OTHER | MOUNTAIN HEIGHTS

TO THE TOP OF THE WORLD

When worldbuilding we are lead to add a couple of big mountains on the map and just call it a day - but, how tall should they be? *cue music*

That's a rather complicated question, to put it on simple terms, it mostly depends on the compressive strength and density of the material used on the mountain, and planet gravity.

The amount of weight above the mountain base shouldn't exceed the compressive strength of the material, ie, stress - or else the entire thing will collapse back to stability.

Let's assume our mountain is made of granite, which have a density of ~3g/cm³, and a compressive strength of 200MPa.

Let's make our mountain a cone of base h and radius r - it's volume will be given by:

Since (1/3)*π approaches ~1 (1,03669), we can get rid of that part for simplicity sake, leaving us with r²h.

The weight of our mountain can then be calculated by multiplying it's density, planet gravity and volume altogether.

Now, our mountain is a cone, which means that it has a round base, to get how much force it exerts on the base, we need to know the base area, which is a circle of area = πr².

So the little count below should give us the stress exerted by our mountain.

Switching to solve for hmax, we get:

Solving for our granite mountain, we get just over 6,79~7,84km, within the density range of pure granite.

If something seems wrong to you, knowing that the Everest is 8,8km high, you're not alone - first of all, this model assumes uniform density, which let's be realist - don't really exist in most natural formations, because the Earth's crust exist in layers, which are twisted by geological processes over the eons.

Another very important factor people seem to glance over is the stable shape of our mountain, we can't just throw any r and h in our equation too.

For any material that's being piled, there's a certain angle base-to-top up to which the pile is stable - the Repose Angle - which is defined by the specific material's static friction and grain-size.

The repose angle of a pile is an important concept in Civil Engineering because it helps saving material and funds when building earthen structures, bases, and grain silos. The smaller the repose angle the better the flow properties of the material used - by opposition, the higher the repose angle the worse the flow properties, and thus, more rigid is our material.

Okay, knowing the angle of repose of our mountain helps us to better understand it's dimensions.

There are two ways to easily get to those values:

    1. Figure out a way to find the exact materials in question, mix them, and then pour over a plane to a determined height or base width, and take the measurements yourself.

    2. Take the data from someone that has gone through method 1.

On relatively small piles, we can assume dry grain size to be the most important factor, such as:

Although mountains are made of grains and crystallized material, it's much more cohesive than a pile of sand or muscovite clay.

Also, the grains within rock are compressed by their neighbors along the very vast majority of the rocky structure. So rather than considering grain size, we will be using static friction coefficients, and determine fault planes along which will form our mountain's slope - ie, an inclined plane problem.

Through a series of transformations, we get to this:

The angle (in degrees, not radians) at which a body/grain will roll downhill is equal to the inverse tangent of the static friction coefficient.

The problem here arises from the fact that it varies a lot from the materials used - you see, a piece of glass will slide more against concrete than a piece of rubber.

Concrete-to-Rock friction coefficients, depending whether it is wet or dry rock and what rock was tested varies from 0,50 to 0,70, with Concrete-to-Concrete friction having an average of 0,53.

With this in hand, our mountains can have slopes between 26,56º and  34,99º.

Now, we get to the though parts.

We know the angle of our slope, let's pick 32,47º, and determine that the base of our mountain is 16km wide, the crest of our mountain is at half that distance so 8km - to figure out the height, we just have to multiply the tangent of our angle by the base length (8km).

We get just about 5km, which, thrown into our stress equation, becomes 150MPa - which is little bellow our granite compressive strength limit, and so, just about okay.

I can't resist but to estimate things on Paart, so using the same constraints, this mountain could be 373m taller before having the same pressure at the base, implying a slope angle of 34,36º.

Calculating for a mountain that's just over the strength limit but with the apparent density of the Everest, we get about 9,5km tall, with a base width varying between 19,0 ~ 13,6km in radius depending on whether it's slope is less or more steep - compared to Everest's approximate 11,4km.

A simple image showing the proportions of the two mountains, the one on Paart has been temporarily named to Caelum (sky) mons.

(relative to immediate surrounding terrain, measured from Everest Camp I at 6km to summit gives 3,7km, following the steepness to sealevel takes that to 11,4km).

