20
$\begingroup$

I want to make a calculator that can determine the range at which spaceships will be detected against an infrared detection system.

Figuring out how "stealthy" a ship is is mostly a matter of apparent magnitude calculations which is not difficult. The calculator will allow me to put in the visible cross sections and temperatures of various parts of the ship and use distance to calculate the apparent magnitude of the ship's IR black body emissions.

The part I am struggling with is trying to figure out the actual sensitivity of pyroelectric detectors. I can find a lot of vague sources like how some specific telescope can detect an apparent magnitude of X, and that it does so by focusing on a narrow section of space for some undefined number of minutes/hours straight. But, I'm having a hard time finding all of the factors I need together to make the basis of an effective range calculator.

I am looking for an equation that defines the relationship been the minimum detectable apparent magnitude of an object, the size of the collection area, the exposure time, and the angle of detection to be able to finish the calculator.


It's not done yet, but here is the working prototype: https://jsfiddle.net/nosajimiki/4kmrf07g/73/

$\endgroup$
1
  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Worldbuilding Meta, or in Worldbuilding Chat. Comments continuing discussion may be removed. $\endgroup$
    – L.Dutch
    Commented May 22 at 2:52

6 Answers 6

17
$\begingroup$
  1. What you need to compute is how many photons hit the entrance pupil of the camera in the allowable exposure time.

    This is one place where the field of view of the camera comes into play. The entire celestial sphere has 129,600 / π ≈ 41,253 square degrees. The telescope is looking at a certain field of view; for example, the James Webb Space Telescope is looking at a field of view of 2 by 2 arc minutes, or 0.001 square degrees. To cover the entire celestial sphere once per week, it cannot dwell more than 16 milliseconds on each direction.

  2. Ideally, against an absolutely black background, if one photon hits the entrance pupil that's it, you are detected. But the background is never absolutely black...

    All practical photosensors have some level of inherent noise, so that in practice they may need more than one photon hitting a pixel. (This is why, for example, why it was so important to keep the James Webb Space Telescope as cold as possible.)

  3. Which is why you will need to compute the signal (interesting photons) to noise (photons from other sources) per sensor pixel, and decide what signal to noise ration counts as detected.

  4. Meaning that detecting a faint point of light on the background of Barnard 68 or the Boötes Void is much easier than on the background of the Sun or Jupiter.

  5. And then, all the telescope would have seen is a point of light. It must then decide that the point of light is a target. This is a completely different topic...

$\endgroup$
13
  • 2
    $\begingroup$ I would not say this exactly answered my question, but +1 for getting me thinking in terms of pixels. I think I'm able to figure it out from here. $\endgroup$
    – Nosajimiki
    Commented May 20 at 19:32
  • 1
    $\begingroup$ One more factor: to locate a spaceship we may also need to know how fast it is panning across the field of view, i.e. how much time we have before it crosses from one pixel into another. This would be relevant if it is so dim and moving relative to us at such speed that it is liable to skip to the next pixel before we can gather enough photons to spot it. $\endgroup$
    – causative
    Commented May 21 at 11:00
  • $\begingroup$ @causative Imagine having a pile of sensors that scan the sky. Some are narrow and some are wider. They look for more IR photons than expected in their region. A moving target would show up in the 3 dimensional plot of this just as much as a stationary one, but instead of being in the same region the region they are in would form a line (curves at this scale require gravitational boost; thrust is too obvious). Finding such lines is no harder than finding a stationary object really. $\endgroup$
    – Yakk
    Commented May 21 at 14:28
  • $\begingroup$ @Yakk You need to do a Hough transform to find the motion-blurred streaks across the camera. It is harder because in addition to the two dimensions of your camera plane, there are also the two dimensions for their apparent velocity across the plane. That means you are trying to find a point in four dimensions rather than just two. That means you need more photons for a statistically significant detection. $\endgroup$
    – causative
    Commented May 21 at 16:10
  • 2
    $\begingroup$ @AlexP You're technically right in your first sentence but wrong in your second. Photons do make perfect points when they hit, but the points are randomly distributed in an Airy disc, even when it's only a single photon at a time. There don't need to be multiple photons to get an interference pattern. That's the big deal about the double slit experiment: a single photon interferes with itself. $\endgroup$
    – causative
    Commented May 21 at 18:11
15
$\begingroup$

