<p>There is no rotation speed which can achieve that - globally. For a local analysis see below. The global equilibrium surface temperature in the absent of any atmospheric greenhouse effect and with an albedo of 0 is around 450K, thus 180°C:</p>
<p>Distance from the Sun: <span class="math-container">$d_{merc} = 0.38$</span> AU</p>
<p>Solar constant at earth radius: <span class="math-container">$S_{earth} = 1360$</span> W/m²</p>
<p>Thus solar constant at mercury distance: <span class="math-container">$S_{merc} = S_{earth}/d_{merc}^2 = 9081$</span> W/m²</p>
<p>Now we can calculate the input power, assuming mercury absorbs with its area cross section:</p>
<p><span class="math-container">$r_{merc} = 2439700$</span>m</p>
<p><span class="math-container">$E_{in} = S_{merc} \cdot \pi \cdot r_{merc}^2$</span></p>
<p>The heat is re-radiated into space via the complete surface <span class="math-container">$A_{merc} = 4\pi r^2$</span>. So we get the equilibrium equation</p>
<p><span class="math-container">$$E_{in} = A_{merc} \cdot \sigma T^4$$</span></p>
<p>solving this for <span class="math-container">$T$</span> we get:</p>
<p><span class="math-container">$$T = \left(\frac{E_{in}}{\sigma A_{merc}}\right)^{1/4}$$</span>
<span class="math-container">$$T = \left(\frac{S_{merc} \cdot \pi r_{merc}^2}{\sigma\cdot 4\pi r_{merc}^2}\right)^{1/4}$$</span>
<span class="math-container">$$T = \left(\frac{S_{merc}}{4\sigma}\right)^{1/4} = 447\mathrm{K}$$</span></p>
<p>Solving for the distance at which a given temperature is fulfilled (again without any greenhouse effects):</p>
<p><span class="math-container">$$ d = \sqrt{\frac{\pi r_{planet}^2 S_{earth}}{4\pi r_{planet}^2\sigma T^4}} = \sqrt{\frac{S_{earth}}{4\sigma T^4}}$$</span>
With T = 20°C = 291K we get d = 0.91 AU, so even outside Venus' orbit.</p>
<p>Assuming a habitable zone between 0°C and 100°C (liquid water), we get a range between 0.56 AU and 1.05 AU.</p>
<p>In reality one will have to take into account two opposing effects which change surface temperature quite a bit: albedo decreases it as it scatters away radiation without it raising the temperature, and atmospheric greenhouse effect which effectively captures thermal radiation, raising the surface temperature.</p>
<p>In essence: earth is quite in a fortunate position, in a delicate equilibrium.</p>
<p>Now, if we are not interested in the planet as a whole, but want to ask the question &quot;is there any zone or place on the planet where the average temperature is 20°C?&quot; we have to look at the lattituda variation. We know that from Earth: The sun is much higher in the sky in the tropics and only shallowly over the horizon polewards of the polar circles.</p>
<p>There is an analysis of the <a href="http://luna1.diviner.ucla.edu/%7Ejpierre/mercury/posters/Poster-04/poster-04.html" rel="noreferrer">surface temperature of Mercury by David Paile from UCLA</a>. It shows the yearly (and also diurnal) variation in surface temperature (solid line in the plot. The other broken lines are for different depths). There is a large amplitude at the equator, and a much more shallow one near the poles. The diurnal temperature variation on Mercury were plotted by <a href="https://reader.elsevier.com/reader/sd/pii/S0019103599961754" rel="noreferrer">VASAVADA, PAIGE, AND WOOD (1999)</a>: &quot;Near-Surface Temperatures on Mercury and the Moon and the Stability of Polar Ice Deposits&quot;. The solid curve is the maximum surface temperature near perihel as a function of lattitude. Near aphel the maximum surface temperature near the equator is about 120°C less, but the maximum temperature near the pole does not change.</p>
<p><a href="https://i.sstatic.net/nrUtv.png" rel="noreferrer"><img src="https://i.sstatic.net/nrUtv.png" alt="Diurnal temperature variation on Mercury. VASAVADA, PAIGE, AND WOOD (1999): Near-Surface Temperatures on Mercury and the Moon and the Stability of Polar Ice Deposits" /></a></p>
<p>The variation is given in the same paper in figure 3 for the equator (left) and 85°N (right). From the graph one sees that the night time temperature is the same, but the maximum quite different with 700K and ~380K respectively. The central line in each plot gives the mean temperature (note that at 85°N it is at ~220K, thus -50°C):
<a href="https://i.sstatic.net/i9vgE.png" rel="noreferrer"><img src="https://i.sstatic.net/i9vgE.png" alt="Diurnal variation at 0°N and 85°N on Mercury" /></a></p>
<p>Now, how exactly is the diurnal temperature variation? That is also given, in figure 2 for 90°W longitude. (The small wiggles are the secondary sun rise and set respectively occuring there):</p>
<p><a href="https://i.sstatic.net/iwP85.png" rel="noreferrer"><img src="https://i.sstatic.net/iwP85.png" alt="Diurnal temperature variation at the equator of Merury" /></a></p>
<p>But lets use this model to estimate a time constant for cooling or heating. We see, especially on these wiggles, that the temperature rise and drops are very quickly (due to the absence of the atmosphere). So to smooth that out, you would need a day-night cycle which is considerably shorter than one hour. Adding an atmosphere and especially water (be that liquid or as vapor) as heat buffer would allow that to become longer.</p>
<p>There is one point which could spoil this: the seasons. However Mercury's rotation axis is very well aligned with its orbital rotation axis (about 2° - compare with Earth's 23°) so that the influence of the obliquity of the ecliptic on Mercury is very low - hower that will still add some minimum to the temperature variation - irrespective how fast you choose Mercury to rotate to even out the daily temperature variation.</p>
<p>In summary: there are two lattitudes (one Northern, one Southern) where the average is temperature is 20°C: The polar mean temperature is below 20°C, the equatorial one is WAY above 20°C, so there must be  a (small) region in lattitude with a 20°C avarage. Thus with some inevitable daily variation an average of 20°C can be achieved for a certain lattitude. If you want the variation to be such that each temperature is bearable, then the day-night cycle would have to last quite a lot less than an hour.</p>