Of course, mountains aren't just giant rocky cones that elevate above plains of terrain, mountains like the Everest originate from mountain ranges full of convoluted systems of canyons sculpted by the drainage of snow over the ages.

surrounding terrain to Everest in the Himalayas

Be sure to give your world's top a couple of sister mountains that branch away into more and more along the entire mountain range.

Below a table of reference materials that you could use to make a mountain out of.

Theoretically, you could define the different layers of material in your planet's crust where the mountain forms, and then, calculate mountains within mountains for every material and thus determine what it's made of (relative to a cut at sealevel). But that's up to you now.

- M.O. Valent, 04/06/2021

BIOLOGY | PART 8 | PLANT COLOR & SUNLIGHT - PART 3

PAART'S AUTUMNAL COLORS

VOTING RESULTS

Winning 2 to 1, for dominant plant color of Paart is...... Pastel Red, well, sorta, theory is beautiful, about it's feasibility - it gets complicated.

The starting conditions (early Cambrian) of Paartian soil is a pH 4,79 (wet soil), or similar to that of coffee left over to sit for 24h - places where the rain is minimum, ie, further inland of Sthalika, the acidity could be as low as pH 6,69 - equivalent to that of milk or Earth normal rainwater.

One way to do make reddish plants would be to use carotenoids such as β-carotene, licopen or lutein, or even use the low soil pH and give them red anthocyanin.

 

VIABILITY

On Earth, green algae wouldn't appear until ~1Gya, and red algae been around for at least 1,5Gya, however carotenoid biosynthesis wouldn't appear until ~580Mya (split between cilliates and diatoms), red anthocyanins until 450Mya.

So far, seems inevitable we would first have blue-green algae - but, while on Earth in it's analogue stages to where Paart is now, the oxygen levels were rising just above 10%, on Paart the oxygen levels have been decreasing from 14% to 10% in the last 310My, which means that the ozone layer of Paart during this period would have been in such high levels it would take another 200My for Earth to reach into the Devonian period.

One great limiting factor for life on land have been the amount of UV radiation incidence, with a thick ozone layer by 450~400Mya, mosses and land plants could finally arise make their way into the barren continents.

SOLUTIONS

Paart has already a great advantage compared to Earth, being so much far away from a low-mass star with a not so intense UV output, and receiving relatively little light - a thick ozone layer seems like the cherry in the cake, the bright lamp in a room full of moths - begging for an early plant-life explosion, short, but early explosion, at least, enough algae biodiversity that could have armed paartene life with the right chemicals for anthocyanin production.

Anthocyanin acts as plant sunscreen, regulating UV stress, the declining levels of oxygen after this small algal bloom would select ones that could produce anthocyanin more efficiently to deal with increasing UV radiation - mosses and terrestrial algae could have appeared as soon as 3950PMA.

MEET, THE CHRYSOPHYLL

Apart from early Earth plant-life, which would have a hard time trying to find magnesium in such acidic soils for the making of chlorophyll b, but would have no problem in finding iron, cobalt, copper or zinc - curiously enough, zinc and copper porphyrins are more efficient than conventional magnesium ones used by plants.

Zinc Protoporphyrin (ZPP) can be found in the bloodstream when by some reason the production of heme is inhibited (lead poisoning) or the body has a lack of iron - plants also could use ZPP as an anti-cancer agent in their body structure to deal with increased UV radiation.

Zinc Porphyrin 8 has a power conversion efficiency of 7,13%, with a major absorption line at ~425nm, 560nm and 610nm - magnesium porphyrin efficiency is around 5,73% - in practice Chlorophyll's efficiency is about 4%, I couldn't find papers on nature's use of Zn(II)P8, but on the same scale of fallibility, about ~4,98% efficiency, or +24,4% as efficient - such an efficiency boost would yield the same as using conventional chlorophyll a under 85% Earth's sunlight - instead of Paart's dull 68% Earth sunlight incidence.

On the other hand, by adding zinc acetate to porphyrin a in acidic medium, you are able to make zinc-chlorophyll, at room temperature (about 25°C), said to have similar potential as magnesium-chlorophyll - similar processes can be made to tweak bacteriochlorophyll to work with a zinc center metal.

Zn-Chl-1 has a strong absorption at 720nm, and slightly less at 420nm, mostly reflecting yellow-green light, but making far better use of these wavelengths than chlorophyll a (noticeable when comparing quantum efficiency yield), having an efficiency of 11,44% - turned down by the same previous fallibility value, maybe around ~8,0% efficiency, or +40% as efficient.

