I was commenting on Talmand's scenario; take "three of the world's largest ships, put them together to form one ship and then placed it in a random spot somewhere within the orbit of Pluto; how long would it take you to find it?"
Using the formula in the "Refrigeration" section of the page linked to by Avshalom, the "maximum range a ship running silent with engines shut down" (assuming diameter = 500 m and temperature = 290K is about 500 million km, or about 3.3 AU. That's well within the orbital radius of Jupiter.
That's assuming the entireasteroid is at 290K. Pluto averages about 40 AU out, given a 0.056% chance that a room-temperature asteroid would be detectable.
I'm also uncertain about the reasoning in the page linked to by Avshalom. I think that assumes a 3K background, and I think you also think there's a big heat differential. However, the average temperature of the moon is something like -25C, and if an asteroid were closer to the sun, then the difference in surface temperature between an internally heated vs. sun heated asteroid is well within the errors in measuring the physical properties of the asteroid.
166C at Mercury orbit: http://www.google.com/search?&q=(((1-0.1)+*+(3.827E26+watts)+%2F+(0.9+*+Stefan-Boltzmann+constant+*+16+*+pi+*+(0.3+*+150000000+km)**2)))+**+(1%2F4)
90C at Venus orbit: http://www.google.com/search?q=(((1-0.1)+*+(3.827E26+watts)+%2F+(0.9+*+Stefan-Boltzmann+constant+*+16+*+pi+*+(0.6+*+150000000+km)**2)))+**+(1%2F4)
5C at Earth orbit: http://www.google.com/search?q=(((1-0.1)+*+(3.827E26+watts)+%2F+(0.9+*+Stefan-Boltzmann+constant+*+16+*+pi+*+(1.0+*+150000000+km)**2)))+**+(1%2F4)
(The values for albedo and emissivity are not well known, so there is a wide error range in this calculation.)
Still, that's enough to show that the heat differentials inside of the Earth's orbit are not that big.
You are using averages where the min and max are extremely wide apart. For example, the Moon is not actually at -25C, it is either +130C or -110C (or transiting between). We know the time when those temps occur, so we can look for other objects not at those temps at that time. Same goes for the other objects.
I am definitely using averages because if you have an ship which is 300+ meters across then you can leave the top few meters for insulation, so the surface of your ship has the expected extreme temperature ranges.
I should restate what I said: instead of "that's assuming the entire asteroid is at 290K", I mean, "except the insulation layer", which you need anyway to keep the near-surface area livable.
If you need 20 meters for that, then your living volume is 21 million cubic meters, or a reduction of about 20%.
That's not to say that there's no contribution. More that I believe the calculations from that linked-to page ignore the difficulties of differentiating between the internal heat (cooled through black-body radiation) from surface heat from the sun.
For example, from Stefan-Boltman, cooling goes as T * * 4, which means if you have a cooling radiator on the sun-side which is at 150C, or 20C above "normal", then it's about 6 times more effective than a -90C radiator on the cold side. I suspect it's easier to stop the 20C delta on the cold side than the warm side.
Using the formula in the "Refrigeration" section of the page linked to by Avshalom, the "maximum range a ship running silent with engines shut down" (assuming diameter = 500 m and temperature = 290K is about 500 million km, or about 3.3 AU. That's well within the orbital radius of Jupiter.
That's assuming the entire asteroid is at 290K. Pluto averages about 40 AU out, given a 0.056% chance that a room-temperature asteroid would be detectable.
I'm also uncertain about the reasoning in the page linked to by Avshalom. I think that assumes a 3K background, and I think you also think there's a big heat differential. However, the average temperature of the moon is something like -25C, and if an asteroid were closer to the sun, then the difference in surface temperature between an internally heated vs. sun heated asteroid is well within the errors in measuring the physical properties of the asteroid.
You can even work it out for yourself, using the equation at http://en.wikipedia.org/wiki/Standard_asteroid_physical_char... . I worked it out to:
(The values for albedo and emissivity are not well known, so there is a wide error range in this calculation.)Still, that's enough to show that the heat differentials inside of the Earth's orbit are not that big.