CC BY-SA 4.0
2021-12-10T14:03:52.073Z
Even worse, so to speak:  I doubt humans could tolerate the irradiance levels for more than a fraction of a second during "daytime" .
Carl Witthoft
CC BY-SA 4.0
2021-12-10T14:06:51.927Z
That surely depends on the atmospheric conditions. Were it not for the temperature, the direct irradiation on Venus' surface would not be harmful. Without atmosphere exposure to the unshielded Sun in an Earth orbit is very quickly harmful. Just an increase to 6 solar constants itself is not what hurts - it's the short wavelength contribution.
planetmaker
CC BY-SA 4.0
2021-12-10T14:26:12.247Z
This seems to be not right in a few senses.  If the temperature extremes range from "way below freezing" to "way above boiling", then there must be places which achieve "just right".  The issue then becomes how stable these zones are: how stable are the "just right" zones (do they wander, do they spike out of the zone?), and how large is a "just right" zone (if it's 3 inches wide that's no good, but a few miles might work)? If Mercury were tidally locked we should have fairly stable temperatures, not accounting for solar activity spikes; but it's not, though.
zibadawa timmy
CC BY-SA 4.0
2021-12-10T15:44:19.770Z
Does this calculation apply only to the equator? The rotation-averaged power reaching a unit area on the sphere will decrease as cos(latitude). Unlike Earth with atmospheric transport and mixing, won't the high latitudes be *substantially* cooler?
uhoh
CC BY-SA 4.0
2021-12-10T21:29:13.920Z
@zibadawatimmy - You could still have intermediate zones which are "just right" only on average, while in reality they fluctuate between untolerable  extremes all the time.
Vilx-
CC BY-SA 4.0
2021-12-10T23:22:45.143Z
And they would do so
planetmaker
CC BY-SA 4.0
2021-12-11T11:47:22.230Z
Thanks for the edit, *looks great!* `+1`
uhoh
CC BY-SA 4.0
2021-12-11T18:47:57.287Z
While the average 20C zone will experience major temperature fluctuations you can dig in and get year-round comfort.  I'd actually go closer to the poles because of the thermal load of the people and equipment.
Loren Pechtel
CC BY-SA 4.0
2021-12-10T11:57:57.307Z
2021-12-11T10:39:42.227Z
2021-12-11T10:39:42.227Z
planetmaker
<p>I wondered what rotational speed would Mercury need to have to achieve a temperature comfortable to humans, let's say 20 °C.</p>
<p><strong>EDIT:</strong>
My idea was that if Mercury is so cold on the night side and so hot in the daylight then if it rotates fast enough maybe it can reach normal temperatures. But now I understand why this cannot be possible.</p>