1. SNR to exposure time

Although your mileage may vary depending on the details of your detector, you can take an approach similar to deriving what is known as the CCD equation. It's derived for the case of CCDs, but similar approaches can be used for any digital array, including the kind you're interested in. Here, I'm going to use the notation from Chromey's To Measure the Sky, but you can find similar derivations elsewhere online, e.g. these notes.

There are several sources of noise in the detector:

  • There's inherent uncertainty in the number of photons detected from the object you're searching for, $N_*$.
  • There's a background from the sky all around the source.
  • Even in the absence of any signal, there's some sort of thermal noise, the dark current, and uncertainty in what constitutes zero counts, the bias.
  • There are several additional sources of noise arising from reading the signal and from estimating the dark current and bias, including something called the read noise.

We can quantify the strength of a possible detection by its signal-to-noise ratio, SNR. If the detector operator is able to properly calibrate the dark and bias, the SNR is approximately $$\mathrm{SNR}=\frac{\dot{N_*}t}{\sqrt{[\dot{N_*} + n_{\mathrm{pix}}a_b(\dot{b}+\dot{d})]t + n_{\mathrm{pix}}a_b\rho^2}}$$ with $\dot{N_*}$, $\dot{b}$ and $\dot{d}$ the rate of photons from the object, background and dark current, $\rho^2$ the read noise, $t$ the integration times, $n_{\mathrm{pix}}$ the number of pixels in the region surrounding the object, and $$a_b\approx1+\frac{n_{\mathrm{pix}}}{p_b}$$ with $p_b$ the number of pixels used to estimate the background.

If you want to detect a source at a certain signal-to-noise ratio, you can then invert the CCD equation using the quadratic equation to calculate how much time you'll have to observe it for; I leave this as an exercise, but it's fairly simple to calculate.

There are a couple limiting cases:

  1. The object you're looking for is very bright. In this case, $\dot{N_*}t$ is bigger than any of the noise terms in the denominator, and the exposure time needed to reach some SNR is $$t=\frac{\mathrm{SNR}^2}{\dot{N_*}}$$
  2. The background is comparatively bright. Then $$t=\frac{\mathrm{SNR}^2}{\dot{N_*}^2}\left[\dot{N_*}+n_{\mathrm{pix}}a_b(\dot{b}+\dot{d})\right]$$ If the background is bright enough, it's possible that $\dot{b}\gg\dot{d}$ and $\dot{b}\gg\dot{N_*}$, making the equation even simpler.
  3. The read noise is large. Then for short exposures, $$t=\frac{\mathrm{SNR}}{\dot{N_*}}\sqrt{n_{\mathrm{pix}}a_b\rho^2}$$ It's quite possible that -- being in space -- the dark current will be minimal, since it should be easier to cool the instrument. Since your world has space travel, perhaps the technology has also made bias estimation more accurate. It's possible that the only sources of noise are the background and the Poisson noise from measuring the number of counts from your source.

2. Magnitudes to number of photons.

I've phrased the above in terms of the number of photons the object emits, but you asked about magnitudes. There's one more step to get there.

Your sensor detects photons across some band of the spectrum, but the device doesn't respond uniformly to photons of all energies. It's characterized by something called the photon response $R_{PP}(\lambda)$, a function of wavelength. If the object actually emits $\phi$ photons over some span of the time, then the number of photons detected per unit time is $$\Phi(\lambda)=\int R_{PP}(\lambda)\phi(\lambda)\;\mathrm{d}\lambda$$ where we integrate across all wavelengths the CCD is sensitive to (and $\Phi$ is essentially $\dot{N_*}$!). You can turn this into a magnitude via $$m=-2.5\log(\Phi)+C$$ with $C$ a constant.

