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Space Forum / Space Flight / October 2003A small, polar-orbiting moon
I'd like input from all you orbital mechanics out there as to this concept's feasibility: What if sometime in prehistory, the Earth had captured a Near Earth Orbit asteroid, say 10 km in diameter, into a nice circular 20310.8 kilometer polar orbit? So every 8 hours the Moon2, let's call it 'Cynthia' [derives from 'woman from Kynthos', a reference to Artemis, sister of Apollo and Greek goddess of the Moon, who was reputed to have been born on the mountain of Kynthos] rises from either the Northerly horizon or the Southerly and, in about 3 hours or less, sinks below the opposite horizon. The ancients would quickly figure out the regular patterns of its orbit and how to use it to determine longitude (using only the local time of day; no accurate clocks synchronized to Greenwich time needed). Accurate maps would appear early. The apparent size of Cynthia would be only about 1/12 that of Moon1 (Luna) and would vary by about 5%, depending on the time and place of the observer. The size variation would let the ancients determine the distance to Cynthia and start giving a sense of the scale of the solar system. The ancients would probably deduce that Cynthia was brighter (relative to size) than Luna because it's closer to Earth. So they'd estimate the distance to Luna and, when they compared the orbital periods of the two moons, would derive the gravitational inverse-square law centuries earlier than OTL. Eclipses (solar transits, really) would happen often, encouraging study of the Sun's surface (maybe using camera obscura, really big pinhole camera minus film). Cynthia often being 25 times closer to us than is Luna, naked-eye observation would show craters and such on Cynthia, an early intro to Galilean ideas of imperfect heavenly bodies or, better yet, the idea of other planets and moons being actual places, that is, destinations. There is some question as to how bright Cynthia would be. I'd appreciate any critiques of the following analysis: … 1) For the non-astronomers out there, the brightness of celestial bodies/stars/whatever is given as 'visual magnitude'. For some reason, early astronomers defined that an object of magnitude 1 was 100 times brighter than an object of magnitude 6. So an object of mag 1 would be 2.512 times brighter than an object of mag 2 because 2.512^(6-1)=100 and 2.512^(2-1)=2.512. Setting magnitudes was an attempt to compare the brightness of stars but later really bright objects were assigned magnitudes as well. For example, Luna has a magnitude of -12.5 (brighter objects have larger negative numbers). 2) Let's wave our hand and make Luna (diameter 3476 km) disappear and be replaced by Cynthia (diameter 10km). The area, and so the amount of light reflected, has shrunk to (10/3476)^2 or 0.000008276 of its former value. Its magnitude has changed by 12.7, meaning it's dimmer by 2.512^12.7 times. Its magnitude is now 0.2 (-12.5+12.7), like a really bright star but still not visible in full daylight. 3) Now let's move Cynthia closer to Earth. At closest, Luna is about 378028 km from an observer on Earth, while Cynthia would be about 13938.8 km away, 27.12 times closer. So it's brighter by a factor of 735.5 (27.12^2) due to the inverse square law. This produces a change in magnitude of -7.2 (2.512^7.2~735.5) so Cynthia, at best, would have a magnitude of -7.0 (-7.2+0.2). 4) By comparison, Venus can have a magnitude of -4.9 while Luna, as I mentioned before, has a magnitude of -12.5. So Venus < Cynthia < Luna. 5) From sci.astronomy.amateur, "The Great Comet of 1744 reached -7.0 magnitude and was visible 12 degrees from the Sun in broad daylight." So I think Cynthia would often be visible during the day and available for use in navigation. My greatest concern is whether/how long Cynthia might maintain a stable, circular polar orbit. Would the influence of Luna disrupt her orbit? I assume the orbit would precess but how fast? Would the precession be at a constant rate, one that ancients could include in their tables/calculations? Is it even possible for a Near Earth Asteroid to take up a circular orbit around Earth? Thanks in advance for any input. > I'd like input from all you orbital mechanics out there as to this > concept's feasibility: > > What if sometime in prehistory, the Earth had captured a Near Earth > Orbit asteroid, say 10 km in diameter, into a nice circular 20310.8 > kilometer polar orbit? I'm guessing the probabilities are slim to none, but then, Earth's past record with moons involves some pretty slim probabilities. :) The reasons for the low probability of capture (as I know'em): 1) Virtually everything in the Inner System (starting at Pluto and working in) has settled down into a single plane, give or take a few degrees. It is very improbable that anything in Earth's neighborhood would be approaching from a steep angle to end up in a polar orbit. 