# Mercury's rotation, Disturbation's caused by thermodynamics

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;

Uranus, the Physical model to change it's rotation axle

Earth's Kinetic energy change

https://physics.stackexchange.com/questions/216411/how-would-earths-atmosphere-change-in-10-x-faster-rotation-speed

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.

• Hi Jokela, pointing Hubble at Mercury may not be the best idea, kinda hot and bright in those parts, as you point out above. regards
– user81619
Nov 4, 2015 at 15:41
• How does the thermal expansion coefficient (of sodium no less) have anything to do with the temperature profile into the planet? Using the thermal diffusivity of rock, you will find that even for a 600K variation, that doesn't go down very far at all (it doesn't take much crust on a lava tube for you to be able to walk on it!). Any thermal expansion is confined to a thin surface layer. So, yes, it is crazy because it isn't happening the way you think it is. Nov 4, 2015 at 15:52
• My feeling is that seasonal temperature variations don't penetrate very far into the surface of Mercury. Consider the Earth: Seasonal temperature variations don't penetrate very far into our planet's surface. I think that a few feet down into the Earth that the temperature is generally a pretty constant value year-round.
– user93237
Nov 4, 2015 at 15:57
• @count_to_10 Well, then we need to have Solareclipse at the same time when Mercury is at Perihelion. And ofcourse it should be in conjunction either. When might such a be possibilty be? 9.3.2016 -27 days, 2017 -25 d, 2018,2019,2020,,, No, it doesn't fit any time soon. Some other ideas are needed. Hubble doesn't have sunglasses? Nov 4, 2015 at 15:58
• @JonCuster Well, with 400 K difference it actually goes guite quickly and to some deepness. I calculated half year ago, and I can do it again. I would like to use the correct material info. Is the Sodium ok? 142 W/mK? With just 102.5 K difference this material can absorb the whole power of sun inside it. Not to forget that the solar day of Mercury is 2 mercurian years, 176 Earth days.. I'll calculate it again. Nov 4, 2015 at 16:07

Check out this chart as very rough baseline. Solar energy, even with 88 days of Mercury level sunshine, Wouldn't reach nearly as far into Mercury as you suggest. A few KM, perhaps 10 or 20, but not 4,000. I could back that up with a thermal energy calculation of Mercury's mantle and compare it to annual solar energy it gets hit by, but I'm quite sure my point is correct. Solar heating doesn't reach very deep into a planet's solid crust, even over 88 days. Now if it was a circular orbit and tidally locked, that might be a different discussion.

Source

we use the Thermal expansion coefficient of Sodium, 71 µm/(m·K)

and

we need evaluate the crusts chemical properties. 29% is Sodium. But Sodium melts at 371 K! Well, This means that it should be at least flexible for thermal changes.

This needs some fixing. Mercury's crust is mostly Silica or Silicon dioxide mixed with Magnesium Oxide and some other elements. Source so for melting point and thermal expansion coefficient, you need to look at Silicon dioxide or magnesium rich basalt. The sodium content isn't nearly what you say it is, besides, thermal expansion is a property of the molecule, not the elements in the molecule. (if you have a source for 29%, please post)

On the thermal expansion of magnesium rich basalt, I'm not an expert in this subject at all, but it's probably pretty low - see here. Silicon dioxide's is 5.6 x 10-7th, or less than 1/100th the expansion of Sodium you posted. It says that in general, the expansion coefficient for oxides is very low.

So, while I agree with you that there likely is a thermal rise on the side of Mercury that faces the sun, it's probably measurable in meters, not even close to 1 KM much less 138 KM.

There's also likely a tidal bulge on Mercury that follows the sun. I found several references to that but couldn't find any estimates to how high it was. My guess is that the Tidal bulge is bigger than the thermal expansion, but without actual numbers, I'm just guessing.

Maybe we need to point the hubble to there the take some video, about the possibly boiling surface?

Mercury's crust is basically basalt-rock that doesn't melt, even with the sun at high-noon and peak temperatures. Mercury's surface preserves craters very nicely.

Like this biggest crater of the solar system; Caloris Basin It just happens to be positioned on the hottest spot of Mercury, and it's 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

I think your meteor size estimate is off a bit. The Chicxulub crater is some 110 miles in diameter caused by a meteor estimated to be 6 miles in diameter. (18 times the size). and Barringer crater in Arizona is thought to be some 30 to 50 times the size of the impacting meteor.

Velocity is a key factor in object of impact to crater size and with Mercury's proximity to the sun, very high impact velocity certainly possible. Also, I'm not sure what rotational energy of mercury has to do with that. It's entirely possible that Mercury used to have a much greater rotational energy, which slowed over time due to it's proximity to the sun and the solar-tidal effect.

• It's a confusion, if you think that I claimed that the surface heat would reach the depth of 585 km. NO. I just calculated how high "Arch bridge" the thermal expansion would cause. Of course this idea is fully theoretical. I simply wan't to point out the the impact scale of the thermal expansion. Nov 4, 2015 at 16:14
• Yes, with silicon dioxide thermal expansion of 4880000 m and 400 K, is only 1093 m, which would produce only 52 km high Arch bridge in flat surface and simple triconometry. The 29 % sodium-mercury was found in wikipedia. the impactor diameter scale 12 was also chosen from wiki-source. Yes, the velocity makes a difference, but the energy amount should be in scale Nov 4, 2015 at 16:21
• I don't follow your math. 4,880 KM is the diameter of the planet. To calculate any rise on the sun side or fall of the night side, you have to start with the radius, not the diameter and even the radius is a crazy place to begin this calculation cause 88 days isn't enough to warm the planet to it's core. Solar heat won't get nearly that far in that amount of time. I also don't follow how you convert 1093 meters to a 52 KM high arch. Nov 4, 2015 at 16:30
• On the 29% sodium, I see that now, but that's it's atmosphere, not it's surface. space.com/18644-mercury-atmosphere.html Overall I think it's an interesting question but with some poor calculations. Nov 4, 2015 at 16:33
• @JokelaTurbine I did a bit of google searching on atmospheric sodium and I found this and several similar articles. Apparently it's not uncommon in trace amounts. albany.edu/faculty/rgk/atm101/sodium.htm Nov 5, 2015 at 10:27