With relatively acid oceans (pH 6,6~7,0) Paart's zinc is mostly abundant in the form of ionic Zn+2 (~60%), and then in the form of citrates, phosphates and sulfates - in that scenario, despite being 40x less abundant than magnesium, it could have proportioned a great advantage over chlorophyll users - one example of this rare but efficient is vitamin B12, using cobalt which is 3x rarer than zinc in Earth's crust and is essential to DNA protection.

I will dub this form of photosynthetic pigment, which reflects yellow and red, Chrysophyll (Chy), or "golden-leaf" pigment.

When land plants appear, we would have a variety of yellow-orange mosses depending on their content of anthocyanin, purple and blue plants may appear the further away from acidic soil, and green plants may be a rarity, or, reduced to low-lying plants such as grassland and shrubland even in equatorial areas, whereas reddish plants occupy the dense forest spaces.

THE COLORS WE COULD HAVE

Honestly, I have to admit that I made the palettes before exploring the actual photosynthetic chemistry that's viable on the planet, so my original options doesn't include any of that material.

Potential 6 flavors of pigment plus different anthocyanin levels

I made an additional table for those extra 2 options that seem to be more viable than conventional chlorophyll.

If we assume every scenario is equally likely, red will appear 28,6% of the time, green 24%, yellow 19,5%, blue and purple 8,6% and 7,5%, respectively - the average of the values is a yellow-brown color (HEX 876150).]

I mapped out the distribution of acidity in soil colored by anthocyanin response to it, for the early Cambrian maps of Paart.

Regions A, B and C to track the mean plate movement of such regions

 

By the Paleocambrian period, 62,97% of the land area is neutral, 13% of the land have a pH around 6, 9,12% around pH 5, and 10,4% with pH of about 4,5.

In turn what this all mean, is that about 1/3rd of the land favors red/purple pigments, while the rest favors blue plants - what would not have happened if Paart's continents where smaller and tropical, what would have avoided a large neutral area to be protected by the polar cap.

 

OZONE LAYER AND ANTHOCYANIN RESPONSE

2,8% of the volar spectrum is UV radiation, on Earth the UV part of the solar spectrum is about 3,2%. Paart has proportionally half as much oxygen than Earth, but actually 2,52x more of the gas, being of very similar size to Earth, we can ignore distribution differences for now - Earth has an ozone layer that's ≤10ppm in ozone (surface levels are 0,3ppm).

Ozone is created when UV dissociates an Oxygen molecule, the subsequent reaction creates Ozone which absorbs UV light and dissociate as result.

Using the graphs 1 and 2, we are assuming a healthy ozone layer as ~300 Dobson Units thick.

Paart receives only 87,5% as much UV as Earth, and does have lower oxygen proportions than Earth's atmosphere, the total Ozone that would be naturally occurring is about 4,76ppm, or a layer that's about ~124,9DU thick (similar to ozone layer thickness above Antarctica in 2008), the global UV radiation levels would be ~1,2x that of UV with 300DU - weighted UVI for an Earth clear-sky is about 10,6, while a Paart clear-sky UVI is 11,1. With wet soil reflecting about 20% of UV light, the effective UVI for someone on the surface of the planet on a clear day is about 13,32, and 10,6 when it's cloudy.

Anthocyanin in plants increase with UV intensity, this may yield in some minor size/mass-losses in comparison with plants under lower UV levels (at least, with Earth-plants).

Exposure to UVI for UVB of 12,0 showed to increase Anthocyanin levels ~4x in certain plants, and exposures mixed with blue-light and UVB mixed (UVI ≤1,8), increases of 6~7x - gene expression for mixed UVB/Blue light are 400~5000x stronger. Wit,h our 330mW/m² UV incidence, we should expect the anthocyanin gene expressions to be at least 100~500x stronger with UVB/Blue light of 475 mW/m², in what case including blue light the exposure index is about 19,0.

Assume the "normal" anthocyanin expression is about 32ug/g of the fresh-weight (100th of the levels in peppers), if our expressions are around 200~250x times that, we should expect plants with 8mg/g of fresh-weight, or 1,3x more anthocyanin than in some grape peels, with the maximum expressions about 16mg/g, nearly as much as in black grapes (20mg/g).

