If they tow a big chunk of ice to LEO, shielding from the Sun will be required. My question is, would they have to shield from Earth as well? 

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    $\begingroup$ Ice cannot really melt in LEO, it requires a bit of pressure in order to get liquid. $\endgroup$
    – fraxinus
    Commented Mar 1, 2023 at 10:01
  • $\begingroup$ Adding to @fraxinus comment, ice doesn't melt in space. It instead sublimates (turns directly from solid to gas). $\endgroup$ Commented Mar 1, 2023 at 11:20
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    $\begingroup$ "Boiling is conversion of liquid state into gaseous state by reaching its boiling temp., whereas Sublimation is conversion of solid state into gaseous state without being changed into liquid." - Would you have any boiling going on in comets closing up to the Sun - for another question. $\endgroup$ Commented Mar 1, 2023 at 12:42

1 Answer 1


Would the reflected sun's radiation melt ice in LEO?

This is an elegant question and an interesting challenge because though very simple to ask requires a lot of real, practical spaceflight considerations to answer thoroughly.

But I'm not going to do that. Instead I'll ballpark it in this answer.

tl;dr: on the back of a spherical cow's envelope I get 255 K / 21/4 or -59°C, so no.

Heating from the Sun and from the Earth are both issues for temperature sensitive payloads in LEO. To do the calculations thoroughly one would have to do a full-blown radiative transfer calculation, including detailed radiated spectra from the Sun and Earth, the spectral reflectivity or albedo $a(\lambda)$, emissivity $e(\lambda)$ and transmission $t(\lambda)$ of the ice, geometrical factors like shape and surface roughness, etc.

However we are in luck! The ice is shielded from the Sun where almost all of the radiation is in near-UV-vis-near-IR (hereafter called "visible" to distinguish from thermal IR) where ice can be pretty transparent and so one would have to deal with how dirty the ice was and how full of optical scattering defects it was (like bubbles and cracks)

While the Earth does reflect some visible sunlight on our very weakly absorbing bright white snowball, it's primary heat load will come in the thermal IR, and that makes our task much simpler because ice is opaque and essentially black in this range!

If you could see in thermal IR the surface of ice would look like obsidian or black marble - fairly low real and imaginary indices of refraction, so only weak average Fresnel reflection and sufficiently strong bulk absorption that it can be treated as opaque.

With transmission out of the way, we can assume the emissivity $e(\lambda)$ and reflectivity or albedo $a(\lambda)$ sum to unity.

So our white snowball is actually a black snowball, and the Earth is radiating toasty thermal IR at it from almost a complete hemisphere!

So let's assume the ice is in equilibrium, meaning heat in = heat out.

Let's also assume we're not going to do this the right way using detailed spectral calculations using Earth's mean thermal IR spectrum $I(\lambda)$ and the ice's emissivity $e(\lambda)$ and albedo $a(\lambda)$, but instead use some average values to ballpark this.

Shamelessly googling I find a value of 255 K for Earth's effective radiating temperature (here's one example) and and effective emissivity in the thermal IR of roughly 0.6 to 0.8 though because of the way radiative transport works in Earth's atmosphere it's strongly spectral.

Note: that 255 K is -18°C, already below the freezing point of ice, so we have a pretty big clue where this is going.

Since I need to finish this post and get on with making coffee, let's simplify a bit.

If we assume Earth's emissivity is unity (it's a bit lower, like 0.7-ish), and it radiates towards the ice from one complete hemisphere (it's a bit less, depends on several things) then the equilibrium temperature of an object of unity emissivity (it's a bit lower, like 0.96-ish) will be simply $2^{1/4}$ lower than that of the radiator since it receives incoming radiation from one hemisphere and radiates into two.

That would be 214 K or -59°C.


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