I can't answer the first, but I can say with some certainty that unless we had our own satellite around the moon, it would be extremely difficult for us to notice it. Space is, after all, massive, and 2 km ships, even if there are 20 of them, would be difficult to notice unless they either purposely made themselves seen, or were using extremely noticeable explosives for their mining purposes. Even then, it would have to be a truly MASSIVE explosion.
On the contrary, the fact that there is no stealth in space already applies to todays technology. Those ships and mines are going to show up on any random scan of the area in the IR spectrum, because they'll be a couple of hundred degrees hotter than everything else. Finding a single LED in a football stadium at midnight might be hard, but is not if it's switched on. Although they might not be noticed for a while, simply because no-one looked in the right direction. As far as i'm aware, there is no program systematically scanning the sky for alien activity, so they might stay unnoticed until someone looks in the right direction.
As for what to mine, well there's no gravity switch that's off for asteroids and on for moons. Especially the many small gas giant moons aren't that much larger than asteroids. The best idea would probably be to simply mine those objects that have the desired materials present in the most easily accessible form. Whether that is a moon, an asteroid or on Oort Cloud object.
Any reason they wouldn't contact Earth?
We might be primitives compared to them but if they are mining for something that we MIGHT have, they could trade or take it from us.
Would save them the trouble of mining it themselves.
Nooo, this is way too simplistic. It's only easy to spot the LED in your example because its intensity, even when spread out over an area that corresponds to our visual resolution, is still significantly higher than that of the background. It's not obvious at all that this criterion is met in this case, IMO.
Let's consider the visual detection case, for the sake of simplicity. Hubble has a resolution of something like a tenth of an arcsecond. Jupiter is about 5 AU away, which gives us a corresponding area on the order of
A ~ ((5 AU) * (0.1 arcsec))^2 ~ ((10^12 m) * (0.1*(2 pi)/(360*60*60)))^2 ~ 100,000 km^2
So, we don't just need our 1 km^2 (or so) spacecraft to be significantly brighter than the background, we need it to be significantly brighter than 100,000 times brighter than the background. Totally different story.
Which is not at all to say that it's not doable. They may well emit radiation in some bands at intensities that high. Then again, they may not. The extra margin due to being small relative to the resolution of the detector does indeed make stealth in space entirely possible, at this range.
Soon the Cold One took flight, yielded Goddess and field to the victor: The Lord of the Light.
Eye'm the cutest!
^ Not to mention a lot of instrumentation isn't sensitive enough (or worse will think of it as background noise or an instrumentation anomaly) to detect such things.
A 100C object in Jupiter orbit is effectively invisible if it doesn't want to be found. How so? 99.99% certainty that not only is that sector of sky unwatched by instrumentation that could plausibly detect such an object, few people are going to give it much thought.
It took Voyager 1 basically getting lucky to find a volcanic eruption on Io at a distance of 20,000 kilometers. We had technology that could detect such things for 30+ years at the time. (If you include optical telescopes we had such technology for 400 years.)
If we could not detect a volcanic eruption of several square kilometers in size until we were 20,000 kilometers away, what makes you think a 100C object will be lit up like a Christmas tree visible to every detector on or near Earth?
Worse, Io isn't always cold owing to its volcanism. Yet we miss most of the volcanic eruptions there even today.
"Allah may guide their bullets, but Jesus helps those who aim down the sights."
Once we get past Mars though, we get into the kind of territory where solar output is so low that PV cell arrays start to look like solar sails, so in all probability we're going to see nuclear power, and the issue with nuclear power is that it's either low power, or high thermal output.
Also, Hubble isn't that good for IR (its scale is mostly visual with a bit of IR at one end and a bit of UV at the other). Now when the James Webb goes up, then we'll see how good dedicated IR can be.
![[up]](https://static.tvtropes.org/pmwiki/pub/smiles/arrow_up.png "[up]"){kind=icon}
Longer wavelength -> worse resolution. I was being generous.
Bigger primary mirror (6.5m composite mirror, as opposed to 2.4m monolithic mirror), plus you're probably not looking at anything bigger than about 5 µm, plus that's using only modern equipment, which would be about like criticising Bleriot because he could barely make it across the channel. Hells, look at how fast sensors have developed in comparison to spacecraft, are you really telling me that a civilisation capable of launching an interstellar mission is going to be incapable of going beyond the James Webb?
