How detectable is a 30 kW source?
Hum.
There's a formula for Equilibrium Temperature for Non-Ideal Radiative Heat Transfer for a Sphere in Free Space.
It goes:
T = [ (E • α • AS + PINT) / ( ε • σ • ARAD) ]^1/4
where
E: Irradiance in Watts/m2 from a solar source
α: Absorptivity.
ε: Emissivity.
PINT: Internal heat/power generation in watts.
ARAD: Radiative area in square meters. (Surface area of your ship basically)
AS: Absorption area in square meters. (The area of the side that faces the sun).
σ: Stefan-Boltzmann’s constant, which is: 5.67 x 10^-8 W/m^2 K^4
Some assumptions. This will take place near the center of the asteroid belt; the best place for skullduggery in a solar system.
The amount of solar incidence at a given AU can be discerned through a simple calculation:
E = C • (1/DIST)^2
Where
E: Incidence in W/m2.
DIST: Distance from the Sun in AU's.
C: Solar Constant at 1 AU (1,366 W/m2)
At 2.5 AU; the Solar Incidence is 218.6 W/m2.
We'll assume that your fictional ship is a sphere.
(Spheres make my life easier and are the most efficient in terms of volume/mass/area.)
We'll assume your main hab/weapons/whatnot sphere is 20m in diameter; and it's covered in graphite epoxy or some composite compound like that that is not only dark, but adds to radar stealthiness.
Graphite Epoxy has the following stats for absorption and emmisivity:
α: 0.950
ε: 0.750
Your sphere would have the following specs:
ARAD: 1,256 m2 of surface area that radiates heat
AS: 314.15 m2 of area pointed towards the sun that absorbs heat.
We'll do the calculations for a bunch of power profiles.
30 kW of internal power: 205~K (51-76 times hotter than space)
10 kW of internal power: 193.7~K (48-72 times hotter than space)
6 kW of internal power: 191.1~K (48-71 times hotter than space)
3 kW of internal power: 189.1K (47-70 times hotter than space)
0 kW of internal power: 186.9K (47-69 times hotter than space)
Space is between 2.7 and 4K.
Even flying completely 'dead' or with a super-insulated capsule that soaks up all the heat from your electronics preventing 'leak through', the skin of your spaceship is going to be 47-69 times hotter than space itself due to solar irradance.
Of course, you can get that temperature down through two ways:
- Invest in an active cooling system for the side that's currently pointing towards the threat vector (which will drain your heat sink of it's capacity even faster).
- Paint it in something that has a high α/ε ratio.
If we painted your ship with white paint (α: 0.25 / ε: 0.853); the temperature for 0 kW of PINT would be 129.7K; still some some 32-48 times hotter than space. And by doing this, you've become much more visible to visible light detectors, since you're TEH SHINY.
So how far can we effectively detect an enemy ship?
A good rule of thumb for the resolution of a device is a modified form of the Rayleigh Criterion:
θ = (λ / D)
Where:
θ: Angular resolution of the imaging system in radians.
λ: Wavelength of observed radiation used by the imaging system.
D: Diameter of the lens aperture / Length of the maximum physical separation of devices in an Interferometric Array.
Feeding into that is a modified Angular Diameter Equation:
D = d / [ 2 • TAN(δ/2) ]
Where:
δ: Angular Diameter of object in radians (aka resolution of your imaging system)
d: Diameter of Object.
D: Distance between observer and object.
NOTE: d/D must be expressed in the same unit.
So lets compute two different types of imaging devices.
The first one is a spherical spacecraft 20m in diameter. On each side, we have a truss system that can extend 17 meters; allowing us to create a baseline of 54 meters (54,000,000 μm). We're looking for a target 20m in diameter (the enemy spacecraft); and cranking the formulas results in these maximum ranges:
Visible Light (0.562 μm): 1.9 million kilometers
Near Infrared IR (1.075~ μm): 1 million kilometers
Mid-Wavelength IR (5.5~ μm): 196,300~ kilometers
Long Wavelength IR (11.5~ μm): 93,900~ kilometers
Of course, these ranges are for a 'head on' target that is not generating any appreciable angle-off. If the target is moving at an angle to the sensor; it will be detectable at longer distances, depending on it's own off-angle and speed, by 'smearing' long exposures.
What's truly interesting is if we take two spacecraft and have them flying in "formation" 50 kilometers apart. Their sensors would form a very long interferometric baseline.
At 50,000 meters baseline (50,000,000,000 μm), sensors would have against a 20m target the following ranges:
Visible Light (0.562 μm): 1.78 billion kilometers
Near Infrared IR (1.075~ μm): 930~ million kilometers
Mid-Wavelength IR (5.5~ μm): 181~ million kilometers
Long Wavelength IR (11.5~ μm): 86.9~ million kilometers
If we take your assumption that our weapons can open fire at 2.5 million klicks; and that our ships are moving at a speed of 10 km/sec towards each other for a closing speed of 20 km/sec; even on Long Wavelength IR; we'll detect you 48~ days out with a pair of ships cruising together.
That doesn't even get into the possible absurdities you could create with fairly large sensor arrays at the Earth-Moon L4/L5 lagrange points.
Of course; this is basically imposing very conservative assumptions onto detection technology -- for one, you don't need to resolve something to detect it. However, without resolving it; you will be limited in how much track data you can generate on it.
Amusingly enough; these calculations make small escort ships feasible from the standpoint of acting as a giant sensor for a 'shooter' platform.
A group of six ships would be able to generate a baseline in almost every direction with minimal manouvering needed; and you could have a central ship in the center of the sphere that keeps them stocked up with manouvering fuel and supplies.
You could even have the escorts be completely unmanned with advanced enough technology -- they just dock with the central ship every few months to be checked over in an enclosed hangar, fuelled back up, and then sent out again.
Plus, they become ideal as expendable defense craft -- they can manouver into the way of a missile salvo to help absorb it with their own point defense weapons, or self-detonation.
*give award to the AI of FSC #231231 who sacrificed itself bravely*