# Where on a tidally locked planet with a 25 °C maximum is the 0 °C isotherm?

I am writing a novel. In that novel, humans have settled a tidally locked planet. That planet is positioned at just such a distance from its sun that the sun heats the planet to an average of 25 °C at the substellar point, that is, at the center of the side facing the sun. The planet is like our Earth in size, composition, and atmosphere. As that planet is tidally locked, the permanently dark side is much colder than the side that is in perpetual daylight, perhaps even freezing, with the coldest point on the opposite side of the substellar point.

How far around the planet from the substellar point is the 0 °C isotherm?

Looking at average temperatures on Earth, either for a specific month or the whole year, the distance between a place with 25 °C and a place with 0 °C always seems to be somewhere around 40° to 50° (e.g. from Mexico to Northern Canada in July). So I would suspect the 0 °C isotherm to be about 45° from the substellar point.

• It depends on the efficincy of heat transport. – JDługosz Oct 9 '16 at 18:53
• Heat transfer would be nothing like Earth, because you wouldn't have ocean currents on most of the planet. You need to stay above 0 on most of the globe for that, and that's contradicts main idea of your question. – Mołot Oct 10 '16 at 8:03
• Interesting question, it feels like one we should be able to answer here but not one that anyone can answer without doing some work on it :) – Tim B Oct 12 '16 at 12:18
• Note that England and Labrador are at the same lattitude. The land and sea pattern and large scale circulation patterns will mix this up so much that you cannot expect a single lattitude isoterm, but a complex pattern of livible regions. – JDługosz Oct 12 '16 at 14:52
• While not exactly an answer to this exact question, I think this link might be relevant enough for someone trying to figure this out for their world. – Tezra Oct 13 '16 at 17:36

This is just a basic answer, but I used Samuel et al. (2014), who in turn cited the climate model of Léger et al. (2011) for a tidally-locked planet with no atmosphere. They give the formula for surface temperature as $$T_s=\left(\frac{\epsilon_5}{\epsilon_2}\right)^{\frac{1}{4}}\left(\frac{R_*}{a}\right)^{\frac{1}{2}}T_*\cos^{\frac{1}{4}}\theta\tag{1}$$ Substituting in $T_s=298$ (in Kelvin) and $\theta=0$ means that $$\left(\frac{\epsilon_5}{\epsilon_2}\right)^{\frac{1}{4}}\left(\frac{R_*}{a}\right)^{\frac{1}{2}}T_*=298\text{ K}$$ Therefore, we get an easy formula for $\theta$: $$\theta=\arccos\left[\left(\frac{T_s}{298\text{ K}}\right)^4\right]\tag{2}$$ If we set $T_s=273$, we then get $\theta=45.25°$, which happens to be almost exactly what you predicted.

A very similar derivation of $(1)$ equation can be found here, although that $\theta$ is latitude, not the same as our $\theta$, and so $\sin\theta$ should be replaced by $\cos\theta$, leading to the above formula. A related equation is given such that the change in temperature from the greenhouse effects can be added in. I think we can do the same thing here: $$T_s=\left(\frac{\epsilon_5}{\epsilon_2}\right)^{\frac{1}{4}}\left(\frac{R_*}{a}\right)^{\frac{1}{2}}T_*\cos^{\frac{1}{4}}\theta+\Delta T_s\tag{3}$$ where we can compute $\Delta T_s$ assuming that it comes from radiative forcing: $$\Delta T_s=\lambda\Delta F$$ given the climate sensitivity $\lambda$ and radiative forcing $\Delta F$. I wrote about this in a lot more depth in this answer. To accurately calculate both $\lambda$ and $\Delta F$, we need to know a lot more about the planet, its atmosphere, and its crustal composition. Drawing from my previous answer, if we knew the concentrations of various greenhouse gases, we could use a table from the IPCC Second Assessment Report to calculate the relevant $\Delta F$s. However, that would require a lot of guesswork on my part.

Now, perhaps we could neglect the greenhouse effect entirely. Yang et al. (2013) write

The greenhouse eﬀect for tidally locked planets is much smaller than those for non-tidally locked planets (Fig.2d). This results from a low-level temperature inversion on the nightside of tidally-locked planets (see also Joshi et al. 1997; Leconte et al. 2013). The inversion is due to eﬃcient radiative cooling by the surface on the nightside and strong atmospheric energy transport from the dayside to the nightside (Merlis & Schneider 2010). The outgoing infrared radiation to space is therefore similar to the near-surface upward infrared radiation, resulting in a small $G$.