In practice

You can then use these equations to go between brightness and the required exposure time.

  1. Let's say you know how bright the object is. You can compute the number of photons that should reach the source per unit time (and convert that to a magnitude, if you so desire) and then use $R_PP(\lambda)$ to estimate how many will actually be detected, $\dot{N_*}$.
  2. Use a quantity called the point spread function, or PSF, to estimate how many pixels the object will appear to be smeared out over, which gives you the appropriate $n_{\mathrm{pix}}$.
  3. Take the above and calculate the SNR you can reach with an exposure time of choice, given the background level and the detector's dark current, bias, read noise, etc.
$\endgroup$
11
  • $\begingroup$ Smear caused by the ship's movement is a good point. I'm adding it to the calculator. $\endgroup$
    – Nosajimiki
    Commented May 21 at 16:05
  • $\begingroup$ What SNR would you need to use to detect an enemy ship that could be anywhere in the whole sky, without getting too many false positives? $\endgroup$
    – causative
    Commented May 21 at 22:11
  • $\begingroup$ @causative In the context in which the instrument is used, I think there's not a clear line between what constitutes a detection or not -- just varying degrees of confidence. If you can determine the statistics of the background, you can use Monte Carlo simulations to calculate the false alarm probability; the question is then what's an acceptable false alarm probability. For what it's worth, in my work, $5\sigma$ typically constitutes a possible detection, but I don't do imaging, and certainly not real-time computations. $\endgroup$
    – HDE 226868
    Commented May 22 at 15:27
  • $\begingroup$ @causative To try to solve for this, I've created an function that estimates how many objects there are of a given Apparent Luminosity, and then calculate the minimum apparent luminosity your sensor should be able to detect so that you can see how many objects there are to rule out. I then use a star chart fidelity coefficient that basically lets you ignore all charted celestial bodies so that you only have to track/ID the N brightest uncharted objects that your sensor can pick up. From what I can tell, a good star chart may be more important than a good scanner. $\endgroup$
    – Nosajimiki
    Commented May 24 at 16:01
  • $\begingroup$ @Nosajimiki I think the way to go about it that is less susceptible to errors and criticism is to make a photon-level telescope simulator. You make a solar system object with background light and a number of placed radiation sources. You have a telescope object that can look at a small part of the sky and collect photons into a pixel array over a period of time (using the Poisson distribution and Airy discs). Then you write statistical detection routines to decide whether and where the objects are based on the observed photons. $\endgroup$
    – causative
    Commented May 24 at 16:31
3
$\begingroup$

This is a Frame Challenge

A quick note: a question I asked on Astronomy, which wasn't well received, does make a few good points that are brought up here.

You're detecting light in the infrared spectrum. This obvious statement is necessary because it's important to realize the universe is full of light in the infrared spectrum and differentiating your ship from the quantity of available light is important — and it will have very little to do with the sensitivity of your detectors. The problem is a photon is a photon... you can't tell the difference between the enemy ship and the remnants of a supernova a hundred-thousand light years distant just by the photons alone.

Velocity: The likelihood of space-based inter-ship combat at anything less than breathtaking velocities is pretty much zero. You need to get the ship up to speed just to maneuver around and there's no value to expending energy to lower the velocity. You could hand-wave this, but if you do... what's the point of the question, right?

Processing Speed: This means you need to care about signal processing speeds. As detector sensitivity increases, processing speed decreases. But you also need to care about pattern matching. How do you tell the difference between the minimal splotchy distribution of thermal radiation from your enemy and a suspicious-looking cluster of red/brown dwarfs much, much further away? The whole point of the detection system is that you don't initially know what you're looking at, which means you need to separate your enemy from everything in the background. Occasionally you can do this by determining that the object you're looking at is nearby (light-seconds to light-minutes), but most of the time you'll be pattern-matching in a manner similar to sonar tracking. And you need to do that fast enough to matter.

