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Assuming all these companies are interested in launching their own constellations of ~10K-100K satellites into L/MEO, how many companies could actually do this
by embedding-shape 10mo ago
Assuming all these companies are interested in launching their own constellations of ~10K-100K satellites into L/MEO, how many companies could actually do this before cascading collisions starts becoming a real worry?
- m4rtink 10mo agoIf they put their sats low enough (like Starlink already mostly does) any collision debris should be quickly deorbitted by drag, before a cascade can happen.
- JumpCrisscross 10mo ago> how many companies could actually do this before cascading collisions starts becoming a real worry? Twenty of them at 100,000 birds each to start approaching the density of planes in the sky [1]. Not around an airport. In all of the sky. Oceans and all. Practically speaking, this is not a pressing concern for our generation. [1] https://news.ycombinator.com/item?id=46711405 https://news.ycombinator.com/item?id=46711405
- manacit 10mo agoIt's interesting that people have a hard time visualizing this. The area in Earth's LEO is, definitionally, bigger than the Earth itself. The SEA parking garage fits 12,000 cars in it. Two of those spread over the entire planet would be an imperceptible amount of space. You could drop a pin on a map your entire life and probably never hit one.
- paulb73 10mo agoSEA parking garage? Unfamiliar with this size reference.
- deleted 10mo ago[deleted]
- mlmonkey 10mo agoHere, "SEA" = "Seattle Tacoma International Airport" in the state of Washington, USA.
- doublesocket 10mo agoThat is ... an oddly specific reference?
- deleted 10mo ago[deleted]
- skyo 10mo agoIt's one of the largest parking garages in the world
- y1n0 10mo agoThis is fair but the cars are stationary, occupying a parking space. Satellites occupy a ring.
- Gravityloss 10mo agoSpeed matters a lot. You can fit a lot more walking people than speeding motorcycles in the same space. Satellites need to travel at 8 km/s to not fall down.
- JumpCrisscross 10mo ago> Speed matters a lot Not really. You're correct inasmuch as it increases collision energies. But it also increases momentum, which maintains orbital integrity within predictable bounds. Nobody is maneuvering around satellites, they–and their debris–stay where the math tells them to.
- notahacker 10mo agoOrbits are predictable, but they intersect and decay [at different rates] and occasionally get perturbed by space weather. This already needs periodic conjunction avoidance manoeuvres, and whilst orbits are fast satellite manoeuvres are slow, so the notice you need to avoid a conjunction is measured in hours rather than seconds. Can't imagine a scenario in which it would be sustainable for LEO to even approach the density of commercial aviation, except perhaps for a hypothetical where a single entity actually managed all the satellites. The other underestimated dimension is that satellite manoeuvres use up a finite supply of expensively-launched propellant. That's tolerable when Starlink is doing 50k conjunction avoidance manoeuvres in six months across its constellation, but once it becomes each satellite moving at least weekly you either need bigger satellites carrying more propellant or have to accept significantly higher collision risk than they currently do.
- dylan604 10mo ago> and whilst orbits are fast satellite manoeuvres are slow This is something people unfamiliar tend to misconceive in their limited thinking on the subject. You can't just tap the breaks to slow down. Changing altitude of satellites is done by speeding up to increase altitude and slowing down to lower altitude. Once you change the velocity and reach the desired altitude, you have to then undo that acceleration to get back to orbital velocity. Fuel is required in both directions. The less fuel used the better for the maneuver. Most satellites EoL is defined by remaining maneuvering fuel vs functionality of the hardware. My first understanding of accelerating in space was from the old Asteroids game. To slow down, you had to rotate 180° and start accelerating in that direction. Others might learn it from Kerbal.
- suncore 10mo agoAccording to https://www.youtube.com/watch?v=b66ZZ05wKC0 https://www.youtube.com/watch?v=b66ZZ05wKC0 this might end very badly very soon.
- JumpCrisscross 10mo ago> According to https://www.youtube.com/watch?v=b66ZZ05wKC0 https://www.youtube.com/watch?v=b66ZZ05wKC0 this might end very badly very soon The paper they cite [1] estimates "the no-manoeuvre collision time" for various orbits. It has no alarming results. That paper cites another paper [2], which raises the possibility of runaway conditions. It, in turn, runs a model developed in this paper [3]. [1] https://arxiv.org/pdf/2512.09643 https://arxiv.org/pdf/2512.09643 [2] https://conference.sdo.esoc.esa.int/proceedings/sdc9/paper/305/SDC9-paper305.pdf https://conference.sdo.esoc.esa.int/proceedings/sdc9/paper/3... [3] https://www.researchgate.net/publication/234449150_Critical_number_of_spacecraft_in_low_Earth_orbit_using_satellite_fragmentation_data_to_evaluate_the_stability_of_the_orbital_debris_environment/link/541ca2a90cf241a65a136d22/download?_tp=eyJjb250ZXh0Ijp7ImZpcnN0UGFnZSI6InB1YmxpY2F0aW9uIiwicGFnZSI6InB1YmxpY2F0aW9uIn19 https://www.researchgate.net/publication/234449150_Critical_...
