It’s not like it can just evaporate into the atmosphere...
Vapour doesn’t need an atmosphere to take it up. Empty space does just as well.
So, how long would a snowball in high orbit last? Sounds like a question for xkcd. A brief attempt at a lower bound that is probably a substantial underestimate:
How much energy has to be pumped in per kilogram to turn ice at whatever the “temperature” is in orbit into water vapour? Call that E. Let S be the solar insolation of 1.3 kW/m^2. Imagine the ice is a spherical cow, er, a rectangular block directly facing the sun. According to Wikipedia the albedo of sea ice is in the range 0.5 to 0.7. Take that as 0.6, so the fraction of energy retained is A = 0.4. The density of ice is D = 916.7 kg/m^3. Ignore radiative cooling, conduction to the cold side of the iceberg, and time spent in the Earth’s shadow, and assume that the water vapour instantly vanishes. Then the surface will ablate at a rate of SA/ED m/s. Equivalently, ED/86400SA days per metre.
For simplicity I’ll take the ice to be at freezing point. Then:
E = 334 kJ/kg to melt + 420 kJ/kg to reach boiling point + 2260 kJ/kg to boil = 3014 kJ/kg.
For a lower starting temperature, increase E accordingly.
3014 916.7 / (86400 1.3 * 0.4) = 61 days per metre. Not all that long, but meanwhile, you’ve created a hazard for space flight and for the skyhook.
I suspect that ignoring radiative cooling will be the largest source of error here, but this isn’t a black body, so I don’t know how closely the Stefan-Boltzmann law will apply, and I haven’t calculated the results if it did. (ETA: The black body temperature of the Moon is just under freezing.)
(ETA: fixed an error in the calculation of E, whereby I had 4200 instead of 420 kJ/kg to reach boiling point. Also, pasting in all the significant figures from the sources doesn’t mean this is claimed to be anything more than a rough estimate.)
This is vacuum—all liquid water will boil immediately, at zero Celsius. Besides I’m sure there will be some sublimation of ice directly to water vapor.
In fact, looking at water’s phase diagram, in high vacuum liquid water just doesn’t exist so I think ice will simply sublimate without the intermediate liquid stage.
Here is the proper math. This is expressed in terms of ice temperature, though, so we’ll need to figure out how much the solar flux would heat the outer layer of ice first.
Vapour doesn’t need an atmosphere to take it up. Empty space does just as well.
So, how long would a snowball in high orbit last? Sounds like a question for xkcd. A brief attempt at a lower bound that is probably a substantial underestimate:
How much energy has to be pumped in per kilogram to turn ice at whatever the “temperature” is in orbit into water vapour? Call that E. Let S be the solar insolation of 1.3 kW/m^2. Imagine the ice is a spherical cow, er, a rectangular block directly facing the sun. According to Wikipedia the albedo of sea ice is in the range 0.5 to 0.7. Take that as 0.6, so the fraction of energy retained is A = 0.4. The density of ice is D = 916.7 kg/m^3. Ignore radiative cooling, conduction to the cold side of the iceberg, and time spent in the Earth’s shadow, and assume that the water vapour instantly vanishes. Then the surface will ablate at a rate of SA/ED m/s. Equivalently, ED/86400SA days per metre.
For simplicity I’ll take the ice to be at freezing point. Then:
E = 334 kJ/kg to melt + 420 kJ/kg to reach boiling point + 2260 kJ/kg to boil = 3014 kJ/kg.
For a lower starting temperature, increase E accordingly.
3014 916.7 / (86400 1.3 * 0.4) = 61 days per metre. Not all that long, but meanwhile, you’ve created a hazard for space flight and for the skyhook.
I suspect that ignoring radiative cooling will be the largest source of error here, but this isn’t a black body, so I don’t know how closely the Stefan-Boltzmann law will apply, and I haven’t calculated the results if it did. (ETA: The black body temperature of the Moon is just under freezing.)
(ETA: fixed an error in the calculation of E, whereby I had 4200 instead of 420 kJ/kg to reach boiling point. Also, pasting in all the significant figures from the sources doesn’t mean this is claimed to be anything more than a rough estimate.)
This is vacuum—all liquid water will boil immediately, at zero Celsius. Besides I’m sure there will be some sublimation of ice directly to water vapor.
In fact, looking at water’s phase diagram, in high vacuum liquid water just doesn’t exist so I think ice will simply sublimate without the intermediate liquid stage.
Right, I forgot the effect of pressure. So E will be different, perhaps very different. What will it be?
Here is the proper math. This is expressed in terms of ice temperature, though, so we’ll need to figure out how much the solar flux would heat the outer layer of ice first.