CC BY-SA 4.0
2021-12-10T21:53:11.710Z
A rotational period equal to its orbital period, ie : tidally locked.  Somewhere around the permanent terminator it will be a comfortable temperature.  An atmosphere is out of the question.
J...
CC BY-SA 4.0
2021-12-10T22:55:13.123Z
@J... no, not really. The terminator is much less permanent than one would think due to the orbit's ellipticity. See the daily temperature variation I added to my answer.
planetmaker
CC BY-SA 4.0
2021-12-11T01:01:43.240Z
@planetmaker It moves slowly.  Habitats on wheels? ;)
J...
CC BY-SA 4.0
2021-12-11T02:43:03.583Z
See: https://www.drroyspencer.com/2016/09/the-faster-a-planet-rotates-the-warmer-its-average-temperature/
Nilay Ghosh
CC BY-SA 4.0
2021-12-11T13:45:40.873Z
@J... [There's a sci-fi novel with exactly that principle.](https://en.wikipedia.org/wiki/2312_(novel))
zibadawa timmy
<rotation><mercury>
What rotational speed would Mercury need to have to achieve a temperature comfortable for humans
Piotr Golacki
# What rotational speed would Mercury need to have to achieve a temperature comfortable for humans

(Asked by: username_0 on 2021-12-10)

I wondered what rotational speed would Mercury need to have to achieve a temperature comfortable to humans, let's say 20 °C.

**EDIT:**
My idea was that if Mercury is so cold on the night side and so hot in the daylight then if it rotates fast enough maybe it can reach normal temperatures. But now I understand why this cannot be possible.

**Tags:** rotation, mercury

> **Comments:**
> **username_1**: A rotational period equal to its orbital period, ie : tidally locked.  Somewhere around the permanent terminator it will be a comfortable temperature.  An atmosphere is out of the question.  
> **username_2**: @username_1 no, not really. The terminator is much less permanent than one would think due to the orbit's ellipticity. See the daily temperature variation I added to my answer.  
> **username_1**: @username_2 It moves slowly.  Habitats on wheels? ;)  
> **username_3**: See: https://www.drroyspencer.com/2016/09/the-faster-a-planet-rotates-the-warmer-its-average-temperature/  
> **username_4**: @username_1 [There's a sci-fi novel with exactly that principle.](https://en.wikipedia.org/wiki/2312_(novel))  

---

## 1 answers

### Answer by username_2 on 2021-12-10. Score: 22 | Accepted answer

There is no rotation speed which can achieve that - globally. For a local analysis see below. The global equilibrium surface temperature in the absent of any atmospheric greenhouse effect and with an albedo of 0 is around 450K, thus 180°C:

Distance from the Sun: $d\_{merc} = 0.38$ AU

Solar constant at earth radius: $S\_{earth} = 1360$ W/m²

Thus solar constant at mercury distance: $S\_{merc} = S\_{earth}/d\_{merc}^2 = 9081$ W/m²

Now we can calculate the input power, assuming mercury absorbs with its area cross section:

$r\_{merc} = 2439700$m

$E\_{in} = S\_{merc} \cdot \pi \cdot r\_{merc}^2$

The heat is re-radiated into space via the complete surface $A\_{merc} = 4\pi r^2$. So we get the equilibrium equation

$$E\_{in} = A\_{merc} \cdot \sigma T^4$$

solving this for $T$ we get:

$$T = \left(\frac{E\_{in}}{\sigma A\_{merc}}\right)^{1/4}$$
$$T = \left(\frac{S\_{merc} \cdot \pi r\_{merc}^2}{\sigma\cdot 4\pi r\_{merc}^2}\right)^{1/4}$$
$$T = \left(\frac{S\_{merc}}{4\sigma}\right)^{1/4} = 447\mathrm{K}$$

Solving for the distance at which a given temperature is fulfilled (again without any greenhouse effects):

$$ d = \sqrt{\frac{\pi r\_{planet}^2 S\_{earth}}{4\pi r\_{planet}^2\sigma T^4}} = \sqrt{\frac{S\_{earth}}{4\sigma T^4}}$$
With T = 20°C = 291K we get d = 0.91 AU, so even outside Venus' orbit.

Assuming a habitable zone between 0°C and 100°C (liquid water), we get a range between 0.56 AU and 1.05 AU.

In reality one will have to take into account two opposing effects which change surface temperature quite a bit: albedo decreases it as it scatters away radiation without it raising the temperature, and atmospheric greenhouse effect which effectively captures thermal radiation, raising the surface temperature.

In essence: earth is quite in a fortunate position, in a delicate equilibrium.

Now, if we are not interested in the planet as a whole, but want to ask the question "is there any zone or place on the planet where the average temperature is 20°C?" we have to look at the lattituda variation. We know that from Earth: The sun is much higher in the sky in the tropics and only shallowly over the horizon polewards of the polar circles.