- WarmWash 10mo agoMake the US land area ~20% larger. Randomly place 50,000 shoe boxes up and down the entire eastern seaboard. Randomly place 50,000 shoe boxes up and down the entire western seaboard. Send them in straight lines towards the other side of the country. See if any collide. Almost certainly none of them will. Edit: They will almost certainly For reference, if you placed all 50k boxes next to each other on the same beach, it would be about 10 miles wide. The total shoreline on either side would be ~1800 miles wide. And that's only 2D.
- thomascountz 10mo agoThis is funnily oddly specific.
- woah 10mo agoSpoken like someone who's never placed a shoebox on a seaboard
- hatthew 10mo agoBy my calculations there will be an average of 500 collisions, no? Each shoebox has an effective width of 2 feet, and with 50k of them that's about 1% density. With 50k in the other direction, and about a 1% collision rate, that's 500 collisions.
- wmf 10mo agoAFAIK they're in separate shells so the probability of collision is basically zero.
- idiotsecant 10mo agoThings you put in orbit at a certain elevation don't stay at that elevation forever.
- dylan604 10mo agoPresumably, there would be a corridor for traveling through elevations whether that was for reaching orbit or de-orbiting. The people placing things in orbit are not doing this with out coordination.
- idiotsecant 10mo agoThere are in fact many objects that deorbit in an exceedingly uncoordinated manner. It's a statistical inevitability that kessler syndrome is in our very near future if we allow higher orbits to be polluted.
- chris_va 10mo agoWhat your describing is called Kessler Syndrome https://en.wikipedia.org/wiki/Kessler_syndrome https://en.wikipedia.org/wiki/Kessler_syndrome ... It is a very real possibility, but less of a problem below 550km altitude because the decay time is much shorter (and why all of these mega constellations tend to stay at lower altitude, even though ~1000km is generally better for a communications satellite).
- JumpCrisscross 10mo ago> It is a very real possibility It's really not. Not in the popularly-portrayed manner. Militaries have been researching how to intentionally cause such a cascade in even a limited orbit. To my knowledge, there isn't a solution.
- chris_va 10mo agoIt's a very real possibility, it's just a very slow exponential. If you have 100,000 satellites, and each collision produces 50,000 pieces of shrapnel with some distribution of altitudes and atmospheric drag, it's not that hard to do the math. The cinematic portrayal of cascading failure (ala the movie Gravity) is completely insane, but that doesn't mean this isn't a real problem on a 100 year timescale.
- JumpCrisscross 10mo ago> If you have 100,000 satellites, and each collision produces 50,000 pieces of shrapnel with some distribution of altitudes and atmospheric drag, it's not that hard to do the math One, it is. And two, you need insane distributions to get over the energy requirements of plane changes. > on a 100 year timescale Irrelevant in LEO.
- torginus 10mo agoEven if these sats do collide and produce debris, they are in decaying orbits, so that stuff will eventually fall back down.
- dylan604 10mo agoDebris moves in 3D. Debris moving up will continue moving up. There is no force acting on it to bring it back down. Your comment makes it sound like an explosion would only be in 2D along the same orbit as the original object.
- toast0 10mo ago> There is no force acting on it to bring it back down. Gravity? But also orbital dynamics (at least as I understand it) means debris that debris that is flung up is going to have a more oval orbit, so the high point (apogee) increases and the low point (perigee) decreases. And a lower perigee means more atmospheric drag, which will help deorbit the debris.
- dylan604 10mo agoThat'd have to be one slow explosion to give it less than 1G of acceleration.
- KiwiJohnno 10mo ago>means debris that debris that is flung up is going to have a more oval orbit, so the high point (apogee) increases and the low point (perigee) decreases. And a lower perigee means more atmospheric drag, which will help deorbit the debris. Not quite. If you are at apogee and accelerate, your perigee will be raised. If you are at perigee and accelerate, your apogee will be raised. You can't increase your apogee and perigee at the same time. If the impulse is in the direction of orbit, then the altitude of your orbit 180 degrees from your current position will raise. If the impulse is against your orbital direction, your height 180 deg away will be lowered. Once you complete an entire orbit (360 degrees) you will pass through your current position again. If you wish to move to a higher, circular orbit two impulses are required, 180 deg apart.
- zikduruqe 10mo ago> cascading collisions https://en.wikipedia.org/wiki/Kessler_syndrome https://en.wikipedia.org/wiki/Kessler_syndrome