Mercury rotates sychronized like two-stroke machine. As its solar day is 2 mercurian years.(Orbital periods) Or should we say 3:2-stroke machine, as its Orbital period is 87.969 Earthdays, and its sidereal day is exactly 58.646 Earth days, shortly its orbital period is 3/2 Sidereal days. I think most of readers has fallen out about what's happening; so here's a picture;
It's said that this synchronization is caused by eccentric orbit, as it has the most eccenttric orbit of all planets. But my question is, how does the thermal expansion influence on the gravity center of Mercury, and its orbit? (The G measurement on picture below)
Mercury's surface Temperature varies 600 K, from 100 K to 700 K. It's mainly constructed from magnesian basalt. If we use the Volumetric Thermal expansion coefficient of Basalt; 1.6x10^-5 /K And we assume that the mantle is heated ie. about one third of the circumference, for the easy of calculation we use the value 2 x radius,2 X 2440 km = 4880 km. The peak temperature is also not all over, but as the Poles are already constant 180 K, so with 2/3 from the peak we could calculate with 400 K change inside a circle of 4880 km diameter, and down to depth of 0.5 km; Volumetric thermal expansion is 50 x 2440000^2 x pi x 400 x 0.000016 = 6x10^13 m3; If we consider this volume as a cylinder, it would mean an vertical raise of a 0.32 m. Or if it goes to horizontal push, the diameter of this cylinder would grow 15.6 km. HEY! this is cracy. Such an movement would surely make a hundreds of km long faults. Like Santa Maria Rupes.
We could also calculate the depth how far the temperature causes this expansion. For this we need the thermal concuctivity, for basalt it's in order of 2 W/mK, its Spesific heat is in order of 1 kJ/kgK. this means that the max radiation power of 14500 W/m2 cant be absorbed; with maximum Temperature difference of 600 K, only 1200 W can be transferred to stone through conductivity, and as the distance grows, even less. So it's not Probable that surface would be heated very deep at all, through conductivity. This must lead to an exessive overheating of the surface. And it could be the explanation to the High amounts of water related ions like O+, OH−, and H2O+ surprise.
Typical basalt is constructed as follows (%); SiO2 — 49.97 TiO2 — 1.87 Al2O3 — 15.99 Fe2O3 — 3.85 FeO — 7.24 MnO — 0.20 MgO — 6.84 CaO — 9.62 Na2O — 2.96 K2O — 1.12 P2O5 — 0.35
So it can be seen that all the stuff found from the atmosphere of Mercury, can be resulted by the melted and boiled basalt-stone; which shoots the atoms towards the sun like rocket engine. We May calculate the rough amount of this "rocket engine"; the average solar radiation is 10412 W on Mercury; If 10 % is absorbed, the rest will "burn the rock". Let's calculate with 10 kW/m2, and with silicon. HEat of fusion and Vaporisation 50+383 =433 kJ /mol, Molar 28 g, W = J/s, so it means with 10 kW we are shooting atoms away with a 10/433 = 0.023 mol and 0.023 mol X 28 g/mol = 0.64 g/s per m2 of Mercury. This means that the Mercury's sun side is shooting atoms with an average rate of 12 mio tons in second. If this is true, it would mean that Mercury has burned away in just 865 000 Years.
HEY! This is too crazy, the surface is full of craters! Like this biggest crater of the solar system; Caloris Basin It just happens to be positioned on the hottest spot of Mercury, and its diameter; 1550 km means that the impactor should have been 100-150 km in diameter; and thus have an impact energy of might have had an impact energy of 4.32x 10^26 J; It's 715 times more than the rotational energy of the Mercury; I calculated it to be 6.04x 10^23 J
This doesn't sound plausible. Maybe the 1500 km diameter "crater" is the area where the planet melts under the sun?
This question is a part of my Series of Questions;
And to be honest, I don't even quite know what I should actually ask! Maybe we need to point the hubble to there the take some video, about the possibly boiling surface? -Unluckily the Hubble forgot his sunglasses to home.