Built on Facts

Stealth in Space

While doing some poking around online, I came across a website called Project Rho, which tries to provide some science background for science fiction writers who want some degree of technical accuracy in their imaginative work. Generally it looks like they’re on the right track.

In their section on stealth in space, they explain with the weary air of repetition that there’s no such thing. The flare of a rocket is bright enough to be seen from basically anywhere, and the thermal signature of even a spacecraft with rockets off is visible from clear across the solar system. The first I can believe (though of course it’s worth checking), but the second sounds fishy. But so do lots of true things. Why not run the numbers ourselves? Before we do, let’s see what they say:

“Well FINE!!”, you say, “I’ll turn off the engines and run silent like a submarine in a World War II movie. I’ll be invisible.” Unfortunately that won’t work either. The life support for your crew emits enough heat to be detected at an exceedingly long range. The 285 Kelvin habitat module will stand out like a search-light against the three Kelvin background of outer space.

They go on later in the article:

The maximum range a ship running silent with engines shut down can be detected with current technology is:

Rd = 13.4 * sqrt(A) * T^2
where:
Rd = detection range (km)
A = spacecraft projected area (m^2)
T = surface temperature (Kelvin, room temperature is about 285-290 K)

If the ship is a convex shape, its projected area will be roughly one quarter of its surface area.

Example: A Russian Oscar submarine is a cylinder 154 meters long and has a beam of 18 meters, which would be a good ballpark estimate of the size of an interplanetary warship. If it was nose on to you the surface area would be 250 square meters. If it was broadside the surface area would be approximately 2770. So on average the projected area would be 1510 square meters ([250 + 2770] / 2).

If the Oscar’s crew was shivering at the freezing point, the maximum detection range of the frigid submarine would be 13.4 * sqrt(1510) * 273^2 = 38,800,000 kilometers, about one hundred times the distance between the Earth and the Moon, or about 129 light-seconds. If the crew had a more comfortable room temperature, the Oscar could be seen from even farther away.

The equation given isn’t derived. We have no idea where they’re getting that 13.4 proportionality constant. Dimensionally it’s correct, and it’s pretty easy to derive the equation up to that constant which will depend on the sensitivity of the detector. That equation modulo some uncertainty with respect to that constant is accurate as far as it goes given a spacecraft of hull temperature T and cross-sectional area A.

I would take you through the steps of the derivation, but it would be pointless because the assumption that the hull temperature has anything to do with the interior temperature is simply flat wrong. We can prove this with a potato.

Switch your oven to the “Bake” setting at a temperature of 350 F. After preheating, put in the potato. The interior of the oven, and eventually the potato, are maintained at a constant temperature of 350 degrees. How hot is the exterior surface of the oven? Depends on how well insulated your oven is, but I can guarantee it’s a lot less than 350 degrees.

The key is the understanding the relationship between heat and energy. Put hot coffee in a thermos – the hot coffee is hot because it contains thermal energy. If the energy can’t leave, the coffee will stay hot because the energy stays inside the thermos. The outside of the thermos stays at the temperature of the surroundings. Now neither the thermos nor the oven is a perfect insulator. Some energy leaks out of the oven’s interior, cooling it down. The oven thus has to pump energy into the heating elements to make up for this loss. Equilibrium is reached when the rate of energy being put into the oven equals the rate of loss through the insulation.

For a spacecraft in a vacuum, the pretty much the only way to lose energy from the interior is by radiant heat. The higher the temperature of the outside, the higher the rate of energy loss via radiation. But the temperature itself is irrelevant, since just like the oven and the thermos it’s not necessarily related to the actual temperature inside the cabin at all. It is always and everywhere a function of the total power passing through the hull. If the temperature inside the cabin is constant, the power leaving the hull by radiation is exactly equal to the power being generated inside the hull.

So how far away can we detect a given amount of emitted power? According to Wikipedia, a telescope of 24″ aperture can detect stars of magnitude 22 after a half-hour exposure. I think this is a pretty good realistic limit for detection with reasonable equipment in a reasonable time frame. Now we need to compare this magnitude to something of known power output. How about the Sun? The sun has magnitude -26.73 as seen from the Earth’s surface (smaller magnitude is brighter), for a difference in magnitude of 48.73. The exponent used for magnitude is 2.512, so the difference in power per unit area of telescope is 2.512^48.73 = 3.1 x 1019. Since the Sun radiates about 1000 watts per square meter at the distance of the earth, the smallest radiant power we can reasonably detect in our telescope is about 3.123.1 x 10-17 watts per square meter.