There is an analysis of the [surface temperature of Mercury by David Paile from UCLA](http://luna1.diviner.ucla.edu/%7Ejpierre/mercury/posters/Poster-04/poster-04.html). It shows the yearly (and also diurnal) variation in surface temperature (solid line in the plot. The other broken lines are for different depths). There is a large amplitude at the equator, and a much more shallow one near the poles. The diurnal temperature variation on Mercury were plotted by [VASAVADA, PAIGE, AND WOOD (1999)](https://reader.elsevier.com/reader/sd/pii/S0019103599961754): "Near-Surface Temperatures on Mercury and the Moon and the Stability of Polar Ice Deposits". The solid curve is the maximum surface temperature near perihel as a function of lattitude. Near aphel the maximum surface temperature near the equator is about 120°C less, but the maximum temperature near the pole does not change.

[![Diurnal temperature variation on Mercury. VASAVADA, PAIGE, AND WOOD (1999): Near-Surface Temperatures on Mercury and the Moon and the Stability of Polar Ice Deposits](https://i.sstatic.net/nrUtv.png)](https://i.sstatic.net/nrUtv.png)

The variation is given in the same paper in figure 3 for the equator (left) and 85°N (right). From the graph one sees that the night time temperature is the same, but the maximum quite different with 700K and ~380K respectively. The central line in each plot gives the mean temperature (note that at 85°N it is at ~220K, thus -50°C):
[![Diurnal variation at 0°N and 85°N on Mercury](https://i.sstatic.net/i9vgE.png)](https://i.sstatic.net/i9vgE.png)

Now, how exactly is the diurnal temperature variation? That is also given, in figure 2 for 90°W longitude. (The small wiggles are the secondary sun rise and set respectively occuring there):

[![Diurnal temperature variation at the equator of Merury](https://i.sstatic.net/iwP85.png)](https://i.sstatic.net/iwP85.png)

But lets use this model to estimate a time constant for cooling or heating. We see, especially on these wiggles, that the temperature rise and drops are very quickly (due to the absence of the atmosphere). So to smooth that out, you would need a day-night cycle which is considerably shorter than one hour. Adding an atmosphere and especially water (be that liquid or as vapor) as heat buffer would allow that to become longer.

There is one point which could spoil this: the seasons. However Mercury's rotation axis is very well aligned with its orbital rotation axis (about 2° - compare with Earth's 23°) so that the influence of the obliquity of the ecliptic on Mercury is very low - hower that will still add some minimum to the temperature variation - irrespective how fast you choose Mercury to rotate to even out the daily temperature variation.

In summary: there are two lattitudes (one Northern, one Southern) where the average is temperature is 20°C: The polar mean temperature is below 20°C, the equatorial one is WAY above 20°C, so there must be a (small) region in lattitude with a 20°C avarage. Thus with some inevitable daily variation an average of 20°C can be achieved for a certain lattitude. If you want the variation to be such that each temperature is bearable, then the day-night cycle would have to last quite a lot less than an hour.

> **Comments:**
> **username_5**: Even worse, so to speak:  I doubt humans could tolerate the irradiance levels for more than a fraction of a second during "daytime" .  
> **username_2**: That surely depends on the atmospheric conditions. Were it not for the temperature, the direct irradiation on Venus' surface would not be harmful. Without atmosphere exposure to the unshielded Sun in an Earth orbit is very quickly harmful. Just an increase to 6 solar constants itself is not what hurts - it's the short wavelength contribution.  
> **username_4**: This seems to be not right in a few senses.  If the temperature extremes range from "way below freezing" to "way above boiling", then there must be places which achieve "just right".  The issue then becomes how stable these zones are: how stable are the "just right" zones (do they wander, do they spike out of the zone?), and how large is a "just right" zone (if it's 3 inches wide that's no good, but a few miles might work)? If Mercury were tidally locked we should have fairly stable temperatures, not accounting for solar activity spikes; but it's not, though.  
> **username_6**: Does this calculation apply only to the equator? The rotation-averaged power reaching a unit area on the sphere will decrease as cos(latitude). Unlike Earth with atmospheric transport and mixing, won't the high latitudes be *substantially* cooler?  
> **username_7**: @zibadawatimmy - You could still have intermediate zones which are "just right" only on average, while in reality they fluctuate between untolerable  extremes all the time.  
> **username_2**: And they would do so  
> **username_6**: Thanks for the edit, *looks great!* `+1`  
> **username_8**: While the average 20C zone will experience major temperature fluctuations you can dig in and get year-round comfort.  I'd actually go closer to the poles because of the thermal load of the people and equipment.  

---