2) Earth already has a big, jealous companion. I'm not sure Cynthia could settle down into a circular orbit (it wouldn't be circular to begin with) before Luna destabilized Cynthia's orbit to intersect Earth, Luna, or ejected it from the area all together. I'd put my money on "intersect Earth." Interesting concept for accelerating early astronomical theories, though. Mike Miller, Materials Engineer >1) Virtually everything in the Inner System (starting at Pluto and >working in) has settled down into a single plane, give or take a few >degrees. It is very improbable that anything in Earth's neighborhood >would be approaching from a steep angle to end up in a polar orbit. It doesn't have to approach from a steep angle. You can't ignore Earth's motion around the Sun when thinking about such things. The incoming object only has to pass Earth a few tens of thousands of kilometers to (say) the north, and then lose some velocity while there. To pass, say, 50,000km north of Earth, the inclination of its orbit needs to be only about 0.02deg -- remember, all this is happening 150Mkm from the Sun, so the angle between the orbits needed to get 50,000km difference is very small. It's the "lose some velocity" part that's hard. The Earth *can* capture objects from heliocentric orbit, as witness the temporary capture of object J002E3 (which is almost certainly Apollo 12's S-IVB!) last year. But as witness that case, the resulting orbits tend to be very large -- well beyond the Moon's -- and rather precarious (J002E3 is gone into solar orbit again). >2) Earth already has a big, jealous companion. I'm not sure Cynthia >could settle down into a circular orbit (it wouldn't be circular to >begin with) before Luna destabilized Cynthia's orbit to intersect >Earth, Luna, or ejected it from the area all together. I'd put my >money on "intersect Earth." Most likely is to eject it, in fact: both Earth and Moon are rather small targets, in the celestial scheme of things, and a near-miss that changes your path into an escape trajectory is much more likely than an actual collision. In general, three-body systems which are not "hierarchical" -- one close pair plus a distant third -- over time have a strong tendency to lose one body by ejection.  SignatureMOST launched 30 June; first light, 29 July; 5arcsec | Henry Spencer pointing, 10 Sept; first science, early Oct; all well. | henry@spsystems.net > >1) Virtually everything in the Inner System (starting at Pluto and > >working in) has settled down into a single plane, give or take a few [quoted text clipped - 17 lines] > be very large -- well beyond the Moon's -- and rather precarious (J002E3 > is gone into solar orbit again). But an object _could_ (very small chance, I admit) be in heliocentric orbit and yet pass over the Earth at just the right speed to enter a circular polar orbit at 20310.8 km radius, could it not? Without having to shed any velocity at all? (I feel like a cross-examining attorney;"You admit that my client _could_ have been carrying that plutonium for perfectly innocent reasons?") > >2) Earth already has a big, jealous companion. I'm not sure Cynthia > >could settle down into a circular orbit (it wouldn't be circular to [quoted text clipped - 10 lines] > pair plus a distant third -- over time have a strong tendency to lose one > body by ejection. And yet Jupiter has a number of moons in pretty stable orbits, over millenia. > But an object _could_ (very small chance, I admit) be in heliocentric > orbit and yet pass over the Earth at just the right speed to enter a > circular polar orbit at 20310.8 km radius, could it not? Without > having to shed any velocity at all? I don't think so. If the object is passing over the Earth at the right speed to "enter" an orbit, it's already *in* orbit. To put it another way: Imagine filming this object and then running the film backwards: you'd see an object in orbit around the Earth suddenly leave orbit without gaining any velocity, which is impossible. Since Newtonian mechanics works the same whether time runs forwards or backwards, it would also be impossible if you run the film forwards. Therefore entering orbit without shedding velocity is impossible. SignatureDave Empey > > But an object _could_ (very small chance, I admit) be in heliocentric > > orbit and yet pass over the Earth at just the right speed to enter a [quoted text clipped - 11 lines] > would also be impossible if you run the film