The OP specifies Next Sunday A.D., though. I thought we were talking about us detecting the aliens, so what does their interstellar capability have to do with anything? You're confusing me! *scowls*
edited 26th Aug '12 3:42:01 AM by kassyopeia
Soon the Cold One took flight, yielded Goddess and field to the victor: The Lord of the Light.
@Kassyopeia You're assuming sensors would take a picture of the whole sky at once and search for objects, which is not only impractical, but also physically impossible for any sensor on or close to earth.
Anyway, we have the resolution to find objects that size, for example: https://en.wikipedia.org/wiki/25143_Itokawa
(which is not as far as Jupiter, put also not even one km in size)
And the ships will be highly visible. Except for a tiny fraction, all energy produced internally will be radiated into space as IR (unless they have extremely inefficient reactors, in which case some large-ish fraction might be radiated in shorter wavelengths). Something at a temperature allowing typical organic life (not to mention all that waste heat from the machinery) is a LOT hotter than the background. Which is the nice thing for sensors in space, there is pretty much nothing in the background, so things don't get written off as background noise. Basically it's a question of how soon someone takes a look in that direction, not if they could see the ships.
If they're in front of Jupiter, they'd be much less visible (still much hotter though, and i suspect there's a lot more telescopes pointed at Jupiter than at the asteroid belt). But that the good thing about a mining operation compared to a volcanic eruption. The eruption happens ... and is gone. The mining operation just sits there. If you take a look at the place a couple of days later the eruption is over and you'll see nothing. But the mining operation is still sitting pretty.
Addendum: Note that i was not talking about the picture of Itokawa on wiki, that was taken by a space probe close to it, but the asteroid was of course detected from earth before the probe was send to it. ;)
edited 26th Aug '12 3:52:44 AM by McKitten
Okay, but the James Webb is still Next Sunday A.D., and it's got a far higher resolution (from what I've read) than Hubble. Out in the Asteroid Belt and beyond Photovoltaics are next to useless, thus the aliens will be using nuclear power of some sort, and nuclear power generates a lot of heat, thus big radiators, which are going to show up nicely. Of course space is a big place, so while we'd spot them easily if we knew sort of where to look, we'd have to be extraordinarily lucky to spot them at random.
Say what?
Linear size has almost nothing to do with anything. That's the point. There's no problem finding something a micrometer across, if it's sitting under a microscope, or if it's a few parsec away and emitting a few solar luminosities. I'm not sure how to explain this any better than I did previously. Isn't it completely obvious?
Yes, this is true. And this is why I said above that the ratio required for readily detecting something of that size at that distance, i.e. well in excess of 10^5 times the background levels, could easily be met by the craft in question. But this is an assumption which you have to make and support (as you did in the post I'm replying to now), not something which can be taken as a given.
Soon the Cold One took flight, yielded Goddess and field to the victor: The Lord of the Light.
We do have the technology to achieve better resolution at IR wavelengths than Hubble has at visual wavelengths, but it requires interferometric linking of several telescopes. As IR detection is a real chore from within the atmosphere (akin to trying to do visual astronomy during daylight), this would require at least two space telescopes working in concert. Which there are no concrete plans for, as far as I'm aware, though it'd certainly be awesome. 
Eye'm the cutest!
To quote Armageddon, it's a pretty big sky. Radio telescopes and sensor systems of any kind can only look at very small slivers of it at a time. A thousand sensor stations might not even cover 5% of the total sky as visible from Earth.
It's why there's so much concern about NEO*-type asteroids and objects. Despite their closeness, many slip by our planet undetected because our observation capabilities are extremely little and limited despite thousands upon thousands upon thousands of optical telescopes, radio telescopes, radar systems, orbital observatories, and sensor systems of every kind.
Also, this thread is rapidly devolving into (once again) showing Atomic Rocket is (once again) full of shit.
edited 26th Aug '12 6:40:32 AM by MajorTom
"Allah may guide their bullets, but Jesus helps those who aim down the sights."