Here, $G$ is a parameter - a temperature difference - between the upward infrared fluxes at the planet's surface and at the top of its atmosphere; it's essentially $\Delta T_s$. The difference in how important $G$ is is shown in the author's Fig. 2:

The 1:1 data points are for a tidally locked planet with clouds; the 2:1 and 6:1 points are for slightly faster rotators and the other two tracks are variations on the tidally locked model. See how the track without clouds has an extremely large greenhouse effect.

Perhaps, then, at most stellar fluxes, we can neglect the greenhouse effect on our tidally locked planet.

Something assumptions I'm still making:

• There is little to no atmospheric circulation.
• The surface is approximately homogeneous.

Take this answer (v1) with a grain of salt. It's only a really, really basic approximation.

• Since emissivity from a planet's surface (epsilon in Samuel's equations) follows the Stefan-Boltzmann law, the factor of 1/4 is correct and 45 degrees vice 23 should be the estimate for an atmosphere-less planet. – kingledion Oct 12 '16 at 15:33
• I see convincing arguments for the exponent 1/4 to the cosine, my initial model was too naive. – jknappen Oct 12 '16 at 16:02
• @kingledion That's what I thought, coming from the planetary effective temperature formula; I've removed that caveat. – HDE 226868 Oct 12 '16 at 16:15
• In those charts you posted, does surface temperature (top left) account for greenhouse effect (bottom right chart) or not? – kingledion Oct 14 '16 at 14:36
• @kingledion I believe it does not, no. – HDE 226868 Oct 14 '16 at 14:40

As of right now, this is a partial solution

I set up a solution using a non-linear, non-separable first order differential equation. I ran out of brainpower midway through this post, and can't solve that right now, but I will revisit it later. If someone else can solve first, please edit this post. I wanted to post what I had, in case anyone can find errors now, and to re-type my notes more legibly.

# Principles

Assume that heated air rises at the sub-stellar point and proceeds at high altitude equivalent to Earth's stratosphere towards the poles, where it enters the 'dark side', a cool, infinite heat sink and heat source. Cool air from the 'dark side' returns in the surface layer equivalent to our troposphere. It moves towards the sub-stellar spot, heating up along the way as it absorbs solar radiation.

We will calculate the energy balance of this surface layer of the atmosphere to calculate a delta-T as a function of angular distance from the sub-stellar spot. The angular distance where delta-T = -25 is the answer to the question.

# Data and simplifying assumptions

Data on radiation and atmopsheric circulation taken from here. This reference will be referred to as Fig. X of data.

No land-sea interface; that is, the planet is either all land or all ocean. No heat transport from ocean circulation.

Insolation at the sub-stellar point is equivalent to earth's equator.

Heated air rises at sub-stellar and travels towards the dark side in the upper atmosphere. Cold air returns from the dark side in the lower atmosphere.

The 'dark side' is an infinite, constant low temperature heat source for returning winds.

Ignore effects of expansion and contraction of air when approaching and departing sub-stellar point.

We will calculate atmospheric delta Ts from radiation balance in two bands, an upper atmosphere band, and lower atmosphere band.

Atmosphere is mass $5.15\times10^{18}\text{kg}$, surface area of earth is $5.10\times10^{14} \text{m}^2$ for air column mass of $1.01\times10^{4} \frac{\text{kg}}{\text{m}^2}$.

Lower layer is equivalent to troposphere, everything from sea level to 10 miles, with 75% of the atmosphere's mass, with air column density 7.5 $\frac{\text{kg}}{\text{m}^2}$, Upper layer is rest of atmosphere, air column density 2.5 $\frac{\text{kg}}{\text{m}^2}$.

From figure 1.3 of data, 0.67 of energy absorption is from surface, 0.33 by atmosphere, of which 0.25 is lower layer (by above definition) and 0.08 by upper layer. We will simplify that all absorption is in lower layer.

Assume solar energy absorption and radiation to space are the dominant means of energy transfer. Ignore heat transfer between upper and lower layers.

Assume isometric heating and cooling, specific heat of air is constant 718 $\frac{\text{J}}{\text{kg}\cdot\text{K}}$.

