How could I calculate the habitable bounds, in relation to temperature, near the twilight zone of Gliese 667 Cc? Assuming an Earth-like atmosphere and no tidal working.

Here are the numbers that I have been using:

  • Parent star Temperature = $3350\text{ K}$
  • Distance from star = $25.4\times10^6\text{ km}$
  • Radius = $9\,800\text{ km}$
  • Surface area = $1.2\times10^9\text{ km}^2$

I have little experience in astrophysics, or basic physics for that matter, so my understanding of heat dissipation through space is poor at best. In my mind, since I have the temperature of a heat source, a distance to the center of the planet, and the planet's farthest points from the center (25.4 million km ± 9,800 km), it wouldn't be difficult to calculate if you understood radiant flux and such.

  • 2
    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – HDE 226868 Jan 15 '18 at 22:41
  • 1
    $\begingroup$ But here is a site that does the calculating, $\endgroup$ – Justin Thyme Jan 15 '18 at 22:45

Radiant Power Equilibrium

Per Stefan-Boltzman $$ {P \over A} = \sigma T ^4 $$


  • P/A = watts per meter squared of radiated power
  • $\sigma$ is the Stefan-Boltzaman constant, $5.67 \times {10^{-8}} {W m^{-2}} {K^{-4}}$
  • T is temperature (in Kelvin). 3,350 K in your original post

That is the power emitted ($7.1 {MW \over m^2}$) at the surface of the star, which for Gliese 667C is $.42 R_{sun}$, about $2.92 \times 10^8 m$

Area expands as a factor of $r^2$, but at $25.4 \times 10^6 km = 2.54 \times 10^7km = 2.54 \times 10^{10} m$. Dividing the distance by the radius to the surface of Gliese667C(c) gives ${2.54 \times 10^{10}} \over {2.92 \times 10^8}$, or very roughly 100 $R_{surface}$

There is a shortcut that will eliminate our need for calculating that wattage.

Get the cross-sectional area of Gliese 667C(c) $ A_{cross} = \pi r^2 $

Get the surface area of Gliese 667C(c) $ A_{surf} = 4 \pi r^2 $

We don't actually need to calculate either of those values because what's important is $ P_{absorbed} \over P_{radiated} $ which is a function of cross sectional area on the top half, and surface area on the bottom half. So, the equilibrium Wattage is one-quarter the absorbed amount.

Putting it all together : $ {1 \over {4 \dot (100)^2}} \sigma T_{star}^4 = \sigma T_{planet}^4 $. Lose the $ \sigma $ and calculate $ \sqrt[4] {(3,900K) ^4 \over 40,000}$. The result I get is about 295 degrees Kelvin

Gravity and Gasses

The next important thing is to calculate the gravity of Gliese667C(c) and figure out, for the temperature what kinds of gasses the planet can hold on to. Like Mars and Venus, the planet may be in the process of venting away lighter material. The escape velocity is $\sqrt { 2 G {M \over R} },$ where $$\begin{align} \text{G} &= 6.67 \times 10^{-11}\\ \text{M} &= 3.7 M_{earth} (5 \times 10^{24} \text{ kg}) = 1.85 \times 10^{25} \text{ kg}\\ R &= 9800 \text{ km} = 9800000 \text{ m} \end{align}$$.

Escape velocity would be 17 km/s.

Since that is much than the escape velocity for Earth, but your temperature is roughly the same, it stands to reason Gliese 667C(c) has a generous atmosphere of whatever you'd like to imagine it having.


Given a tidal locked planet with a twilight zone, but having a good atmosphere for carrying convection currents. Winds will moderate the temperature by driving up and over the night-side cool air. You need a good carrier of thermal energy (moisture) for convection to work. With good convection, you will get a nearly constant cooling breeze on the ground level from the night side, which might pick up speed as it moves to fill in the higher high pressure regions facing the star.