forwards. Therefore > entering orbit without shedding velocity is impossible. How about entering orbit and _gaining_ velocity? Aren't some moons of Jupiter suspected to be captured asteroids? How can a planet capture an asteroid if the asteroid makes no velocity changes? I'm proposing that Cynthia is an asteroid/comet that is orbiting the Sun between the orbits of Venus and Earth when it happens to pass near Earth, is accelerated and captured in a circular orbit. >>> But an object _could_ (very small chance, I admit) be in heliocentric >>> orbit and yet pass over the Earth at just the right speed to enter a [quoted text clipped - 13 lines] > > How about entering orbit and _gaining_ velocity? A.) That would make it even more likely to escape; B.) What the heck do you think would it gain velocity _from_ ?!? > Aren't some moons of Jupiter suspected to be captured asteroids? Yes, but they aren't "permanent" moons. They are in highly eccentric orbits near the edge of Jupoter's "sphere of influence" (the rather non-spherical region in which Jupiter's gravity dominates the Sun's, which is bounded by a surface known as the "separatorix" that passes through L1 and L2), and eventually, they will re-escape. > How can a planet capture an asteroid if the asteroid makes no velocity > changes? Jupiter is on a slightly eccentric orbit, which means that the size of its "sphere of influence" varies depending on how far it instantaneously is from the sun. The "captured" moons have orbital periods on the order of a terrestrial _year_; they were "captured" because they passed just inside L1 or L2 on a trajectory almost alone the separatrix; by the time they got back to the separatrix, it had expaned slightly, "trapping" them on highly eccentric, precessing orbits. Eventually, they will again happen to have an apojove close to L1 or L2 at a time when the separatrix is small, and they will escape. > I'm proposing that Cynthia is an asteroid/comet that is orbiting the Sun > between the orbits of Venus and Earth when it happens to pass near Earth, > is accelerated and captured in a circular orbit. orbital mechanics doesn't work that way. It will have to interact with something, such as the Earth's atmosphere of via tidal friction to become "trapped;" this will leaving it in a highly eccentric orbit, that will require further interactions to circularize it so that it does not simply continue to lose energy via atmospheric or tidal friction and come crashing down to leave a large, hot, ecosphere-destroying hole in the ground. -- Gordon D. Pusch perl -e '$_ = "gdpusch\@NO.xnet.SPAM.com\n"; s/NO\.//; s/SPAM\.//; print;' >> It's the "lose some velocity" part that's hard... > >But an object _could_ (very small chance, I admit) be in heliocentric >orbit and yet pass over the Earth at just the right speed to enter a >circular polar orbit at 20310.8 km radius, could it not? Unfortunately, no, because Earth's gravity will accelerate it as it approaches. If it arrives from infinity, its speed must be at least escape velocity (for that distance), which is about 1.4x circular-orbit velocity. The only exception to this, which is what got J002E3 captured temporarily, is if the whole thing is happening at the outer fringes of Earth's sphere of influence, where the Sun's gravity is quite significant and three-body complications invalidate simple concepts like "escape velocity". But that's out around 900000km radius. (Oh, there's one other exception, if it chances to make a lunar flyby that robs it of some energy. But that will necessarily leave it in an orbit that goes out, at least, nearly to the Moon's orbit.) >> In general, three-body systems which are not "hierarchical" -- one close >> pair plus a distant third -- over time have a strong tendency to lose one >> body by ejection. > >And yet Jupiter has a number of moons in pretty stable orbits, over >millenia. Those cases are not really three-body systems, because Jupiter dominates the situation so overwhelmingly. Interactions between the moons are minor by comparison. (Sometimes they are non-trivial -- e.g. the resonance with Europa and Ganymede that maintains the slight eccentricity of Io's orbit and hence its internal tidal heating -- but not to the extent of actually altering another moon's orbit substantially.) SignatureMOST launched 30 June; first light, 29 July; 5arcsec | Henry Spencer pointing, 10 Sept; first science, early Oct; all well. | henry@spsystems.net > >> It's the "lose some velocity" part that's hard... > > [quoted text clipped - 6 lines] > escape velocity (for that distance), which is about 1.4x circular-orbit > velocity. So it doesn't arrive from infinity but rather from an orbit closer to the Sun so Earth accelerates