Combat Tactics: Between knowing which photon matters and figuring that out quickly enough to do something about it I conclude you're approaching this from somewhat the wrong perspective. Combat tactics would include, even at high velocities, the need to take "the high ground" by positioning the ship such that it's difficult to distinguish it from the background infrared light already present. That may be a nearby sun, planet, nebula, or the galactic center. Anything that's emitting heat. Picking where your battle takes place is important as that ship with the right amount of background infrared light will be hardest to find and hit using thermal detection.

enter image description here
Range finder on the HMS Revenge courtesy Naval Gazing. Click to enlarge.

Triangulation: It doesn't matter if you're trying to triangulate a radio broadcast or the visible location of a target — photons are photons and you need two points of reference to know both the bearing and the distance. (You'd actually need three in outer space....)

Unlike other issues I bring up here, triangulation is massively dependent on detector sensitivity (look at the size of the optic ports on that WWII-era range finder...). The further away something is, the more precisely the detection system must discriminate between the two angles of incidence. Said another way, the wider the distance between the two detectors, the greater the detected distance. Of course, space is really, really, really big, which means that detectors separated to the greatest possible distance on a single space ship might not be enough to accurately detect even one light-second of distance. But this can be overcome (to a degree) by the sensitivity of the detectors.

Conclusion

Without the ability to filter out background noise (all that infrared you don't want to detect) that your enemy is intentionally trying to hide within and the ability to triangulate, the issue of sensitivity isn't particularly relevant. Frankly, while individual ships may have range finders for close combat (at hopefully ridiculously slow speeds), it would make more sense to use every ship in the fleet to act as an array for the greatest distance and accuracy possible.

What am I getting at? You don't mention solving these issues, so I wonder if you're looking for the wrong variable. You're trying to fix the detector sensitivity so you can determine range. But for the sake of a story (or the believable infrastructure of a world), you need to fix the range and determine the sensitivity — and then declare your Johannson Thermatic Stellar Combat Range System Mark IV with Pattern Recognition Software Version 121.99.8.45 (released just yesterday and guaranteed bug free!) sensitive enough to meet that need... or not.

In worldbuilding, weaknesses are as important as strengths.

So, I think you're looking for too much detail when you haven't solved bigger problems first. As I said, maybe you have and just didn't mention it. But keep in mind we use antenna arrays because a handful of (*ahem*) low-quality detectors will act like a single, large, high-quality detector. In other words, you don't need to detect all of the photons (not even close) to get an accurate fix on a small object. Your fleet of ships in the correct formation would be capable of detecting a much smaller apparent magnitude than any single ship can.

enter image description here
Image courtesy Cadence Design Systems. Click to enlarge.

$\endgroup$
6
  • 1
    $\begingroup$ I am not trying to use IR to determine range, I'm trying to determine at what range IR can detect a ship. Once the ship is detected, other technologies with a narrower focus should be able to take over the job of precise identification, triangulation, and tracking. $\endgroup$
    – Nosajimiki
    Commented May 21 at 16:11
  • $\begingroup$ That said, I think I will add a variable under the IR scanner specs that lets you set an array size. $\endgroup$
    – Nosajimiki
    Commented May 21 at 16:13
  • $\begingroup$ @Nosajimiki I don't believe your first comment negates my frame challenge, which is offered only to see the bigger picture. I think your basic goal is a great idea. But a realistic determination of how much IR can be detected is almost meaningless if you can't tell which IR is a ship and which isn't. Combat aircraft try to position themselves between the enemy and the sun for a reason - the best stealth (indeed, the very nature of camouflage) is when your enemy can't tell the difference between you and what's behind you. $\endgroup$
    – JBH
    Commented May 21 at 18:50
  • 1
    $\begingroup$ You can tell the diffrence between light from diffrent sources using spectroscopty, just not a single photon. also you should be looking at a map of space not just noticing every star brand new every time you point a detector at it. And not you can't intentionally hide your emisions unless you know exactly where the detector is all the time for a long time in advance AND you can perfectly match the emessions spectrum and brightness of the background thing the entire time you are between it and the detector AND you can somehow only emit for the breifest fraction of a second. $\endgroup$
    – John
    Commented May 21 at 21:48
  • $\begingroup$ I agree those are important. I've already included a feature that lets you define how much of local space is charted, to determine when your apparent magnitude is too low to differentiate a potential target from an overwhelming number of uncharted background objects of similar or greater apparent magnitudes as well as calculations for when speeds become high enough that motion blur and relativistic propulsion systems have a meaningful impact. Right now I have background radiation at a fixed 2.7 K, but I might change that so I can also calculate oddball tactical maneuvers. $\endgroup$
    – Nosajimiki
    Commented May 21 at 22:08
3
$\begingroup$