Indeed. The better your resolution, the more sampling you need to do to cover the whole sky. But how does showing that a, by our current standards, good resolution may still be insufficient to observe something a kilometre across near Jupiter translate into "assuming sensors would take a picture of the whole sky at once"? Either, there's a connection there that I'm not seeing, or I didn't explain properly what I was calculating... or something.
Eye'm the cutest!
^ It's the assumption that anything warmer than the nearest local body will shine like a Christmas tree and be instantly detected.
Decades of volcanic eruptions on Io being missed even today is proof that Stealth in Space is a lot more credible than Atomic Rocket will tolerate.
"Allah may guide their bullets, but Jesus helps those who aim down the sights."
@kassyopeia I have explained several times why starships radiate significant amounts of heat. Heck, even a human body fits your criteria of 100.000 times the background. (Humans radiate at ~5W/m², the cosmic background at 0.000005 W/m². And linear object size translates perfectly into angular size if the distance is the same. If we can find a .5km Asteroid that's at 3K and only visible via reflected sunlight in the asteroid belt, we can certainly find a 2km spaceship parked next to it, that's radiating megawatts of energy. Hell, we routinely map the asteroid belt to discover new asteroids, there is no way a mining operation wouldn't be noticed sooner or later.
@Major Tom Atomic Rocket maybe be wrong about some things, but they are right about the stealth part. The reason NEO asteroids are hard to find is because they are cold and small. Finding the tiny amount of sunlight reflected off them is not easy unless one knows where to look at. The point about no stealth in space is not that space is small, but that absolutely everything you do requires you to radiate tons of highly visible energy. That we may only observe 5% of the sky doesn't really matter much, unless those telescopes are never ever moved. If they'd all be coordinated, they could do a complete sky survey in 20 days, while still having a couple hours worth of pictures of every single section. So what if it takes a little time, anyone setting up a mining operation is going to be there for quite a while. I never said the aliens would be found immediately, but even if it takes a year or two, eventually someone is going to point a telescope in the right direction and see them.
Addendum: It's not about being warmer than the nearest local body, it's about being warmer than the background. Which, at 2.75K is pretty damn close to "as cold a physically possible". You can hide in front of a local body (as i also mentioned) but that doesn't give you a lot of options movement-wise. And it requires you to know precisely where observers are located.
edited 26th Aug '12 7:55:38 AM by McKitten
Eye'm the cutest!
Small most often yes. Cold not always. The Moon is 200F on the sunny side. Any NEO is going to be similar at this far out from the Sun. The sunny side will be hotter than Earth's atmosphere. Yet we don't detect that. Ever. 99.99% of the time we find a NEO it's because a radio or optical telescope was incidentally looking in its direction as it moved.
edited 26th Aug '12 8:12:03 AM by MajorTom
"Allah may guide their bullets, but Jesus helps those who aim down the sights."
Good. So your assumption is that the craft can't be significantly more stealthy than a human body scaled to its size. I can think of several ways in which that assumption could be toppled, but that's not really the point at all. It's just a good idea to, firstly, actually state such an assumption, and, secondly, show that the blanket statement you want to make actually follows from the assumption, if you're going to make a blanket statement, IMO.
Yep, and when angular size is less than angular resolution, it doesn't matter in the least, as I keep repeating.
Even I have to admit that that's solid reasoning, but the data looks dodgy to me.
For one thing, asteroids exposed to sunlight would typically be in thermal equilibrium, which means above 100K at that distance. For another thing, the sunlight reflected from a kilometre-sized body would already be well into the megawatt range, again at that distance.
Mainly, though, Itokawa is a near-Earth asteroid, and it seems likely that it was indeed "near Earth" rather than in the belt when it was detected, doesn't it? I haven't had any luck tracking down a figure, but I did find this orbital schematic
◊, which shows its position in July 2004, i.e. two months short of six years after the detection date in September 1998. The asteroid's orbit is prograde (counter-clockwise, here), like Earth's, and has a period of 1.52 years.
Thus, four complete orbits would take about six years and one month, hence, we need to move the Earth backwards (clockwise) by two months (a sixth of its orbit) and the planet backwards (also clockwise) by three months (also a sixth of its orbit) to reconstruct the orbital configuration at the time of detection. It may not have been quite as close as in the graphic, but it certainly was a lot (an order of magnitude or so) closer than the asteroid belt.