it to just the right velocity for a circular polar orbit. >snip< > >> In general, three-body systems which are not "hierarchical" -- one close [quoted text clipped - 10 lines] > and hence its internal tidal heating -- but not to the extent of actually > altering another moon's orbit substantially.) But if Cynthia is 25 times closer to the Earth than to Luna, wouldn't Earth's influenece overwhelm Luna's? >>>> It's the "lose some velocity" part that's hard... >>> [quoted text clipped - 10 lines] > the Sun so Earth accelerates it to just the right velocity for a > circular polar orbit. Orbital mechanics doesn't work that way. If it did, then as others have noted, it would also be equally possible for polar satellites to spontaneously jump into the corresponding outbound trajectory, which one never observes to happen (unless the object fires a kick-motor). Your only hope is for your "Cynthia" (or part of it) to lose energy somehow, e.g., by making a pass through the atmosphere, or via tidal friction. The former option requires an additional interaction with some other body to raise "Cynthia's" perigee out of Earth's atmosphere to prevent subsequent passages from de-orbiting it entirely. The latter option is not likely to work in a single pass unless "Cynthia" is tidally disrupted into fragments, some of which are moving at less than escape velocity relative to the Earth. This is _barely_ possible if "Cynthia" is a highly elongated object whose center of mass is moving moving almost _exactly_ at escape velocity, so that the "inboard" fragments are moving at _just_ under escape velocity, while the "outboard" fragments are moving at a bit over escape velocity. This rather unlikely and rube-goldbergian situation will leave _parts_ of "Cynthia" in a belt of highly excentric orbits --- but the fact that "Cynthia" (and therefore its remaining fragments) must necessarily have had a perigee inside Roche's Limit in order for tidal disruption to have occurred makes it extremely unlikely that the fragments that remain in orbit about the Earth will re-accrete into a single body. Most likely, you will be left with a large number of fragments in highly eccentric orbits that will gradually pulverize each other to dust. -- Gordon D. Pusch perl -e '$_ = "gdpusch\@NO.xnet.SPAM.com\n"; s/NO\.//; s/SPAM\.//; print;' >And yet Jupiter has a number of moons in pretty stable orbits, over >millenia. So does the Sun, but that's because the parent body is *much* larger than the satellites. So the gravitational effects of the small bodies on each other are much smaller than the gravitational effect of the parent. OTOH, Earth-Moon almost has to be treated as a double planet. Any third body in their near vicinity will be influenced more or less strongly by both of the other bodies. That's a much more complicated situation. Gary > >And yet Jupiter has a number of moons in pretty stable orbits, over > >millenia. > > So does the Sun, but that's because the parent body is *much* larger > than the satellites. So the gravitational effects of the small bodies on > each other are much smaller than the gravitational effect of the parent. But Earth masses about 81 times Luna and Cynthia will be about 25 times closer to Earth than Luna so the Earth's influence over Cynthia should be about 50,000 [81*25*25] times that of Luna, no? Are the orbits of polar-orbiting artificial satellites disrupted by Luna? Maybe Luna would simply cause Cynthia's orbit to precess? >OTOH, Earth-Moon almost has to be treated as a double planet. Any >third body in their near vicinity will be influenced more or less strongly >by both of the other bodies. That's a much more complicated situation. Depends on the time scale. On a short time scale, you can pretty much ignore the Moon unless your path takes you quite close to it. It's only 1/80th the mass of Earth, after all. Longer term, yes, it definitely matters, e.g. it affects geostationary comsats' orbits despite it being ten times farther out than they are. SignatureMOST launched 30 June; first light, 29 July; 5arcsec | Henry Spencer pointing, 10 Sept; first science, early Oct; all well. | henry@spsystems.net > But an object _could_ (very small chance, I admit) be in heliocentric > orbit and yet pass over the Earth at just the right speed to enter a > circular polar orbit at 20310.8 km radius, could it not? Without > having to shed any velocity at all? (I feel like a cross-examining > attorney;"You admit that my client _could_ have been carrying that > plutonium for perfectly innocent reasons?") No. Orbits run backwards more or less as well as they run forwards. So if something is now in a circular orbit at 20310.8 km radius, if time is reversed it's not going