NETD vs Pixels

I think I worked it out. Thermal sensors basically work like cameras; so, to figure out how much I can see in 360 degrees, I first need to figure out how many pixels it would take to represent my scan area. My total pixels at a range can be derived from the Area of a Sphere equation. So to fit an object that is a size of N into 1 pixel at a range of R, I would need 4π(R/N)² pixels.

The next important consideration is the IR cameras minimum detectable contrast. For older style IR cameras, this was about 2 Kelvins, but newer IR Cameras, it is about 0.025-0.04 Kelvins. Since I would have to fill an entire pixel with the target to detect a minimum contrast, and each pixel is an average of how much light hits it, I can use a brighter point filling only part of a pixel to detect an object; so, if I take the square root of the temperature to background contrast and divide it by my NETD, then I get the width of a pixel a detected ship of a given heat needs to take up. Let's call this √(T/K)... granted this is super idealized since I'd also have to figure out background noise, but I think this makes for a good best case scenario.

So, if I combine these factors I get an equation that tells me how many pixels I need to scan the whole sky and see a ship of a given P = 4π(R/N)²/T*K.

So, to turn this into a detection range equation, I just need to balance it to make R the unknown. R = √(P*T/K/4/π)*N

So, if I have a 5 megapixel IR system with a 0.025 NETD, then I could see a 300K 100x100 m ship from a range of about 6,912km. I'll still need to do a bit more work to get the finer details of my calculator working, but unless I did my math wrong, or if I'm missing any critical variables that significantly change things, I think I've figured out the missing piece.

$\endgroup$
4
  • $\begingroup$ Your camera's minimum detectable contrast won't be a fixed number. It should get better the longer you look, because that allows you to average over a greater number of observations, reducing noise. (Assuming your telescope is designed to do such long-term averaging, which in a space combat scenario it would be.) Also you need to account for the Airy disc. $\endgroup$
    – causative
    Commented May 22 at 0:34
  • $\begingroup$ @causative standalone, it is not fixed, but one of the other parameters I am collecting is the camera's framerate. Unless I am mistaken, your NETD is more or less fixed at any given exposure time. $\endgroup$
    – Nosajimiki
    Commented May 23 at 16:57
  • $\begingroup$ You don't appear to be considering emissivity. A bright, shiny spaceship will be harder to see from its thermal emission than a blackbody at the same temperature. $\endgroup$
    – jeffB
    Commented May 23 at 17:01
  • $\begingroup$ @jeffB Yes and no... if the emissivity keeps it from heating up, then a shiny ship may be harder to detect, but if you already have to heat the ship up to a given temperature to keep the crew alive, or to radiate waste heat, then reflected light will only add to your IR signature... But it is a good idea. I will add reflected light as an input parameter. $\endgroup$
    – Nosajimiki
    Commented May 24 at 15:47
1
$\begingroup$

Rd = ( 17.8E6 * sqrt( MsAsIsp(1-Nd)(1-Ns) ) )

I hate creating what is basically a copy paste answer but the link is buggy and the relevent part is buried, but it covers everything. So I included all the relvent parts because I want the whole thing available. The work has alreay been done by someone else so please don't upvote, I just want the information available. It uses a known detector so it works perfectly as a baseline, use it to check you equation. The whole thread covers things like coasting , burn time and even intensity. The short version is you have to change direction and speed which is a massive IR flare that has to last long enough to be easily detected in a sweep.