No matter, though - with your assumption of human-equivalent energy output, it sounds at least marginally plausible for the same instrument to detect the craft in question quite a long way farther out. I never said otherwise, to be sure.
Soon the Cold One took flight, yielded Goddess and field to the victor: The Lord of the Light.
For one thing, asteroids exposed to sunlight would typically be in thermal equilibrium, which means above 100K at that distance. For another thing, the sunlight reflected from a kilometre-sized body would already be well into the megawatt range, again at that distance.
Most of the stuff that's being found 2000 and later is really small, 500m and smaller. Also note that even asteroids at the distance of Jupiter are being found quite regularly.
edited 26th Aug '12 12:36:12 PM by McKitten
I think the detection distance was hashed out, in all places, in some OTC thread and I think it was somewhere around 1 AU for a 300 K object the size of a dreadnought. This would be using a near-zero temperature IR high-resolution scanning device to see a blip (but without the resolution to see anything but a heat dot). To actually see the ship you need at least two sensor probes of such awesomeness a significant distance away from each other (maybe like the opposite points of Earth's orbit) to see anything out that far. Otherwise your range is hardly a few hundred thousand km to see what is out there.
As for mining asteroids versus mining natural satellites, I guess this more depends on a number of factors.
Favouring natural satellites is the possibility of gravitational orbits. However, I don't know if you can manage geo-synchronous orbits which would be what you want for mining. You'll have to ask a physicist or an engineer who has studied rocket science to know.
Space based industry around a planet (like Jupiter) where you ship in stuff, would obviously be much better if you only had to ship in from the natural satellites of Jupiter versus the mining distance with asteroids. However, if you can tow and anchor an asteroid around Jupiter, then you'd be set. Plus, mining accidents (while costly) would only nuke-bomb Jupiter by accident. Nobody lives there, so who cares.
Hubble is already far from being the most awesome thing humans sent into sapce.
Fair enough.
Heh, that's a nice little puzzle.
The straightforward criterion would be that that orbital distance at which the period is equal to the moon's rotational period has to be both (a) outside the moon's surface and (b) inside the moon's Hill sphere (the region within which its gravity, rather than the gas giant's, dominates). I'll use lower-case letters to refer to the orbit about the moon and upper-case letters to refer to the orbit about the gas giant:
(1) mass of moon = m ~ (4/3) pi r^3 rho
(2) centripetal acceleration about moon = a_cp ~ v^2 / d ~ 4 pi^2 d / p^2
(3) gravitational acceleration by moon = a_grav ~ G m / d^2
(4) gravitational acceleration by gas giant = A_grav ~ G M / D^2
(2) and (3) must be equal, so d ~ ((G/(4 pi^2)) m p^2)^(1/3) ~ ((G/(3 pi)) r^3 rho p^2)^(1/3), using (1)
(a) means d_min ~ r
(b) means (3) and (4) are equal, so d_max ~ (m/M)^(1/2) D ~ ((4/3) pi r^3 rho D^2 / M)^(1/2), using (1)
(a) d > d_min ===> (G/(3 pi)) rho p^2 > 1 ===> p^2 > 10 / (G rho)
(b) d < d_max ===> G^2 p^4 < (4^3/3) pi^5 r^3 rho D^6 / M^3 ===> p^4 < (6,400/G^2) r^3 rho D^6 / M^3
(ab) d_min < d_max ===> 1 < 64 r^3 rho^3 D^6 / M^3 ===> r rho D^2 > M/4
Let's look at (a) first. From the wikipedia articles, it seems that the density range is about 0.3*10^3 kg/m^3 < rho < 3*10^3 kg/m^3 for gas giant moons, with the smaller moons near the lower end and the bigger moons near the upper end. The lower value gives us a higher (stricter) value for d_min, so we start with that one:
criterion: p^2 > 10 / (G rho)
p^2 > 10 / (7*10^-11 m^3/kg/s^2 * 0.3*10^3 kg/m^3) ~ 5*10^8 s^2 ===> p > 6 hours
From the lists, none of the moons with diameters above 10 km have a shorter period than that, though there are several which come remarkably close (below 8 hours). So, this criterion is always satisfied, but in some cases the stationary orbit is within arm's reach (figuratively speaking) of the moon's surface.
Now, let's skip (b) for the moment and look at (ab), just because it's simpler. Here, the "worst case" would be a small moon (low r and rho) in a close orbit (low D). From the lists, that means Atlas, which orbits just outside Saturn's main ring system:
criterion: r rho D^2 > M/4
(15 km) * (0.5*10^3 kg/m^3) * (140,000 km)^2 ~ 10^23 kg
This is well below the limit of a quarter of the gas giant mass, which is a few 10^26 kg for Jupiter and Saturn. So, the closest moons fail this criterion by a wide margin. Next, I'll try the closest of the major moons for each, Io for Jupiter and Mimas for Saturn:
(1,800 km) * (4*10^3 kg/m^3) * (400,000 km)^2 ~ 10^27 kg (Io passes)
(200 km) * (1.1*10^3 kg/m^3) * (190,000 km)^2 ~ 10^25 kg (Mimas fails)
Since Io passes (narrowly), all of Jupiter's other major moons should be fine, too. For Saturn, Titan can be assumed to be fine as well based on that result, since it's as massive as and a lot farther out than Io. Thus, going inwards from there rather than outwards from Mimas should be more efficient:
Rhea: (800 km) * (1.2*10^3 kg/m^3) * (500,000 km)^2 ~ 2.5*10^26 kg (Rhea passes, M/4 ~ 1.5*10^26 kg for Saturn)
Dione: (600 km) * (1.5*10^3 kg/m^3) * (400,000 km)^2 ~ 1.5*10^26 kg (borderline)
Beyond the major moons, there is a large gap for both gas giants. The next moons don't show up until one is beyond ten million kilometres, and they only become numerous again beyond fifteen million kilometres. These can all be covered by one generic calculation, I think:
(10 km) * (0.5*10^3 kg/m^3) * (1.5*10^7 km)^2 ~ 10^27 kg
Those few below that distance are known or at least assumed to be quite dense, as it happens, so pretty much all outer moons pass easily.
Having established that there are two classes with pass (ab), the major moons (or at least half of those, for Saturn) and the small outer moons, we can now tackle (b). I'll begin by trying out the generic values used just above, using the mass of Jupiter to get the lower (stricter) limit again:
criterion: p^4 < (6,400/G^2) r^3 rho D^6 / M^3
p^4 < (6,400/G^2) (10 km)^3 (0.5*10^3 kg/m^3) (1.5*10^7 km)^6 / (2*10^27 kg)^3 ~ 10^18 s^4 ===> p < 9 hours
Oho. (a) gave us a lower limit of 6 hours, if you'll recall, so the allowed range is rather narrow here: The moon has to rotate just fast enough for the stationary orbit to be barely above its surface, otherwise the gas giant's gravity takes over. As I mentioned above, there are several moons whose orbital periods indeed fall squarely between those limits. Of course, any moon which is bigger or denser or farther out than the generic values I used will have a somewhat wider allowed range, as will all Saturnian moons due to the lower value of M. However, the final step in the calculation is always to take the fourth root (p^4 -> p), which means that those differences won't amount to all that much.
Now, for the major moons, I'll start with Io and our Saturnian (ab) borderline case Dione, just to get a first impression:
(6,400/G^2) (1,800 km)^3 (4*10^3 kg/m^3) (400,000 km)^6 / (2*10^27 kg)^3 ~ 10^16 s^4 ===> p < 3 hours (Io fails)
(6,400/G^2) (600 km)^3 (1.5*10^3 kg/m^3) (400,000 km)^6 / (6*10^26 kg)^3 ~ 7*10^15 s^4 ===> p < 3 hours (Dione fails)
Not so much. Moving on to the "best cases", i.e. the major moons which are biggest and/or farthest out:
Ganymede: (6,400/G^2) (2,600 km)^3 (1.9*10^3 kg/m^3) (1.1*10^6 km)^6 / (2*10^27 kg)^3 ~ 7*10^18 s^4 ===> p < 15 h
Callisto: (6,400/G^2) (2,400 km)^3 (1.8*10^3 kg/m^3) (1.9*10^6 km)^6 / (2*10^27 kg)^3 ~ 2*10^20 s^4 ===> p < 36 h
Titan: (6,400/G^2) (2,600 km)^3 (1.9*10^3 kg/m^3) (1.2*10^6 km)^6 / (6*10^26 kg)^3 ~ 7*10^20 s^4 ===> p < 48 h
Iapetus: (6,400/G^2) (700 km)^3 (1.1*10^3 kg/m^3) (4*10^6 km)^6 / (6*10^26 kg)^3 ~ 7*10^21 s^4 ===> p < 96 h
Unfortunately, all of these are tidally locked, and for that reason have rotation periods in the Earth-week to -month range. So none of them actually pass (b), even though the range between (a) and (b) is quite wide for the farthest of them. Crikey.
In short, based on these calculations, the only moons which can support stationary orbits are those on the more massive side of the small outer moons. I'll finish by doing the full set of steps for one of those for Jupiter and Saturn each, to show that they really do exist:
Jupiter's Elara:
criterion: p^2 > 10 / (G rho)
LHS ~ (12 hours)^2 ~ 2*10^9 s^2
RHS ~ 10 / (G * (2.6*10^3 kg/m^3)) ~ 5*10^7 s^2
criterion: p^4 < (6,400/G^2) r^3 rho D^6 / M^3
LHS ~ (12 hours)^4 ~ 4*10^18 s^4
RHS ~ (6,400/G^2) (43 km)^3 (2.6*10^3 kg/m^3) (1.2*10^7 km)^6 / (2*10^27 kg)^3 ~ 7*10^19 s^4
passes both criteria -> stationary orbit is available at a height (derived earlier) of ((G/(3 pi)) r^3 rho p^2)^(1/3)
((7*10^-11 m^3/kg/s^2)/10 (2.6*10^3 kg/m^3) (12 hours)^2)^(1/3) r ~ 3.5 r ~ 150 km (100 km above surface)
Saturn's Albiorix:
criterion: p^2 > 10 / (G rho)
LHS ~ (13 hours)^2 ~ 2*10^9 s^2
RHS ~ 10 / (G * (1*10^3 kg/m^3)) ~ 1.5*10^8 s^2
criterion: p^4 < (6,400/G^2) r^3 rho D^6 / M^3
LHS ~ (13 hours)^4 ~ 4*10^18 s^4
RHS ~ (6,400/G^2) (16 km)^3 (1*10^3 kg/m^3) (1.6*10^7 km)^6 / (6*10^26 kg)^3 ~ 10*21 s^4
passes both criteria -> stationary orbit is available at a height of
((7*10^-11 m^3/kg/s^2)/10 (1*10^3 kg/m^3) (13 hours)^2)^(1/3) r ~ 2.5 r ~ 40 km (24 km above surface)
Whew!
edited 27th Aug '12 11:03:20 AM by kassyopeia
Soon the Cold One took flight, yielded Goddess and field to the victor: The Lord of the Light.
I've been doing some research on asteroid mining for a story I'm writing. It's a fairly important plot point, so I want to make sure there's at least some logic behind it.
The setup is thus: Alien miners stumble upon the Solar System (with Earth circa Next Sunday A.D.) and decide to set up shop. The aliens, presumably having read the Planet Looters article on this wiki, realize that mining Earth is impractical, so they focus elsewhere. At the moment, I have them set up around the outermost moons of Jupiter.
My question about the setup is twofold:
1) Would it make more sense to have them mining asteroids over natural satellites? In my mind, it seems like mining a satellite would be easier, since you could just park on the surface and let gravity take its course, rather than having to worry about staying attached to an asteroid. And it'd be easier to find the satellites around a big planet, rather than hunting down tiny asteroids in the asteroid belt.
2) How difficult would it be to detect an operation like this from Earth? Let's say, for example, that the ships are around 2km long, and there's about 20 of them. Would a flyby probe or Earth-bound telescope be able to detect them at all? (Let's assume they just happened to be investigating that one moon for some reason.)
edited 25th Aug '12 12:27:59 AM by WarxePB