to go flying back out into interplanetary space - it'll still be in a circular orbit. With close flybys in 3-body systems you can either eject or capture something, but that's not the situation you're describing. -jake > > But an object _could_ (very small chance, I admit) be in heliocentric > > orbit and yet pass over the Earth at just the right speed to enter a [quoted text clipped - 9 lines] > reversed it's not going to go flying back out into interplanetary > space - it'll still be in a circular orbit. Doesn't entropy enter into it? Cynthia would steal some energy from Earth and speed up as it falls into circular orbit. Doesn't entropy dictate that we can't just run the tape backwards? If we run my car backwards, the flies aren't going to jump off the windshield and take flight! (Admittedly lame analogy). >>> But an object _could_ (very small chance, I admit) be in heliocentric >>> orbit and yet pass over the Earth at just the right speed to enter a [quoted text clipped - 12 lines] > Doesn't entropy enter into it? Cynthia would steal some energy from > Earth and speed up as it falls into circular orbit. Neither Entropy nor Conservation of Energy work that way. "Cynthia" doesn't "steal" energy. The Earth / "Cynthia" system _HAS_ a certain amount of total energy, which is conserved; because gravity is a pairwise interaction, that total energy can't be attributed solely to either Earth or "Cynthia," but only to the system as a whole. > Doesn't entropy dictate that we can't just run the tape backwards? Only if friction or other dissipative forces are important. Why do you expect your "Cynthia" to run into "friction" ??? > If we run my car backwards, the flies aren't going to jump off the > windshield and take flight! (Admittedly lame analogy). Indeed it is VERY lame. A much better analogy is: Someone throws a superduperball (tm) at your windshield, and it bounces off. Because a superduperball (tm) is extremely elastic, energy is almost perfectly conserved during this collision. Because energy is almost perfectly conserved in this collision, it is virtually impossible to determine whether you are watching the film "forwards" or "backwards," solely from observing the collision. Now: Gravitational collisons are even =MORE= "elastic" than any "superduperball" could ever possibly be. Absent some other dissipative force, such as atmospheric or tidal friction, any purely gravitational trajectory is perfectly reversible. -- Gordon D. Pusch perl -e '$_ = "gdpusch\@NO.xnet.SPAM.com\n"; s/NO\.//; s/SPAM\.//; print;' > >>> But an object _could_ (very small chance, I admit) be in heliocentric > >>> orbit and yet pass over the Earth at just the right speed to enter a [quoted text clipped - 18 lines] > that total energy can't be attributed solely to either Earth or "Cynthia," > but only to the system as a whole. So the sum of Earth+Cynthia energy is constant (actually Earth+Cynthia+Luna) but that doesn't mean Cynthia's energy can't increase at the expense of Earth's. Ex; when a probe swings around Jupiter to take advantage of the 'slingshot effect', is it not speeding up while Jupiter is slowing down infinitesimally? > > Doesn't entropy dictate that we can't just run the tape backwards? > [quoted text clipped - 16 lines] > could ever possibly be. Absent some other dissipative force, such as atmospheric > or tidal friction, any purely gravitational trajectory is perfectly reversible. OK, I think it's finally gotten through to me: Cynthia needs a mechanism to change her velocity as she approaches Earth orbit. I'd rather not use aerobraking because it doesn't eventually raise her perigee to a circular orbit. So, Cynthia is a regolith-covered comet that somehow found itself in a Aten-esque orbit (within the Earth's orbit around the Sun), an orbit that just reaches the Earth. Cynthia happens to vent at just the right times and in just the right directions to circularize it's orbit around Earth. > With close flybys in 3-body systems you can either eject or capture > something, but that's not the situation you're describing. If Cynthia originally had a sister, a double asteroid like Hermes was just found to be, then you have a three-body system that might allow capture. It would take the same hand-of-god that placed Luna at just the right size and distance to give us nice solar eclipses, but it's possible. Especially with an aerobrake to bleed off some energy, then the second component circularizes Cynthia's orbit before being flung into the utterdark. (A single asteroid aerobraking gives an orbit that passes inside the atmosphere on subsequent passes, which quickly leads to lithocapture and a nice iridium layer for the next intelligent species to find.) Of