Referent to color expressions, while my initial guesses where based on the argument that soil pH would directly influence plant anthocyanin color expression - it seems that the presence of aluminum and iron in the soil helps to stabilize the color expression, aluminum blue-shifts the anthocyanin expression while iron does red-shift it. Anthocyanin has also shown to turn yellow-ish when the plant pH is alkaline.

On Paart aluminum and iron are in identical levels and relatively low abundance when compared to Earth. However, iron is more resilient to acidity changes, inside the paartene soil pH range - the aluminum should be between 2,5ppm and non existent, whereas there could be about 75~100mg/Kg of iron in the soil depending if the soil is more acidic or near-neutral, respectively.

So, actually, the more neutral the soil gets - the purpler the plant, in some cases where the soil happens to be basic, there would be a variety of, yellow/green/blue plants depending on how basic and on iron deposit levels, whereas on neutral and acid soils, plants would be magenta towards the red, respectively.

 

THE TRUE COLORS OF PAART

Finally, after all this research, we get a color field that looks like this:

 Excel color map for reference

 

And then we get, 44% magenta plants, 27% red plants, 27% yellow plants, within variety, green and blue plants may appear where there is yellow plants. If we cluster colors together, 3/4 of plant-life are some tone of magenta/red - the average color of this distribution, ie, the color that would mainly apparent from space, is very close to Pantone 2342 C (HEX b65a65), that's Paart's official plant color from space, among other autumnal colors.

Paleocambrian Paart IF it was fully covered by plant life

Paartene plants would still be classified as photo-litho-autotrophs - but a rather special type, one that is very tolerant to UV radiation, and has learned to thrive without the magnesium and aluminum abundance Earth has - they are, the Ensarkophytes (meat-colored plants).

WHAT IF IT WASN'T RED?

Having in mind everything we've done, seems there wasn't any way we could have blue to start with, how would it go for blue/cyan plants?

If we had to use blue, we would use early algal phycocyanin with a zinc center and manganese ions to make the plant material basic enough to bump the anthocyanin towards blue - as mechanism of defense against the acidity, over time, as the soil incorporated these basic substances, the plants could resort to calcium ions as an evolutionary leap in combating acidity in internal fluids, they would still appear green and yellow depending on the iron content and anthocyanin expression - that's in a very, very summarized way of saying it.

- M.O. Valent, 25/09/2020

PLANETARY MODEL | PART 4 | A MORE DETAILED APPROACH TO CLIMATE AND ATMOSPHERE MODELING

DETAILING APPROACHES TO CLIMATE AND ATMOSPHERE MODELING IN HARD SCI-FI

Over the course of this blog's history, I've made several attempts to tackle climate and atmospheric modeling, often involving some Greenhouse Gas considerations and other comparatives with Earth.

Basic Climate Model (comparing CO² levels)

Energy Budget (comparing energy availability)

Venus Zone (comparing atmosphere escape)

Atmospheric Modeling (trying to dimension an atmosphere)

So today, and over the course of next weeks, I'll try rectify and sum up everything tackled in those posts in a more formal, scientific and less speculative way. Which means this post may get long.

Alright, let's SCIENCE properly this time.

ENERGY BUDGET PROBLEM

Imagine an aquarium, the fish inside can only grow as large as the amount of food you give them, or a garden, where plants will only grow as large as the minerals available in the soil allow.

Our ENERGY BUDGET, is basically the measurement of how much is allowed to happen in a certain place given the available amount of energy.

No sunlight over Earth? No temperature gradient, no wind, no air, no circulation, and a lot of other processes which life needs to exist - ceases to exist, for the exception of the Earth's internal temperature, of course.

That means that in high energy places life would be more active, and in low energy places, less active. For an example of that, is the antarctic ice sheets or anaerobic environments, where only few species of microbes live of very low energy chemical reactions, compared to the African savanna or amazon rainforest, where the abundance of sunlight and carbon compounds allow for a very high biodiversity to be sustained.

This, of course would imply that the relatively closer to the star or the more output light a star produces, the better the planet, but, there are other variables to that, which we will explore later.

Let's just consider what Earth receives of sunlight a being 1E (Earth energy unit).

As sunlight spreads according the Inverse Square law, a planet twice the distance Earth-Sun would not receive 0,5E, but about 0,25E:

1solarLuminosity / 2AU^2 = 1/4 = 0,25

Planets at varying distances to their suns and varying suns would render very different available energies to their systems.