You can use the pipe equation to estimate how much the global maximum and minimum temperature of Gliese 667C(c) will vary from the mean. ${{\Delta P} \over L} = {128 \over \pi} \dot {{\mu Q} \over D^4}$. Q, the volume flow rate, is ${1 \over 2} \pi D^2 v$; so you can simplify to ${{\Delta P} \over L} = {32 \over \pi} \dot {{\mu v} \over D^2}$. This is the pressure loss over distance. Plugging in the 9,800 km radius of Gliese 667C(c), neglecting the velocity term, and picking an arbitary 2.7 meter height (not completely arbitrary, used to obtain 150C swing seen on Earth), $\Delta P$ = 3,564 Pascals.

Use the Bernoulli equation to convert that to temperature: $ PV = \rho R T $. The gas constant is $8.314 J (mol \dot K)$. The density of air is $1.225 kg / m^3 $. Assume V (volume) is a unit volume (1.0). A maximum pressure loss of 3,564 Pascals is equal to about 250 K.

Given the mean temperature of 236K $\pm $ 125K, the temperature should range no lower than 111K (-162 C/-259 F) at the coolest, and no higher than 361K (88 C/190 F) at the hottest. With a range that wide, steady density is not a valid assumption - reality will be a lower swing, because phase transitions (solid to liquid to gas) will smooth things out.


For the rest, it helps to know what part of Gliese 667C (c) we're talking about. I'll choose to call the point that receives the most direct heating the tropics, and run circles of latitude 171 kilometers per degree from the tropic to the equator/twilight zone, calling the tropics 90 degrees north latitude and the middle of the night zone 90 degrees south.

Effect of Water and Other Greenhouse Gasses

If there is plenty of water, it can moderate the temperature of the planet. According to this article, water on Earth provides 33 degrees Celsius of additional warmth by trapping and transporting thermal energy. At the tropics radiant power is 710 Watts/sq.meter. Around 86% of this (610 W/sq-m) makes it to the surface. Provided enough water vapor in the atmosphere, greenhouse gasses (of which water vapor is 60%) trap nearly 90% of what would be re-radiated into space from the ground.

Per here, the emissivity of air and water vapor mixed is 0.3128. More carbon dioxide (which has a partial emissivity of 0.04) can further decrease the amount of energy being re-radiated into space, nudging all temperatures up.

If you plug the radiant heat equation into the equation for specific heat capacity, as someone else has already done here you can get a more precise cooling model, and wind speeds. Wind speed is important, because it establishes the upper limit for this model (wind speed can not be supersonic $ a = \sqrt(\gamma R T)) $ )

Noting that you can substitute N k in the model with $ \rho R $, the equation becomes $ t = (\rho R_{water} / (2 \epsilon \sigma A)) * (1/T_{final}^3 - 1/T_{start}^3) $

Unfortunately, I could not find a nice model to just determine the change in temperature, so you have to try values. Wind speed is $ \sqrt(2 R \delta T) $, which should be lower than the speed of sound.

Some of the following below is a first-draft pass at the model, which produces a lower-resolution answer than the final answer given.

A 10 km high column of cool water vapor traveling from the equator/twilight zone at 0 degrees north latitude to the tropics at 90N at 10 m/s (36 km/hr), then turning around back to the equator would take 855 hours to make the trip, picking up ~57 degrees C of temperature along the way (simultaneously cooling the ground by that same amount). This global air current would then travel to the southern latitudes past the equator/twilight zone, depositing this ~57 degrees before turning back around.

Similarly, an ocean current in the daylight region (200 m deep) making the same global circuit would warm by two degrees in the warming part of the trip and drop by two degrees in the southern part of the circuit.

But how would currents develop on a planet with no tides? There is more than enough temperature difference for Rayleigh-Benard convection cells to develop. These cells are unstable, so you might have something like weather - more moderate temperatures when convection is strong and more extreme temperatures when convection is weak.