Our hypothetical spacecraft is radiating that power into space, evenly distributed over the surface of a sphere of radius r, where r is the distance to the detector. When that power-per-area is the same as the limit of our telescopic capability, that gives us the maximum detection range. Mathematically,

i-1a518f2655b02f9d30ddb0c8226ad8eb-1a.png

Where rho is the sensitivity of our detector. Solve for r:

i-ec6170d6ae0330319bd539b5264503e8-2a.png

So what’s the power? Well, each human on board is going to produce about 100 W just from basic bodily metabolism. Computers, life support, sanitation, and all the rest will contribute more. We might assume 10,000 watts total for a futuristic ship that’s specifically designed to emit as little power as possible. It might well be significantly lower. Plugging in, I ger r = 5.98 x 109 meters. This is pretty far, but it’s only around 4% of the distance from the earth to the sun. Practically nothing in terms of solar system distances. Even a ship dumping a megawatt of power should only be visible from a third of the earth-sun distance.

The reason for this divergence in our estimate versus the Project Rho estimate is that it takes a huge amount of energy to maintain a hull exterior at cabin temperatures. But insulation means that’s not necessary, all that’s necessary is that the power out equals the power generated in the interior.

Or at least this is my impression. I could be wrong. Thoughts?

Comments

  1. #1 CS
    March 10, 2010

    The excess heat does not need to be shed evenly over the entire outer surface. Cool one side of the craft (liquid helium?) and radiate the excess heat only on the other side. You can still have “stealth”, just not for every direction.

  2. #2 Anonymous
    March 10, 2010

    All math aside, these detections would only be valuable at (relatively) incredibly close range. Assuming you’re wandering through deep space, you’re going to want to be moving as fast as possible because it’s a really long drive. Stealth seems less valuable than raw speed.

    Hiding your submarine from sonar makes a lot of sense, given that your boat moves significantly slower than the speed of sound. Once I’ve spotted you, I have plenty of time to ready, aim, and fire or to man the life boats or launch a pursuit.

    Hiding your spacecraft from a heat detector seems a bit absurd if you plan to travel faster than the heat particles you emit. Any particles you do throw off (that aren’t light) likely won’t be going much faster than you. After all, your ship has propulsion – anything that falls off your ship doesn’t.

    If you can reach superluminal speeds, that makes the problem even more trivial. By the time you’ve been detected, you’ll be long gone. Who cares if I can see where you’ve been clear across the solar system? It takes us over 8 minutes just to see the Sun.

  3. #3 Blaise Pascal
    March 10, 2010

    Heat particles?

  4. #4 Zifnab
    March 10, 2010

    Hot particles. Shit moving away from your ship that isn’t really cold.

  5. #5 Markk
    March 10, 2010

    One other thing is that you are describing equilibrium conditions. If you want stealth it is not likely you need it forever. You can combine this idea with the directional heat area and pump heat back in the direction you came and keep almost all the other directions with a surface very close to ambient space for some period of time. Or just heat something up inside for a period. Even so, given your distances it might not matter. Think of how many detectors one would need to cover even a fraction of say, a sphere with a radius of Mars orbit.

  6. #6 Pierce Nichols
    March 10, 2010

    Your detection standard is far too generous to Project Rho — you can only detect the sort of dim object you are talking about if you know where to look. A real spacefaring warship has to be able to scan the skies… and for that problem, the detection range is *MUCH* shorter because the integration time available is on the order of a few seconds instead of half an hour. You should definitely be able to sneak within laser range of an opponent with a little care of the type described by CS.

    And while chemical rockets do produce exhaust plumes visible the better part of a system away, they’re not the only kind of propulsion available. Low-power ion thrusters, among others, should be far less visible.

    I wrote a short essay (that seems to have vanished from the net) entitled ‘On Killing Things In Space’ which argues, based on the points above, that near-term space combat is likely to look an awful lot like submarine warfare. Lots of time lying doggo, very sparing use of propulsion, and one-hit one-kill weapons.

  7. #7 meichenl
    March 10, 2010

    Matt is considering only light. The potential observer is trying to detect the spacecraft using a telescope. There are not other “hot particles” coming into play here.

    You might decrease your visibility by building a large shell around your ship and carrying it along with you. Increasing this surface area allows you to radiate the same power at lower temperature.

