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