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Every satellite I've seen crosses the sky in seconds. How does an orbiting mirror provide consistent illumination? Can it orbit as show as the rotation of the E
by everdev 8y ago
Every satellite I've seen crosses the sky in seconds. How does an orbiting mirror provide consistent illumination? Can it orbit as show as the rotation of the Earth and still maintain an orbit?
- bspammer 8y agoSee below
- georgecmu 8y agoGPS satellites are not in geosynchronous orbits. In fact, they would be close to useless if they were (all in one line orbiting over equator). https://space.stackexchange.com/questions/10837/why-are-the-gps-constellation-satellites-in-such-a-high-orbit https://space.stackexchange.com/questions/10837/why-are-the-...
- M_Bakhtiari 8y agoWhy would a geosynchronous satellite have to be in one line over the equator? Why not a higher inclination?
- rtkwe 8y agoIf the orbit is inclined the satellite will no longer stay above a single point it will move north and south throughout the day. All orbits pass over the equator twice during each period.
- M_Bakhtiari 8y agoThat's true, but it doesn't answer my question. Geosynchronous satellites can and do modify their inclinations and eccentricities to increase their coverage beyond one spot over the equator.
- QuotedForTruth 8y agoThey can have inclination other than 0. If inclination is 0 thats a "geostationary" orbit, a special case of geosynchronous. In fact these are not stable and degrade into mere geosynchronous orbits without station keeping. In general though geostationary orbits are more useful since they are fixed competely in the sky. Ground antenna dont have to track the satellite as it moves north/south in the sky.
- M_Bakhtiari 8y agoIn this case a pair of satellites in a Tundra orbit [1] inclined to spend most of its time over this particular city seems like the right orbit for the job (as silly as it is to light a city by satellite), which is why I was curious to get an explanation behind the sweeping statement that geosynchronous (not geostationary in particular) orbits had to be in a line along the equator. 1. https://en.wikipedia.org/wiki/Tundra_orbit https://en.wikipedia.org/wiki/Tundra_orbit
- QuotedForTruth 8y agoOh thats interesting. I hadn't heard of Tundra orbits. Wiki says 2 can provide continous coverage over an area. Wouldnt 1 be enough to act as a moon then? You only need it for half of the day and their period is 1 day right?
- M_Bakhtiari 8y agoThe problem with using just one is that geosynchronous orbital periods are one sidereal [1] day, not one solar day. So depending on the time of the year, the satellite would be in position in the wrong half of the day. On the other hand a satellite in Tundra orbit spends not half but more than half of its time in the designated area. Maybe a good compromise orbit could be found. 1. https://en.wikipedia.org/wiki/Sidereal_time https://en.wikipedia.org/wiki/Sidereal_time
- georgecmu 8y agoMost people use geosynchronous as a synonym for geostationary. https://en.wikipedia.org/wiki/Geosynchronous_orbit https://en.wikipedia.org/wiki/Geosynchronous_orbit Popularly or loosely, the term geosynchronous may be used to mean geostationary.[2] Specifically, geosynchronous Earth orbit (GEO) may be a synonym for geosynchronous equatorial orbit,[3] or geostationary Earth orbit.[4] Communications satellites are often given geostationary or close to geostationary orbits so that the satellite antennas that communicate with them do not have to move, but can be pointed permanently at the fixed location in the sky where the satellite appears. Certainly the comment I responded to (which has been edited to be non-sensical since) meant geostationary. In any case, GPS are not geosynchronous in the general sense of the term, but rather semi-synchronous (period of half a sidereal day). These orbits were historically chosen for convenience, but the syncronicity is not at all a requirement for global positioning satellites: e.g Galileo, GLONASS.
- samnardoni 8y agoAll GPS satellites are in a geosynchronous orbit.
- georgecmu 8y agoCitation needed? You may find this helpful: http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps.html http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps.... The nominal GPS configuration consists of a network of 24 satellites in high orbits around the Earth, but up to 30 or so satellites may be on station at any given time. Each satellite in the GPS constellation orbits at an altitude of about 20,000 km from the ground, and has an orbital speed of about 14,000 km/hour (the orbital period is roughly 12 hours - contrary to popular belief, GPS satellites are not in geosynchronous or geostationary orbits). The satellite orbits are distributed so that at least 4 satellites are always visible from any point on the Earth at any given instant (with up to 12 visible at one time).
- codewritinfool 8y agoNo, they are not.