    As far as minimum power that can be detected with a microscope, remember that we knew where that faint star was before pointing the telescope at it. If you’re trying to cover the whole sky, it’s going to be a lot harder. I suspect the minimum power also depends on frequency (which is why I think a colder hull might be beneficial).

    I like the idea about cloaking yourself on one side. You can actually do it without actively cooling the hull if you have different emissitivities at different spots.

  8. #8 John Ellis
    March 10, 2010

    Surely the spaceship does not have to become invisible, but merely camoflage itself enough to look like a small asteroid? I.e the heat signature of an asteroid heated by the sun at its orbital radius. Real asteroids must provide enough false positives to a detector that they effectively make a noisey background to hide in.

    I seem to remember in an Iain M Banks culture novel — Excession? — they hid a load of space ships inside a hollowed out asteroid like that.

  9. #9 Anonymous
    March 10, 2010

    Ok, for any Sci-Fi story about space travel there is always a superluminal travel mode (or the story becomes a rather boring one of internal conflict on a spacecraft plodding for years or generations between even relatively nearby stars.)

    Now one would not want to show up in an unexplored neighborhood with a great burst of released energy to announce your arrival. However, about the only thing we got that distorts space-time is mass-energy and I suspect a pile of it is going to be emitted creating the sorts of space-time distortion one might employ for “warp drive”. So a distinct Lack of Stealth might be a more of a concern for entering and particularly exiting “Warp Speed”.

    Frying that small rocky planet that happened to be orbiting some star we visit might not endear us to the natives.

    “Stealth” and “KaBlam!” – We’re Here!” seem irreconcilable somehow.

  10. #10 Max Fagin
    March 10, 2010

    I’m with CS on this one. Compared to interplanetary travel, it seems a rather trivial engineering challenge to simply cool one face of your spacecraft to ~3 K and reject all the heat generated inside the spacecraft out the opposite face of the hull.

    Or taking the idea to the extreme, you could cool 99% of the surface of the hull to ~3 K and reject the rest of the heat in a bright but tightly confined directional beam. Unless you were stupid enough to point this beam directly towards your enemy telescopes, your spacecraft would have exactly the same thermal profile as the cosmic background radiation, and no one would be able to detect you.

  11. #11 Watts
    March 10, 2010

    What about reflected light? We typically detect large, cool objects like asteroids and planets by the light reflected off of them by the Sun. I suppose radar wouldn’t be terribly useful, given how difficult it is to get a reflected signal back from the Moon, but perhaps reflected sunlight would aid detection?

  12. #12 Anthony
    March 10, 2010

    Speaking as someone who knows the site author, the point about ‘detection across the solar system’ has always been an exaggeration, and isn’t even supported by the equation on that page; I believe it’s a knee-jerk reaction to yet another try at ‘space submarines’. A 250 square meter perfect black-body has a power output of 5.67e-8W/m^2/K^4, so at 273K and 1510 square meters it would be 475 kW, which will get our range (by your equation) up to 138 light-seconds, which is well within reasonable margin of error (that 13.4 should probably just be written as 10 to reflect appropriate significant figures). As for why you’d run your surface that hot, sunlight will generally do it for free, a spherical blackbody at 1 AU from the sun will have an equilibrium temperature of 273K. Note that, if you’re beyond the orbit of Jupiter, detection range would be dominated by internal heat.

  13. #13 Matt Springer
    March 10, 2010

    It is an interesting challenge to get around solar heating. You could just paint the hull white, but then you’re bright by reflection. Better options might include blocking sunlight with a flat mirror reflecting light in some convenient direction away from detectors, or constructing as much of the ship as possible out of transparent materials.

  14. #14 Thomas
    March 11, 2010

    One way to detect a stealthed spaceship is to use a bunch of telescopes that watch the known stars and wait for occultations indicating that something passed between the star and the telescope. Pretty useless for a singe ship trying to detect an enemy since the chance of detection is too low but it might be useful for a defender that has had time to set up an array of millions of small telescopes.

  15. #15 Eric Lund
    March 11, 2010

    The solar heating issue Anthony #12 refers to is worse than he indicates. There are engineers at NASA and APL whose job it is to do thermal modeling of spacecraft to make sure that critical parts of the spacecraft don’t get too hot or too cold. I’ve worked with instruments that have routinely seen temperatures exceeding 50C when exposed to sunlight, while parts that are in shadow can get quite cold. And that’s with spacecraft made primarily out of metal; objects made out of insulating material such as rock (e.g., the Moon) have even larger variations. You can cool part of the surface cryogenically, as CS #1 suggests (and there are space telescopes which do this), but it limits your range since you will periodically have to replenish your supply of liquid He or N2.