course, the protagonist realizes this after single-handedly recapitulating the works of Galileo, Newton, Halley etc. from our time line to develop orbital mechanics, Percival Lowell to find the cast off sister, Shoemaker to discover that it will hit Earth in 3 years, Goddard, Tsiolkovskii and Korolev to build a rocket, Oppenheimer and the gang for the payload, and Bruce Willis to get the box office. Bogen: > Luna will give off more light in total because it's larger but Cynthia > is much closer so each solid angle measure (steradian?) should be > brighter. I think Cynthia will lokk like a brighter, fast moving > Venus. No, brightness per steradian depends on illumination (how far from the Sun it is--the same as the Moon is, plus a bit of Earthshine) and how reflective it is, but not on how far away it is. (The inverse square law, an approximation in this case, comes entirely from the solid angle shrinking with distance.) That's why a tree nearby doesn't burn your eyes out while a tree-covered distant mountain is other than black. SignatureDavid M. Palmer dmpalmer@email.com (formerly @clark.net, @ematic.com) Bill> But an object _could_ (very small chance, I admit) be in heliocentric Bill> orbit and yet pass over the Earth at just the right speed to enter a Bill> circular polar orbit at 20310.8 km radius, could it not? Without Bill> having to shed any velocity at all? What you need is some plausible delta-v for this object that occurs well above the atmosphere, closer than the moon. Early in the development of the solar system. How about two objects in heliocentric orbit collide while very close to the earth? Perhaps the two objects are a single object that broke up during a first close encounter with earth, then passed on either side of earth during a later encounter. Doesn't have to happen in plane, but the angles do have to be pretty wide. It's not *likely*, of course. The collision dumps a bunch of kinetic energy so that the resulting mass does not have escape velocity. It explodes, some bits get kicked out of the earth-moon system, some bits enter the atmosphere, and some bits stay in orbit. One particularly big bit is more obvious than the rest. Cynthia. Over time it might even accrete the rest if it were made of, say, particularly sticky green cheese.... :). > The ancients would probably deduce that Cynthia was brighter > (relative to size) than Luna because it's closer to Earth. I thought the brighness is proportional to the solid angle. Signaturehttp://hertzlinger.blogspot.com > > The ancients would probably deduce that Cynthia was brighter > > (relative to size) than Luna because it's closer to Earth. > > I thought the brighness is proportional to the solid angle. Luna will give off more light in total because it's larger but Cynthia is much closer so each solid angle measure (steradian?) should be brighter. I think Cynthia will lokk like a brighter, fast moving Venus. >>> The ancients would probably deduce that Cynthia was brighter >>> (relative to size) than Luna because it's closer to Earth. [quoted text clipped - 5 lines] > brighter. I think Cynthia will lokk like a brighter, fast moving > Venus. You have that exactly backwards. Both Luna an "Cynthia" are receiving the same intensity of sunlight, so unless "Cynthia's" surface has a much higher albedo than Luna's, they will reflect light with roughly equal intensity. Hence, their surfaces will look about equally bright per unit solid angle. How much _total_ light the reflect will depend on how large they are, but not the amount of light per unit solid angle. "Cynthia" will only appear "brighter" if it is highly reflective, or if it is so close that it appears to subtend a larger solid angle than the moon. (Note that the latter would be a Very Bad Thing, as that would imply it would be raising Huge Tides.) -- Gordon D. Pusch perl -e '$_ = "gdpusch\@NO.xnet.SPAM.com\n"; s/NO\.//; s/SPAM\.//; print;' > >>> The ancients would probably deduce that Cynthia was brighter > >>> (relative to size) than Luna because it's closer to Earth. [quoted text clipped - 15 lines] > appears to subtend a larger solid angle than the moon. (Note that the latter > would be a Very Bad Thing, as that would imply it would be raising Huge Tides.) Yes, I was partially wrong: Luna and Cynthia are equally bright per unit of solid angle. Since Luna would appear about 12.6 times wider than Cynthia, or 158.5 times the area, then Cynthia would only have a magnitude (at best) of -7 as compared to Luna's -12.5. But Venus has a magnitude of only -4.9 (at best) so Cynthia, as I said, would like like a brighter, fast moving Venus. It should often be visible in daylight. |
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