# Calculating Energy Balance for the Lower Atmosphere

Simplifying Fig 1.7 of data, Incoming radiation is 300 W/m^2*month at the 0 from substellar, 0 at 90 from substellar, and linear in between.

Temperature change due to energy input is: $$E_{in} = \frac{300 - \frac{10}{3}\gamma\,\frac{\text{W}}{\text{m}^2}}{7.5\frac{\text{kg}}{\text{m}^2}\cdot718\frac{\text{J}}{\text{kg}\cdot\text{K}}} = 0.0557 - 0.000619\gamma\, \frac{\text{K}}{\text{s}}\quad\quad 0^\circ < \gamma < 90^\circ$$ where $\gamma$ is the angle from the sub-stellar point

Simplifying Fig 1.7 of data, Outgoing radiation is 250 W/m^2*month at 0 from substellar, since temperatures are similar to earth's at the equator. By Stefan Boltzmann law, radiative heat transfer is proportional to the fourth power of Temperature. Temperatures must be calculated in Kelvin.

Temperature change due to energy output is: $$E_{out} = \frac{250 \left(\frac{\text{T}}{298}\right)^4\,\frac{\text{W}}{\text{m}^2}}{7.5\frac{\text{kg}}{\text{m}^2}\cdot718\frac{\text{J}}{\text{kg}\cdot\text{K}}} = 5.9\times10^{-12}T^4 \, \frac{\text{K}}{\text{s}}$$

where $T$ is temperature in Kelvin.

# Wind Speed and Pressure assumptions

Changes in temperature drive changes in pressure which drive changes in wind speed which drive the temperature gradient. To avoid solving three simultaneous differential equations, ignore the effect of pressure changes on wind speed.

We will model the expected strong, constant winds at 10 m/s and 20 m/s.

A degree of latitude is 111 km. The ratio between distance and angle is $9.0\times10^{-6}\frac{\text{degrees}}{\text{m}}$. A translation of the above wind speeds to angular velocities is $9.0\times10^{-5}\frac{\text{degrees}}{\text{s}}$ and $1.8\times10^{-4}\frac{\text{degrees}}{\text{s}}$.

# Differential Equation and Solution(?)

Overall energy balance for the atmopspheric circulation is, after converting angle to time using the 10 m/s wind speed: $$\frac{dT}{dt} = 0.0557 - 5.6\times10^{-8}t - 5.9\times10^{-12}T^4$$.

## EDIT:

After attempting to solve numerically using an Euler method, I discovered that this does not work. My problem is ignoring the potential energy imparted to air molecules to raise them from the lower atmosphere to the upper atmosphere. This takes something like 5e8 W at the 10 m/s flow I calculated and needs to be accounted for. Still working.

• Great start. If you retry the numerical solution, by the way, you definitely don't want to use an ordinary Euler method for an energy conservation problem because it violates conservation laws. A symplectic method is preferable - try something like Verlet or leapfrog integration. – HDE 226868 Oct 13 '16 at 16:28
• I was using implict reverse-Euler, which I didn't think had the conservation problem, but I might be wrong. In any case, this method isn't really working, because the potential energy term can't be well defined in terms of 'seconds' as I formulated the problem. So I'm trying a numerical solution with the planet divided into latitude bands exchanging energy with each other (by wind flow) and absorbing/radiating from space. Unfortunately this is starting to move from 'fun math problem online' to 'distracting me from the job for which I am actually paid'. – kingledion Oct 13 '16 at 17:40

I'll try a simple answer. I didn't see the call for hard-science facts, so i thought i can give it a try. I didn't do a master in physics, and i do neither have the time nor the possibility to program a weather simulation that would most certainly be required to give you a correct answer. Let me say this right away: this answer is not really hard-science, but i tried my best.

1) How is temperature distributed on a tidal-locked planet? Do we have real life examples? 2) How does having an atmosphere influence temperature distribution? 3) Which temperatures are required for a planet to carry life? 4) Where is our isotherm point?

## How is temperature distributed on a tidal-locked planet? Do we have real life examples?

Let's start by looking into planet that is close to ours: mercury. Mercury has almost no atmosphere, and the temperatures there range from 100K to 700K, while the poles are constantly <180K. Mercury has almost no axial tilt, and it is actually almost tidally locked. So Mercury gives us a good representation of how the temperature would distribute on our planet without an atmosphere. Mercury is smaller than earth, but since you haven't specified a size for your planet, we will ignore size completely.

