How deep would the frozen atmosphere be?

Question

I have a rough planet with the following specs:

Data	Value
Mean Radius:	4886.4 km (almost 50% larger than mars)
Surface Area:	3.00046x10^8 km^2 (0.58 Earths)
Volume:	4.88715x10^11 km^3 (0.45 Earths)
Mass:	1.73142x10^24 kg (0.29 Earths)
Mean Density:	3.543 g/cm^3
Surface Gravity:	4.83671 m/s^2 (0.49g)

When it was ejected from it's star, it would have had an atmosphere comparable to the current atmosphere of Earth, including the same amount and percentages of gas and water vapor.

I need to calculate how deep that atmosphere would go, if all but the Helium and Hydrogen froze or liquified. (Assume it is cold enough for lakes of liquid oxygen)

Answer 1

In a rogue planet with no external source of heat, the surface of the planet might be expected to have a fairly uniform temperature.

The total mass of Earth's atmosphere is about 5.15×10¹⁸kg. It is made up of approximately

• 78.09% nitrogen
• 20.95% oxygen
• 0.93% argon
• 0.04% carbon dioxide

There are also small amounts of other boring things which aren't really enough to be interesting, so I'll ignore them. They'll make a small addition to the total depth, but their contribution is minimal.

There is also, as you pointed out, water vapour...1.27x10¹⁶ kg of the stuff which is generally excluded from the "dry mass" of the atmosphere.

• If you want liquid oxygen, that temperature will need to be under 90.188K but over 54.36K. This will produce ~1.08x10¹⁸kg of liquid, which at a density of 1141 kg/m³ will produce a volume of 9.46x10¹⁴ cubic metres.

• If you want pretty much everything else to freeze out, you'll need the temperature to be under the freezing point of nitrogen, 63.23K. This will produce ~4.02x10¹⁸kg of nitrogen ice. Its density is a little hard to pin down, but given wikipedia's formula of 0.0134T2 − 0.6981T + 1038.1 kg/m³ it is ~1047kg/m³ at 63K, just below its freezing point. That gives you ~3.84x10¹⁵ cubic metres of ice.

  Note that nitrogen ice is less dense than liquid oxygen... you'll have no surface lakes or oceans of oxygen. You might get transient rains and rivers around sites of active volcanism or meteorite impacts so there may be some surface liquids, but they'll be rare. Once formed they'll persist until disturbed by a fresh eruption or impact, though. Earthquakes may also fracture the nitrogen ice allowing surface oxygen to drain away.

• Argon freezes at 83.81K. Its density as a solid is around 1616 kg/m³, and so will produce ~2.96x10¹⁴m³ of ice.

• CO₂ will obviously be frozen. At a density of ~1700kg/m³ it will produce ~1.21x10¹²m^{^3}.

• The water vapour will freeze first of all, and with a density of 918.7kg/m³ it will have a volume of ~1.38x10¹³m3.

Lets assume the world is a perfect sphere. Lets also assume that each layer of frozen (or liquified) atmosphere forms a perfect spherical shell. The thickness of each successive layer can be found as $$R = \sqrt[3]{\frac{3V}{4\pi} + r^3}$$ where $R$ is the radius from the centre of the planet to the outside of the shell, $r$ is the radius of the planet to the inside of the shell and $V$ is the volume of the shell. The thickness of the shell is therefore $R - r$

The initial value of $r$ will be the radius of your planet (4886.4 km) but each successive layer will increase it ever so slightly.

1. The water vapour will freeze first, with a 4.6cm layer of ice.
2. The CO₂ will freeze after that, forming a ~0.4cm thick layer.
3. The argon will form a ~9.9cm thick layer next.
4. The oxygen will form a ~3.15m thick liquid layer.
5. The nitrogen will form a ~12.8m thick ice layer on the top.

The total thickness of the condensed and frozen atmosphere will be ~16.1 m