    Constructing mirrors on the surface of the ship, as Matt #13 suggests, has an additional problem: the risk that someone will detect you because your brightness varies rapidly. It might work if your mirrors are steerable and you can be sure the enemy is in one small part of your sky. If he has detectors all over the place, he will detect you due to your highly anisotropic brightness.

  16. #16 yud
    March 11, 2010

    If I were trying to detect stealthy spacecraft, I’d just put together an array of WISE-style IR telescopes. (http://www.nasa.gov/wise/)

    According to Wikipedia, “Its detector arrays have 5-sigma sensitivity limits of 120, 160, 650, and 2600 microjanskies (µJy) at 3.3, 4.7, 12, and 23 microns.” I’m not sure how that translates into how many kelvins it can detect at a given distance, but it sure seems pretty sensitive.

    It has apparently already discovered 2,000 new asteroids and 4 new comets

  17. #17 Anthony
    March 11, 2010

    The 273K(0C) figure for solar heating at 1 AU from the sun is for a sphere with perfect heat transfer through the body of the sphere (i.e. its radiative area is 4x its cross-section); depending on the actual heat transfer and the rate of rotation of the body, the sunlit face could reach as high as 386K (113C), the back face can reach almost arbitrarily low temperatures. However, this doesn’t matter a lot: if a ship is being hit by X watts of sunlight, it’s pretty much going to emit X watts worth of photons (unless it has somewhere to store the heat, and it usually only takes hours to days to overwhelm any practical heat sink), and tweaks to albedo and heat distribution across the surface just modify the spectrum and direction of the emissions.

  18. #18 Pierce R. Butler
    March 12, 2010

    You wanna sneak up on Earth from outer space?

    Approach from the far side of the sun, slip around inside the orbit of Mercury, set your rockets at full power and charge straight (okay, really at a slight curve to allow for orbital motion) at the noon point.

    The model is not submarine warfare but WWI aerial dogfighting.

    After the Conquest, don’t forget to reward your local tactical advisor with a plum sub-Overlordship.

  19. #19 Stagyar zil Doggo
    March 12, 2010

    Approach from the far side of the sun, slip around inside the orbit of Mercury, set your rockets at full power and charge straight (okay, really at a slight curve to allow for orbital motion) at the noon point.

    If you start out at* Proxima Centauri, you already need to have an orbital velocity of (2*pi radians/year)*4.2 light years / c = 26.3893783 times the speed of light just to stay behind the sun. Assuming you started from a standstill, you’d have to accelerate really hard to get to 26.4 c without blowing your cover. Add to this whatever radial velocity you can rustle up to actually get closer to the sun.

    So, assuming you are not actually capable of FTL travel or the ginormous accelerations involved, you’ll still need to sneak in much closer, making like one more asteroid and inching your way inwards while venting any radiation in a narrow backwards beam.

    *Or more precisely, at a point on the ecliptic as far away as Proxima Centauri. Anyone know how far the closest star on the ecliptic is?

  20. #20 amphiox
    March 12, 2010

    re: 9

    With superluminal capacity you could probably more or less pick any deceleration/stop point at will to pop in with guns blazing. No subluminal detection system is going to find you fast enough to stop you in that scenario.

    But, in nearly every scifi scenario with superluminal traveal capability, there is almost always superluminal communications ability as well, which is faster than the superluminal travel. If communication is possible, then in theory detection should be equally possible using the same set of scientific principles, whatever they may be imagined to be.

  21. #21 Tacroy
    March 12, 2010

    In one of the later Uplift novels, David Brin uses a heat-pumped communications laser used to cool down the ship (they weren’t trying to be stealthy at all, the enemy was just using something against them that heated up their ship). This would, presumably, also work for stealth – convert your ambient heat to lazors and shoot them out into the depths of space where nobody’s looking.

    As for how you would go about detecting something in space – I would assume that by the time we’ve got space warships, we’ll have self-organizing micro-scale materials. At that point, have some science dudes invent paint with embedded photoreceptors. You slather it on and it hooks up with some visual input traces on the hulls to provide data on what radiation’s impacting the ship. You’d have to “teach” the computer how to see every time you got a new paint job, but that wouldn’t take too long – humans manage to do it, after all, and we have to sleep. There would be no need to “scan” the skies to gather enough data to integrate over – your shipboard computers would be keeping track of everything within line of sight at all times, and because your sensors are actually a thin layer of paint over your entire ship you would have no blind spots until some jerk took a potshot at you.

    As a side benefit, you could sell polka-dot paint to civilians! It’s win-win!

    This might even help with the whole stealth thing; just have your photoreceptors absorb whatever energy hits them, and then reverse the polarity on the ones on the other side and emit it back out (or, you know, include some photon emitters in the paint mix). This doesn’t need to be a perfect invisibility cloak – you just need to diffuse yourself to the point where you’re not much of an occlusion over a distant star, like how we’re doing nowadays with fiber-optic covers on tanks.

  22. #22 amphiox
    March 12, 2010

    re #21,

    Hmm, if you’re able to use photon emitters for stealth, why not go whole hog and do outright deception. Emit brightly in a manner that complete masks/overwhelms your normal signature, but imitate a natural object.

    Is it possible to fake a redshift? Set your emissions so that you don’t look like a spaceship or even an asteroid on approach, but a distant star or galaxy moving away?

    Of something subtler? Let the enemy know you are coming, but fake your velocity so he thinks you’ll arrive at a different time than you actually do, or make your armada appear bigger or smaller than it really is?

  23. #23 Andy P
    March 13, 2010

    Your “22 magnitude” estimate might be leading you to a mistake commonly made when people do this calculation. The wikipedia estimate doesn’t say *where* that 22 magnitude star is in the sky. It’s therefore tough to tell what background we’re looking against. Typically, astronomers look into the darker portions of the sky but one look here will show you that this is not always possible.

    http://www.ipac.caltech.edu/Outreach/Gallery/IRAS/allsky.html

    That blue glow is the zodiacal light- it’s the reflected and emitted radiation from the dust that orbits in the plane of our solar system. If your target spacecraft is in the plane of the ecliptic it MIGHT be able to blend into that background at a certain range from a given detector. Astronomers usually subtract this background from their detections…but their typical targets (stars) are very intense sources (i.e. they put lots of photons into a given solid angle…even if they are very far away). A cooler spacecraft MAY (I’m still not sure) be able to evade detection at strategically, operationally, and/or tactically useful ranges.

    The question for detection of spacecraft becomes: “is the error to which the searcher knows the background larger or smaller than the spacecraft’s signal over the integration period?”

    I’ve struggled on and off for many years to come up with systematic models to answer these questions in my copious spare time. The ones in the current SF literature (like the one on Project Rho) … are too simple to be particularly useful.

    v/r
    Andy P.

  24. #24 Recall
    March 14, 2010

    “Or at least this is my impression. I could be wrong. Thoughts?”

    Moving is going to be a problem.

  25. #25 porno
    March 14, 2010

    nasıl bısey bu answer these questions in my copious spare time. The ones in the current SF literature neden zor bukadra

  26. #26 sex
    March 14, 2010

    seksi cekımler var ama nerde orbits in the plane of our solar system. If your target spacecraft is in the plane of the ecliptic it MIGHT be able to blend into that background at a certain range from a given detector kım bılır kımlerdedır ama ne dıyorum ben ya

  27. #27 Greg
    March 14, 2010

    The detection range that you compute is not that much less than the one given by Project Rho: (5.98 * 10^9 m is ~ 6 million km, which differs from the value they get 38.8 million km by a bit more than a factor of six, still in the same order of magnitude). Moreover, to expand upon the point Recall makes, when you are talking about moving, the power requirements increase markedly: Ad Astra’s mission profile for a 40 day trip to Mars requires 200 megawatts. Even the current version of the VASIMR engine Ad Astra is making consumes 200 kilowatts. As soon as you need to change trajectory, your visibility will increase markedly.

  28. #28 Andy P
    March 15, 2010

    You need to consider the frequency band of the emission too…that can be controlled to a certain degree. Though you pay a price: as Anthony says the total emitted power is still the sum total of your waste heat…if you choose to preferrentially emit in one part of the spectrum, you’ll be brighter there than a blackbody. Only reason to do this would be if the background is particularly bright in that part of the spectrum.

    Also, a good estimate for a high efficiency space power system is 20-25%. That means that our 200 MW power requirement probably comes with 600 or 800 MW of waste heat to your radiators…in addition to whatever portion of the energy that goes into the rocket plume is radiated.