How would your planet behave if it had no atmosphere? We could just put in your maximum temperature of 295K as a max, and downscale everything in relation to that. We would have a maximum of (logically) 295K with a minimum of 42K, and the poles would be at roughly <74K. Brrr.. that's fresh. Given that pluto has a mean temperature of 44K, a temperature where most gases just instantly freeze, our planet would be really really cold on the dark side.

## How does having an atmosphere influence temperature distribution?

Atmospheres are useful. They help spreading the overall temperature more evenly, or even raise it. Venus has an average (!) surface temperature of 735K, which is equal to the max temperature on Mercury. Which is muuuch closer to the sun. Damn greenhouse effects. So the atmosphere, which our planet requires, since humans settled on it, will change the overall temperature distribution, as well as max or min temperatures. Since our max temperature was probably taking atmosphere into account already, we need to raise our minimum temperature. But.. to which level? And how would the temperature distribution change?

I googled a bit and found this document: Document

The document i linked suggests that if the temperature drops too low, the gases in the air freeze, leaving a vaccum. The low preassure causes the air from the warm side to expand, spread to the cold side, and freeze, too. So the planet wouldn't have any atmosphere. To avoid this, we require strong winds that help spreading the temperature so much, that the gases on the cold side can't freeze.

But there are no formulas available to me to estimate how EXACTLY the temperature would behave. That depends on the strenghts of the winds, geological and climating conditions, the exact distance from the sun, gravity, magnetic fields, the evolution of the planet etc.. But i know that the temperatures need to be high enough so our gases don't freeze, even on the dark side.

## Which temperatures are required for a planet to carry life?

Our atmopshere is primarily composed of Oxygen, Nitrogen and Carbon dioxide. The freezing point of these are:

Oxygen freezes at: 54 K

Nitrogen freezes at: 63 K

Carbon dioxyde freezes at: 195 K

All of these should constantly exist as a gas, otherwise we might suffer from atmosphere drain (or life would change very VERY drastically), especially considering that Co2 is a primary catalyst of greenhousing and thus: warmth. We need that stuff in our atmosphere, if we want it to be able to keep our temperatures more civil.

Okay, so 195K seems like a good temperature to set as a minimum. Let's notch it up to 200K, just to be sure, and to allow fantastic "carbon dioxide rain" on especially cold nights on the dark side....just to be sure, 200K is still -73C, and thus VERY cold. This is the temperature we require at least on our dark side. It might be higher on your planet, thus changing all the results of my calculations. I dare say: your question doesn't have a definitive solution.

## Where is our isotherm point?

I think it is safe to say we now have our three points of temperature we require for our calculations:

Max: 295K (as defined by OP)

Minimum: 200K (as calculated above)

Temperature at poles: <212K (as calculated from situation on mercury)

0C is 273K, just by the way.

Now, for the calculations. If the temperature was distributed "evenly", it would drop from 295(substellar point) to 212 (poles) evenly. It probably doesn't, but as i said, the strength of the winds, the unknown geology and other factors make it impossible for me to give a precise estimation. If the substellar point is at latitude 0, and the poles are at latitude 90, and we have above temperatures, our point of 0 degrees Celsius would be at a latitude of:

23,86 degrees....

It depends strongly on the direction of the winds, though. If I assume the winds move around the planet in one direction, the 0-point will be much closer to the substellar point in one direction, and much further out in the other.

• I have a few questions about this, but first, how did you arrive at your final answer of 23.86 degrees? – HDE 226868 Oct 12 '16 at 13:24
• Is this a satire of a science-based answer? – kingledion Oct 12 '16 at 13:31
• This has to be satire.... – Stuart Allan Oct 12 '16 at 13:48
• While the text is almost incomprehensible, the actual answer is correct for a planet with no heat transport at all. The calculation is actually $\alpha = \arccos\frac{273+25}{273}$ – jknappen Oct 12 '16 at 14:08
• @AndreasHeese Actually, while this does not answer the question from the science base required by the question, it nevertheless provides some additional and important detail. Also, I like the way you approached the problem, so I'm upvoting it. What makes me wonder, though, is that temperatures on Earth drop as low as -90 °C, without us loosing our atmosphere (en.wikipedia.org/wiki/Lowest_temperature_recorded_on_Earth). – user8976 Oct 13 '16 at 7:36