Here is the paper he is refrenceing.

Here is the equation given by Dr. John Shilling of Pacific Northwest National Laboratory, Atmospheric Measurement Laboratory.

Rd = 17.8E6 ( MsAsIsp(1-Nd)(1-Ns) )**0.5

Rd = detection range, kilometers

Ms = spacecraft mass, tons

As = spacecraft acceleration, G

Isp = drive specific impulse, seconds

Nd = drive efficiency

Ns = "stealth efficiency", i.e. fraction of waste energy which can be magically shielded from enemy detectors. in the real world this is zero but for a story this can be useful.

Existing astronomical telescopes are not real good at fast scanning, so it's probably best to work out the parameters from scratch rather than trying to extrapolate. Which I have done, using the "Visual and IR payload design" chapter from Space Mission Analysis and Design, by Larson & Wertz, 1993. I'll leave out the math here, and skip to the results.

Telescope - 2 meter aperture, f/2 optics. Dall-Kirkham folded Cassegrain, most likely, but with f/2 it isn't as vital as with some systems.

Detector - Tektronix 2048 x 2048 pixel CCD array, 350-1000 nm response, 80% quantum efficiency, 27 um pixel size.

Angular resolution - 0.0004 degrees (detector-limited)

Field of view - 0.8 degrees

Scan rate - 3600 seconds for full sky assumed, thus 0.7 seconds per FoV

Detection Threshold - 2.5E-17 watts per square meter, at 1:1 Signal to Noise over zodiacal background and 1E-9/pixel false positive rate.

This is optimized for detection of thruster plumes (hence visual/near-IR response) in a distant early warning mode. You'll want something faster for tactical use, but it won't need to be nearly so sensitive - anything that can go from below detection threshold to imminent danger in less than an hour, is going to be putting out plenty of energy. And you'll want a far-IR detector, and a few dedicated target-tracking scopes, as well. I'm only designing one today.

As for detection range, such a system could spot a single Space Shuttle attitude control thruster firing at a range of fifteen million kilometers. Light up the whole package, main engines and SRBs, and the detection range jumps to twenty billion kilometers.

Considering more advanced propulsion systems, a thousand-ton spacecraft accelerating at 0.001g using ion or plasma thrusters should be visible at two hundred million kilometers. We can see the Martian space cruisers as soon as they leave orbit. Replace the cruiser with a ten-ton drone, and magically reduce the signature by an additional 99.9% using some unspecified stealth drive, and the detection range is still close to a million kilometers. Those we pick up during their course-correction burns.

original post

With your engines off the equation is even simpler, its just Rd = 13.4 * sqrt(A) * T2

Rd in km, A is the area of the ship pointed at the detector in meters squared (~23% of a cylinder), and T is in kelvin.

Again this is with current technology, nothing futiruistic.