    Bear in mind that most of the energy radiated in a VASIMR plasma plume is probably in the RF or microwave: not visible. That changes the background for your calculation…if you were thinking of trying one in your copious spare time.

    v/r
    Andy P.

  29. #29 Damien RS
    March 15, 2010

    FTL is irrelevant to this discussion.

    There’s also an implied context, e.g. my first association with “stealth in space” is “are pirates possible?” So we’re not talking about just one ship vs. another ship, but hiding from organized military or law enforcement. Which makes directional radiation or focused rocket plumes problematic, because if the scenario supports commerce or warships in space, the scenario supports militaries having monitors scattered around the system, so you’d have to hide from all angles, not just one.

    And as already noted, even if it’s possible for a cool ship to drift stealthily, how did it get to its trajectory and where is it going? You might be stealthy now, but if you fired rockets to get there, Space Traffic Control knows your trajectory. You don’t have to scan the sky for cool objects, you scan the sky for hot objects and track all the cool objects you know about.

  30. #30 Paul Murray
    March 16, 2010

    “In one of the later Uplift novels, David Brin uses a heat-pumped communications laser”

    Isn’t it impossible to convert heat directly into energy?

  31. #31 ummon
    May 3, 2010

    Are the detectors really dependent on power per area? I would have thought they were dependent on just plain old power. I could be wrong though.

  32. #32 neutrino78x
    July 17, 2011

    I love the Atomic Rockets site, but I disagree with the author on this issue. Of course stealth is possible in space.

    You guys who are so convinced that it is easy to detect something which is not actively dumping heat in the direction you’re looking, and is only heated, if anything, by the Sun: why is it so difficult to find asteroids?? We routinely find out about asteroids which will pass closer to the earth than artificial satellites with only a few days of notice. By your logic, this should never happen, because everything can be detected from infinitely far away, even if it is emitting no more heat than is radiated onto it by the Sun!!! ((roll eyes))

    A surprise attack from the opposite side of the Sun would be very easy; we know how to make something that is nearly invisible to radar, and infrared won’t help you much if you have to look into the Sun to use it; it would saturate your sensor, okay.

    Also, the concept of FTL is not really related to this, nor are spacecraft traveling at nearly the speed of light, because you wouldn’t come out of FTL near a planet if you were trying to be stealthy (unless it was on the other side of a planet from your target), you’re not going to be stealthy in infrared while going at FTL if the enemy can detect you at all in that mode, and if you’re moving conventionally but near the speed of light, you’re not going to maneuver to attack easily at that speed (and if you do, it won’t be stealthy).

    Unless we’re talking about Star Trek or Star Wars (ships can’t detect each other in hyperspace on SW anyway, and I think the Romulan cloaking device can be used at warp), stealth in space would be inside of a solar system, nebula, or similar place, where there are lots of places to hide etc. Good luck detecting someone on the opposite side of a planet (large asteroid, etc.) from you.

    –Brian/neutrino78x

  33. #33 Haji
    October 10, 2011

    First, I’d like to point out you say “future stealth ship” but compere it to todays avarage equipment. Isn’t that cheating a little? For example, Hubble space telescope can see up to 31 magnitude (36 times weaker than in your example) and it isn’t even the best we have designed. In the future, when those “future stealth ships” exist, we should be able to use telescopes like Hubble well enough. That changes you 6*10^9m detection range to 1.8*10^11m detection range, which is about 1.2 AU.

    Second, Atomic Rockets used something with existing paramethers. The 10kW energy have no basis. I mean, you may get even less energy from the drone, but a ship like supercarrier uses 5MW reactor for a reason. Air circulation for a crew for example. Also, radiating this energy at all will generate ‘hot spot’. And how did the ships get there in the first place?

  34. #34 Haji
    October 10, 2011

    I might have made mistake in my calculations. According to this article:

    http://en.wikipedia.org/wiki/Magnitude_%28astronomy%29

    (look lower, at the table of magnitudes)

    magnitude 32 (which is the visible light by Hubble) is 4 orders of magnitude weaker than magnitude 22. This would put the detection range on about 6*10^14m or 4 000 AU.

  35. #35 Leandro
    January 11, 2012

    Hubble is an optical telescope. Project Rho’s point concerns mostly infrared, which is a lot more difficult to detect.

    Also, there is a fundamental limit to how much you can detect, which is how much there is to detect against the background.