"I figure disunirregardless of the low temperatures involved, in 80,000 years it will have evaporated down to a loose assortment of rocks and dust loosely held together only by their own small mutual gravity"
"Virtually zero" loss is not zero loss.
"..average -454 F temperature" is not zero temperature. Remember, temperature is a measurement of the velocity of impacting particles being transferred to the thermometer or object.
There are particles running around loose in "space." Some are bound to impinge on your temperature measuring device, and some are bound to impinge on the comet --imparting velocity to the molecules of the comet or your measuring device. This may be too small an amount to observe, but it's there nonetheless.
Remember the definition of "absolute zero."
Some of those impinging particles may "stick" because of the comet's own gravity, thus tending to
increase its mass again, but my bet is the overall effect of the temperature/velocity of the impact will outweigh this effect. Besides maybe knocking off a couple of other particles to boot.
Your (plural you) assertions are commonly held and sort of pretend the comet will be "lonely."
My assertion is basically theoretical but not commonly held and presumes the comet will not be "lonely."
That's the way I figured it, per quote.
Call me in 80,000 years and let me know whose assertions took precedence in this long-term empirical test of alternatives. My phone number then will be (303) 874
20-939i67399478. Leave a messaage.
Terry, 230RN