$\endgroup$
15
  • $\begingroup$ +1 : I've including a simplified variation of this to determine thruster plume. That said, Ns = "stealth efficiency" is not magic. Different fuels have energy outputs in different parts of the EM spectrum; so, if you are specifically trying to hide from IR scanners, some fuels burn almost entirely in the UV spectrum making them very stealthy against IR detection. There are also additives you can put in fuel that can further cause the afterburn to absorb most of its own IR emissions. Between these, Ns > 99.9% is doable using known science. $\endgroup$
    – Nosajimiki
    Commented May 23 at 17:15
  • $\begingroup$ @Nosajimiki I find 99.9% impossible, since it would mean the egine needs ot be nearly that efficenct in not producieng waste heat. I would be interested ot see about the additive IR absorbtion becasue I am betting that only works in an atmosphere where the fuel plume is not expanding at the speed of sound. And as he says even IR detectors don't detect only in the IR and even with a pure coasting target you are still talking about millions of km. $\endgroup$
    – John
    Commented May 23 at 20:31
  • $\begingroup$ @Nosajimiki hiding waste heat is magic, I am curious what fuels do not produce waste heat? $\endgroup$
    – John
    Commented May 23 at 20:46
  • $\begingroup$ The trick is not to avoid radiating heat, it's to hide it in parts of the EM spectrum that are harder to make a sensor for. Making a high resolution UV sensor is a LOT harder than a high resolution IR sensor, but certain rocket fuels burn almost entirely in the UV spectrum; so, even though you are radiating a lot of heat, very little of it shows up on IR. $\endgroup$
    – Nosajimiki
    Commented May 23 at 22:03
  • $\begingroup$ As for the additive thing, there is a substance that when included in the fuel will absorb most of the IR radiation as it tries to leave your plume. It then and holds onto that heat as it slowly releases it as blackbody radiation, but because the exhaust disperses so quickly, and it radiates so slowly, it does not show up well on an IR sensor. $\endgroup$
    – Nosajimiki
    Commented May 23 at 22:03
1
$\begingroup$

Let us start with the measurements of the cosmic background as measured by the ESA Planck spacecraft. This mapped variations in the cosmic background. The CMB has a thermal black body spectrum at a temperature of 2.72548±0.00057 K. This was measured by Planck using a combination of high-resolution instruments with a resolution of about 10 arc minutes. The telescope aperture was 1.5 x 1.9 meters. The detectors were cooled to 0.1 Kelvin.

The classical limit to blackbody radiation is given by the Stefan-Boltzmann law. The total power emitted goes as the 4th power of the temperature. A contribution to the quoted errors of the CMB temperature comes from unknown objects in the field of view. If your craft can get down to these limits, then it would fool the Planck probe.

The CMB was uniform to about 1 part in 100,000. If your craft is one pixel for the highest-resolution Planck instruments - 600 arc-seconds (10 arc-minutes) across and can match the background temperature to this accuracy then it will not stand out.

Suppose your craft subtended A arc-seconds, and was at a temperature T. The extra radiation for the fraction of a 600 arc-second square pixel would have to be less than 1 part in 100000.

$(T^4 - 1)*(A/600)^2 < 10^{-5}$

..or...

$(T^4 - 1)*A^2 < 3.6$

This may do for now, but this is not the limit. The Planck probe could scan the whole sky. If you knew roughly where to look, a telescope with with a higher magnification could spot smaller details if it knew where to look. Large baseline interferometers can have resolutions of 0.001 arc-second. But it would take forever to sweep the sky at these resolutions.

And then they build the Oumuamua 2 spacecraft.

$\endgroup$
4
  • $\begingroup$ The ESA Planck spacecraft took 7 months to scan the sky. I'd think a scan period this long would make identifying and tracking a moving object nearly impossible. $\endgroup$
    – Nosajimiki
    Commented May 29 at 21:21
  • $\begingroup$ @Nosajimiki 7 months is a long time. But if we are trying to detect a new, faint object, we should already have the previous scan of the whole sky as a reference. It is better to pick up a faint object a long way out, then pick up the same object months later with a quicker but less sensitive scan if the scan was looking for incoming asteroids that may need deflection. $\endgroup$ Commented May 30 at 6:59
  • $\begingroup$ There are certainly settings where this may be fine, but there are also a lot of settings where 7 months is WAY more time than it takes for a ship to take off from one planet in a solar system and arrive at another. My point being that I need this to be a variable for my calculator to be useful across settings; so, I can't just assume a sensitivity of ±0.00057 K is realistic if there is a possibility that someone needs to track a relativistic target using the calculator. $\endgroup$
    – Nosajimiki
    Commented May 30 at 17:16
  • $\begingroup$ I don't think there is a formula for what it visible. You can build a bigger telescope. Or have more telescopes to scan faster. The Planck probe provides a good estimate for what can be resolved above background using infra-red because it had to reject all foreground objects. $\endgroup$ Commented Jun 1 at 7:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .