# Solar System Snow Line: 5AU or 2.7AU?

I am trying to update the Wikipedia article "Frost Line (astrophysics)".

During my last update (by QuantumShadow), I noticed that different sources cite different values for Solar System water ice snow line, most of them mention to 2.7 AU as the water ice snow line distance.

However on the wiki article about Ceres, there is a remark that "Surface water ice is unstable at distances less than 5 AU from the Sun", and indeed the referenced paper "Water in the Small Bodies of the Solar System" gives 5 AU as the distance below which water ice sublimates.

Encyclopedia Britannica also gives 5 AU as the distance beyond which water ice can form.

So which value is correct, I have an idea that maybe 2.7 AU is the value during formation of Solar System, and the 5 AU is the value now. But I can't find any evidence that support this, and with hard evidence i can't proceed. Thanks, Alex.

## 1 Answer

This is a complicated problem but if we make several assumptions, we can an order of magnitude estimate that should address your question.

# Power Source

First, the sun is the source of power/energy, and we know its luminosity is ~ $3.846 \times 10^{26}$ W. Therefore, the power per unit area at various distances can be determined by dividing this result by $4 \ \pi r^{2}$. One astronomical unit or AU is ~$1.496 \times 10^{11}$ m, therefore we can show that the solar power per unit area is: at ~1 AU ~ 1380 W $m^{-2}$, at ~2.7 AU 190 W $m^{-2}$, and at ~5 AU ~ 55 W $m^{-2}$. So that is our power source.

# Properties of Water

It is difficult to figure out how water acts in the (near) vacuum of space because we cannot recreate vacuums on Earth with pressures anywhere near as low as those found in space. Typical ram/dynamic pressures at ~1 AU are on the order of $10^{-9}$ Pa and thermal pressures are on the order of $10^{-12}$ Pa. So let's ignore this issue for the moment, but keep it in mind.

One can look up the properties of water and find the following:

• Molar mass: ~18.01528 g/mol
• Heat of vaporization at $0^{\circ}$ C: ~$45.054 \times 10^{3}$ J/mol
• Mass density at $0^{\circ}$ C: ~999.8395 kg $m^{-3}$

# Energy Balance

Let us make things simple and assume space is at ~0 K and that our ice is a perfect emitter and absorber of energy (i.e., all energy flux incident on ice is absorbed and it radiations as a perfect black body)

If we consider a sheet of ice ~1 $\mu m$ thick with a surface area of 1 $m^{2}$, then we have ~$9.9984 \times 10^{-4}$ kg of ice or ~0.0555 moles of ice. Therefore, it would take ~2500 J of energy to vaporize all of this ice.

We know that the total power radiated by a perfect black body is given by: $$P_{rad} = A \ \sigma \ \left( \Delta T \right)^{4}$$ where $A$ = area of emitting body, $\sigma$ = Stefan-Boltzmann constant, and $\Delta T$ = difference in temperature (degrees Kelvin) between the radiating source and the absorbing media.

Therefore, our sheet of ice would produce $P_{rad}$ ~ $1.6 \times 10^{-5}$ W. Since we were lazy and chose the ice to have a surface area of 1 $m^{2}$, then we can say that the power absorbed, $P_{abs}$, from solar radiation is: ~1380 W at ~1 AU, ~190 W at ~2.7 AU, and ~55 W at ~5 AU. Then we find energy balance by summing all power sources and sinks (Note: I have ignored heat conduction and other sources/sinks here, but they are generally going to be negligible compared to radiated sources and losses.), or $P_{total}$ = $P_{abs}$ - $P_{rad}$.

Then we find that $P_{total}$ = ~1380 W at ~1 AU, ~190 W at ~2.7 AU, and ~55 W at ~5 AU (since $10^{-5}$ ~ 0 compared to values >10). Therefore, ignoring other sources/sinks, we see that the ice would sublimate/vaporize rather quickly (~1.8 s at ~1 AU, ~13 s at ~2.7 AU, and ~46 s at ~5 AU) if these assumptions held.

# Caveats

• The numbers for the frost line likely take the true absorptivity and emissivity of water ice, which would change $P_{abs}$ significantly. We know that ice reflects and/or transmits a lot of the visible spectrum (i.e., it looks clear or white, depending on conditions), which is the main frequency range of solar radiation. Therefore, $P_{abs}$ should be lower than the values I estimated above (which assumed 100% absorption). The emissivity of water ice would decrease $P_{rad}$, but whether the absorptivity decreases $P_{abs}$ enough to make $P_{rad}$ relevant, I am not sure.
• I ignored ice losses due to ablation from impacting particles, whether charged or neutral. In general this should be small compared to other losses at small distances from the sun, but may become dominant beyond Jupiter as the solar flux drops.
• I ignored ice losses due to ionization as well, but again, the ionization rate should be low except for regions close to the sun. By ionization, I mean that a water molecule become ionized and be ejected from the ice solid before undergoing the phase transition to a gaseous state.
• I am sure I have ignored other issues as well, but my answer is meant as a hand waving, order of magnitude calculation to illustrate how these numbers might come about.