multiple times it will break. The DS would have that constant flexing. Even modern airplanes have to take this into account. Presumably this can be managed but the upkeep for the DS would be crazy. Perhaps it is the economics of such a thing that is most unbelievable. ---
**Author:** ![Der\_Trihs](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/der_trihs/32/233_2.png) [@Der\_Trihs](https://boards.straightdope.com/u/Der_Trihs)\ **Post date:** [December 4, 2024, 6:53am UTC](https://boards.straightdope.com/t/roche-limit-vs-death-star/1011141/18 "2024-12-04T06:53:57Z")
> [@Whack-a-Mole](#): > > Perhaps it is the economics of such a thing that is most unbelievable. It makes sense if you think of it _as_ a boondoggle, built by people who believe in rule by fear and _wunderwaffen._ The Empire was pretty blatantly based in large part on the Nazis after all. They could have built a vast number of Star Destroyers instead, but they wouldn’t have been as Big And Scary as the Death Star. Or the various other superweapons the Empire came up with. And if the Emperor wants his giant battle station are _you_ going to tell him it’s a bad idea? Of course not. ---
**Author:** ![Princhester](https://avatars.discourse-cdn.com/v4/letter/p/3e96dc/32.png) [@Princhester](https://boards.straightdope.com/u/Princhester)\ **Post date:** [December 4, 2024, 7:10am UTC](https://boards.straightdope.com/t/roche-limit-vs-death-star/1011141/19 "2024-12-04T07:10:34Z")
> [@Whack-a-Mole](#): > > Does this assume a solid sphere as opposed to a hollow sphere with some structure? Also, the wiggle each time adds up. Same as if you take a metal spoon and bend it back and forth multiple times it will break. As to calculations of the accelerations involved, it doesn’t matter whether it’s solid or hollow. As to structural strength, if I’m understanding the position correctly, 80km (ie radius-length) strands of carbon fibre could support themselves at 1G with strength left over to support other structure. So it would not need to be a solid ball of carbon fibre just to have enough strength to survive the accelerations I have described. As to fatigue (the phenomenon you mention) I’m not really too knowledgeable about the work rate of planet destroying death stars but I would tend to think they would be used relatively seldom and in a reasonably sedate manner - compare that to the gruelling flexing that the structure of a modern aircraft undergoes, operating in turbulent atmosphere and taking off and landing several times a day. And modern aircraft operate for a couple of decades. ---
**Author:** ![md-2000](https://avatars.discourse-cdn.com/v4/letter/m/9d8465/32.png) [@md-2000](https://boards.straightdope.com/u/md-2000)\ **Post date:** [December 4, 2024, 7:18am UTC](https://boards.straightdope.com/t/roche-limit-vs-death-star/1011141/20 "2024-12-04T07:18:33Z")
Rough rule of thumb, water has a density of 1 (10lb/gallon, 6 gal/cu-ft). Rock is density 3 give or take, iron density about 6. Etc. A Death Star, being mainly air (big open spaces with thin metal walls) is likely not very dense. Let’s do a rough calculation - 5,000km up, orbital period is 3.352hr. +/-800km (radius of “that’s not a small moon…”), 3.712hr and 3.005hr respectively. Earth is approx another 6400km radius, So orbit speed should be: 4200km up - r=10600km v=2x10600/3.005=7,054km/h 5000km … v=6800km/h 5800km … v=6680kph The center of the Death Star is travelling in an orbit at 6800kph. The inner and outer edges are respectively closer and further from Earth, but going the same speed. However, to maintain that circle, the edge of the Star closest to Earth should be going about 2,050kph faster than it is. It wants to fall inward. Only the main rivets (and plot-powered artifical gravity) keep the Death Star intact. However, this is based on orbital dynamics. A spacecraft in the Star Wars sense is not orbiting, so much as maneuvering on its own power. so it is not in “free fall” around a planet necessarily. It could be passing by under propulsion (of some form). A ship tied to a space elevator is not really orbiting, but then the space elevator cable has to support the additional weight (pull of gravity) of the ship. Presumably your derelict spaceship is brought down in small pieces. [Next page](https://boards.straightdope.com/t/roche-limit-vs-death-star/1011141.md?page=2)