Past the equator, ice may further insulate against heating loss. Above the ice would be extreme, but like the arctic and antarctic, circulating and insulated running water (and underground) would be kept near 0 degrees C.

Habitable Region

At a guess, the habitable region extends from the tropics (90 North / 30 to 80 Celsius) to the equator (0 N/S, 0 to 15 C). The south pole might be accessible in good weather, but would be mostly ice, subsurface water/ground and extreme surface conditions (90 South / 0 (subsurface) to -111 (surface) Celsius). In the complex setting, you'd have both convection driven weather and climate change.


Tropics : 0 to 30 degrees latitude

  • Surface Area : 301718558 square kilometers
  • Ground Temperature : 418-403K (145-130C / 293-267F )
  • Water Temperature : 299-298K (26-25C / 79-77F )
  • Air Temperature : 312-306K (39-33C / 102-92F )

Temperate : 30 to 60 degrees latitude

  • Surface Area : 220873314 square kilometers
  • Ground Temperature : 403-352K (130-79C / 267-174F )
  • Water Temperature : 298-297K (25-24C / 77-75F )
  • Air Temperature : 306-301K (33-28C / 92-83F )

Daylight Equatorial : 60 to 90 degrees latitude

  • Surface Area : 80845244 square kilometers
  • Ground Temperature : 352-6K (79--267C / 174--448F )
  • Water Temperature : 297-296K (24-23C / 75-73F )
  • Air Temperature : 301-296K (28-23C / 83-73F )

Night Equatorial : 90 to 120 degrees latitude

  • Surface Area : 80845244 square kilometers
  • Ground Temperature : 6-5K (-267--268C / -448--450F )
  • Water Temperature : 296-295K (23-22C / 73-71F )
  • Air Temperature : 296-290K (23-17C / 73-63F )

Night Temperature : 120 to 150 degrees latitude

  • Surface Area : 220873314 square kilometers
  • Ground Temperature : 5-5K (-268--268C / -450--451F )
  • Water Temperature : 295-294K (22-21C / 71-69F )
  • Air Temperature : 290-285K (17-12C / 63-53F )

Night Polar : 150 to 180 degrees latitude

  • Surface Area : 301718558 square kilometers
  • Ground Temperature : 5-4K (-268--269C / -451--451F )
  • Water Temperature : 294-292K (21-19C / 69-67F )
  • Air Temperature : 285-280K (12-7C / 53-44F )
  • $\begingroup$ What keeps the air fro freezing out on the cold side? Does the planet radiate enough heat to keep the far side atmosphere gaseous or, at least, liquid? $\endgroup$ – ShadoCat Jan 15 '18 at 23:12
  • $\begingroup$ There is a lot you can do with greenhouse gasses (water and CO2) to tweak the temperature. $\endgroup$ – James McLellan Jan 16 '18 at 2:43
  • 1
    $\begingroup$ @ShadoCat I tried to figure out a few convection examples to give a better idea. In the coldest part of the world, you may have conditions representing Europa (icy shell with a water interior/subsurface). $\endgroup$ – James McLellan Jan 16 '18 at 13:35
  • $\begingroup$ Excellent work, but you could calculate an actual surface area in your final answer to fully answer the question? +1 $\endgroup$ – kingledion Jan 16 '18 at 13:40
  • 1
    $\begingroup$ @Spencer having looked at it more, the way a planet forms may drive you away from the extreme condition of the atmosphere freezing out. The atmosphere starts with a high percentage of low-emissivity CO2. Higher emissivity O2 is formed, to the best of my knowledge, by life which requires moderate temperatures. If a planet just doesn't get enough heat to keep CO2 from sublimating out of the atmosphere, all of the oxygen will be sequestered with it - creating a very thin nitrogen + trace gasses atmosphere - as best I see it now. $\endgroup$ – James McLellan Jan 22 '18 at 14:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.