Here's my theory :

Since a temperature change is generally smooth when you travel, I was wondering that temperatures would be somewhat decent if you were to stay somewhere in the day/night longitude in Mercury.

I made two little graphics for you to visualize the theory, suppose that it's a 100 km surface seen from above,

Smooth temperature change like on earth :

enter image description here

Radical temperature change:

enter image description here

After all, if you were to zoom in the second picture a lot, you would spot a smooth transition, albeit in a much narrower region.

Does that theory make sense at all?

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    $\begingroup$ This question is hard to answer as it stands. Since Mercury is more or less airless, you have to ask what you are asking for the temperature of? The temperature of surface rocks will probably depend on whether they are in Sun or in shadow (and on how long they have been in Sun or shadow). Maybe you could refine the question to ask how the temperature of typical flat areas of rock in different regions of the planet will vary through the Mercurian day? $\endgroup$ – Steve Linton Jan 7 at 9:22
  • $\begingroup$ Now that I've read your comment I'm having trouble updating my question since you partially answered it! Not really sure about what to ask for now, I mean even though there's no air one would still 'perceive' temperature, no? $\endgroup$ – Aybe Jan 7 at 9:43
  • $\begingroup$ I don't think you really would "perceive" a single temperature. Your spacesuit (if you are not wearing one, then you have bigger problems) will be absorbing heat from sunlight (if it is exposed to it) and from the surrounding rocks by conduction through your boots and radiation, It will be losing heat by radiation to space, and also producing some heat from your body and its systems. That will stabilize at some termperature, but what temperature depends on things like the colour of the suit, the insulation in the boots and whether you are standing on a hill or a valley $\endgroup$ – Steve Linton Jan 7 at 10:05
  • $\begingroup$ A day on Mercury is 175.938 (24 hours) days long, the diameter is 4,879.4 km. You have to move with a speed of 3.63 km/h at the equator of Mercury to stay in a comfortable temperature zone, that seems possible. A very fast transition of surface temperature is not possible, the material needs time to heat up in sunlight. Rise of the Sun over the horizon of Mercury will take very long from the first sunrays to the complete Sun visible, this slows down temperature rise. $\endgroup$ – Uwe Jan 7 at 14:08

We can at least talk about the temperature of the surface of Mercury. There's no atmosphere so we can't talk about the air temperature, and the temperature that an object (space suit, lander) would feel is affected by radiative heating from the hot rocks and any residual sunlight, and by the object radiatively cooling out into space and that requires a detailed model.

It is going to be complicated by the fact that Mercury's surface is quite rough from billions of hears of pummeling by meteors and its own, earlier volcanic activity. That means one side of a crater or a peak will be getting hot in sunlight while the opposite side will still be in darkness and quite cold.

So I think as you zoom in, your diagram will start showing all kinds of hot spots in the cold region, and cold spots in the hot region due to topography. That' region is probably going to be tens of kilometers wide, but it depends a lot on the topographical details of each region.

During the recent total eclipse of the Sun, NASA flew jets with telescopes high above much of the water in the atmosphere in order to make thermal infrared images of the planet Mercury!

From Chasing the Great American 2017 Total Solar Eclipse: Coronal Results from NASA's WB-57F High-Altitude Research Aircraft:

Observing during the eclipse with the WB-57s allowed us to also do some interdisciplinary “bonus” science. Mercury is the closest planet to the Sun and never very far from it in the sky, making it difficult to observe – during the day, the bright sky presents a significant background and equipment must be able to deal with bright sunlight, while observations at twilight (before sunrise, after sunset) are at very low elevation, through multiple airmasses, and thus complicated by significant seeing. At ”thermal” near-infrared (NIR) wavelengths of 3–5 μm, this is a particular problem since atmospheric emission and absorption are that much worse. During the eclipse, the sky is significantly darker (even during partiality) and Mercury is near zenith, providing vastly improved observing conditions.

NIR observations at 3–5 μm probe the temperature of Mercury’s regolith down to depths of a few cm. The cooling timescale of the regolith is poorly known, but is a strong function of the soil composition and density/porosity. MESSENGER X-ray fluoroscopy measurements probed composition only in the top few microns; composition and density below this thin layer are unknown. Measuring the temperature as a function of Mercury local time would reveal the diurnal cooling timescale and, along with modeling, constrain regolith composition and density/porosity to few-cm depths. In turn, this provides insight into how Mercury’s regolith was processed during early planetary formation and bombardment, improving our understanding of rocky planet formation during the early solar system. There are no existing spatially-resolved NIR measurements of Mercury; we attempted to make the first “heat map” of Mercury with the WB-57.

enter image description here

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    $\begingroup$ Temperature can vary widely locally, that's something I would have never thought about! $\endgroup$ – Aybe Jan 9 at 9:50

(this is not an answer; there are no answers)

The temperature change would most likely come from the portion of the sun that's above the horizon. When the Sun is completely below the horizon, the small amount of heat radiated by the ground/rocks should be negligible (there's no atmosphere, so this would just be plain old blackbody type radiation). When the Sun is completely above the horizon, it will be hottest (no atmosphere to refract the Sun). By the Intermediate Value Theorem, there must be a point where the temperature is at a "friendly" temperature, but this assumes that there is a single temperature where a person is standing which is probably not true.

It also assumes the concept of temperature exists in direct sunlight, which may be false or difficult to define. On Earth, we measure shade temperature in the air to avoid measuring temperature where there's direct radiation. Since Mercury has neither shade nor air, this doesn't translate well to Mercury.

  • $\begingroup$ for $T(z, t)$ at the equator of a perfectly smooth ball of basalt, I've gotten as far as $T = T_0 \exp(-\lambda z) \exp(-i\omega t)$ where $\lambda = \sqrt{\frac{\omega}{\kappa}} \frac{1-i}{\sqrt{2}}$ from Eq. 6.11 here but ran out of MathJax steam when I realized I'd have to do the Fourier series of the incident flux $I_0 \text{ceil}(0, \cos(\omega t))$ as well. Since you are handy with math are you interested? $\endgroup$ – uhoh Jan 9 at 8:14

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