- dsr_ 8y agoLet's do some quick math. Geosynch is 36,000 Km out. The Moon's albedo is 0.12. Let's assume we can get a mirror of albedo 0.96 -- this is a little better than polished silver. It's also conveniently 8 times as reflective as the moon. The Moon's angular diameter is about 0.5 degrees. To match that angular diameter at 36,000 Km, we need a 5.4 Km wide object. That doesn't sound like it's within the current state of the art. Maybe in another 10 years? 23 million square meters at 2 g/m^2 (aluminized mylar) is 46,000 Kg -- about two Falcon Heavy trips. That doesn't count any framework or booster or other infrastructure.
- rocqua 8y agoNot sure, but it seems to me that for a mirror, more than albedo matter, because mirrors have a specular reflection rather than a diffuse reflection. Now, reflection won't be perfectly coliminated, but if you assume it is, then the mirror essentially becomes a piece of sun. Since the sun is about 400 000 times as bright as the moon, you'd need to scale the diameter of the sun by sqrt(400 000 / 8) =~ 223. Applying that scaling to your 5.4Km yields about 24m of diameter, which seems waaaay to small to me, so I am probably wrong. That feels really
- dsr_ 8y agoNot the sun, the sun's irradiance at the distance of the Earth-Moon system.
- djrogers 8y ago> The Moon's angular diameter is about 0.5 degrees. To match that angular diameter at 36,000 Km, we need a 5.4 Km wide object. The article doesn’t say it’s going to be as big as the moon - it only says it’ll be 8x brighter. You’re also forgetting about the relative shapes of the objects: the moon, being round, reflects what light it does in all directions, while a designed mirror can be made to reflect it all in one specific direction. All that said, I don’t think this will work, but not be of your math.
- dsr_ 8y ago
- sp332 8y agoYeah, but there's only one such "ring" around the equator and it's pretty crowded. https://en.wikipedia.org/wiki/Geostationary#Orbit_allocation https://en.wikipedia.org/wiki/Geostationary#Orbit_allocation
- ligand 8y agoGeostationary orbit also requires the sattelite to take on an equatorial orbit. I doubt a geostationary orbit will be used in this case.
- RugnirViking 8y agoIs there any kind of funky thing they could do to have it rotate slow enough to take an entire night to cross the sky? something like a lagrange point perhaps (stationary relative to sun?)
- rocqua 8y agoThe closer they are, the easier it is to illuminate. I believe that excludes most Lagrange points.
- sp332 8y agoThe Lagrange points are weird because they orbit two things at the same time. There should be more normal orbits somewhat lower than geostationary that cross slowly, but they would also be out of sight for a long time.
- RugnirViking 8y agoThen I suppose the idea would be to find an orbit where that long time is equal to one day - obviously, the length of one day on a given section of the planet keeps changing, so perhaps you only turn on the light after a certain amount of time or find some orbit that mimics whatever the sun is doing if thats at all possible
- rtkwe 8y agoGeostationary orbits are definitely a thing. Haven't done the math but two should take care of it, one east and one west of the area you're trying to illuminate and one should come out of the umbra before the other one enters. edit: One might also take care of it if it could be far enough east. Would need to do the trig though.
- WalterSear 8y agoAt the equator.
- rtkwe 8y agoYes it would be orbiting above the equator but at GEO it can easily see all of China which is all it need to do to reflect light from the sun to Chengdu.
- deleted 8y ago[deleted]
- SketchySeaBeast 8y agoI guess it would have to be placed in geosynchronous orbit. I didn't see any size comments, but that puppy would have to be quiet large I'd think. To light 14,300-square-meters from 35,786 km seems like you'd need a BIG reflector.
- jkaptur 8y agoAnd, of course, Chengdu is actually 14,300 square KILOMETERS, not meters. (Its 11 million residents aren't quite that cramped).
- SketchySeaBeast 8y agoThat would be 119 m on each side... yeah, that's a little cramped. I copied that totally not critically from the article. I'm beginning to believe that this article may be missing some important information.
- mdorazio 8y agoYes, a satellite can appear to cross the sky as slowly as you want it to if the altitude is correct. This is how geostationary satellites (like the GPS and many communications ones) work - they are at the right altitude so that they are orbiting the earth at the same speed it rotates.
- Vendan 8y agoGPS is not geostationary, it's in a MEO about 12k miles up (GEO is about 22k miles) and they orbit about twice a day.
- samnardoni 8y agoGeosynchronous is the word you’re looking for.
- gwbas1c 8y agoMost communication satellites are in geosynchronous orbit.