This 1E is equal to ~1.368 W/m² at the top of atmosphere, considering a rotating spherical Earth's surface - it is about 1/4th that, or about ~342 W/m².

Several parts of the Earth reflects and absorbs this energy in different ways, this is called Albedo, or how much light/energy does an object reflect back.

Earth's average albedo is about 0,30~0,31 - which means that Earth reflects 30~31% of sunlight back to space, and the closer this value approaches 1, or 100%, the more light it reflects back, and thus the colder it will be.

Even though the amount of energy available is similar to Earth's, the real availability of this energy mostly dependent on the existence of an atmosphere able to trap heat properly.

Considering Earth had no atmosphere, we could use Earth's distance to the Sun, and the Sun's output, to figure out Earth's temperature.

Simplifying this math results in a Temperature formula like this:

T = 279*(1-a)^(1/4) * 1/√d

Where a and d are Albedo and Distance to the Sun, T then  is given in Kelvin, this render us about 254~255K, which is about between -19ºC and -18ºC.

Simplifying further, and assuming the planet has this 0,3 bond albedo, we get

T = L^(1/4)

      D^(1/2)

Where L is the luminosity of your star, and D is the distance between your planet and star, and the final T is a multiple of 254,5K.

The atmosphere creates a greenhouse effect, being made of gases that are transparent to visible light, but pretty much opaque to infrared.

To understand that, we will recur to a simplified model of heat interactions, so, consider that:

- Hot objects lose heat faster than cold objects. And that happens to the 4th power of the temperature. Double the temperature, the rate at which heat is lost is 16x greater than before.

- Planets are found in their equilibrium temperature. They reached a point where the amount of heat energy lost is roughly equal to the energy they receive from their parent star.

Considering  no atmosphere, only oceans, grasslands and forests, and deserts albedo (~0,3), and a rotating sphere, we get ~240W/m².

The relationship (experimentally) between heat loss and temperature can be described by the equation:

T= (F/σ)^1/4

Where F is the rate of heat loss (heat flux), and σ is a fundamental physical constant (Stephan-Boltzmann constant) with a value of 5.67 x 10^-8 Watts/meter² Kelvin⁴s.

Using these values, we also get T=255K, or -18ºC

Now, let's consider a layer of atmosphere that's completely opaque to infrared light, in which case, when it absorbs sunlight, it re-emits it above and bellow, ie, back to the planet and back to space.

In this ideal case, when light enters the system, it warms the planet a little bit, it then also warms the atmosphere, the atmosphere, being this ideal opaque to IR substance re-emits it's heat to space above, and to the planet bellow. The total heat reaching the planet then is twice as before, half coming from sunlight, and half coming from the atmosphere.

T= (480/σ)^1/4 = 303K or 29,85ºC

Of course, this ideal model is an overestimation of the greenhouse effect on Earth, because the atmosphere elements themselves, like clouds contribute to the atmosphere being slightly reflective, that's why we see this blue haze around Earth for instance.

Bellow, a table of different surfaces's Albedo [source]

SURFACE	ALBEDO %
Ocean	2~10
Forest	6~18
City	14~18
Grassland	7~25
Soil	10~20
Desert / Sand	16~20
Ice	35~45
Cloud cover (Thin, Thick)	20~70
Snow (new)	30, 60~70
Snow (old)	75~95

Insolation can also be as low as 37,5% the Equatorial insolation, the further poleward you go.

For an extremely Earth-like atmosphere, the heat energy the atmosphere reflects back to Earth is roughly between +62,3% and +62,5% (assuming perfect black bodies).

The more Earthly your planet appear to be, the more closely it will distribute its energy like Earth does:

1/3 reflected back to space.

1/6 absorbed by the high atmosphere.

1/3 used by the water cycle.

1/6 directly absorbed and radiated to space by ground.

The balance between albedo and atmospheric composition will keep your planet Earth-like in a general way, still, it would be needed to pay further inspection to other aspects such as the...

GREENHOUSE PROBLEM

On Earth, the greenhouse effect is mostly caused by 80~60% H2O, 26% CO2, 5% CH4, 4% O3, 4% CFC/HFC, <1% NO2 and other trace pollutants.

Besides a number of other positive feedbacks, the water cycle is the worst.

Increase the temperature, more water in the atmosphere, the water will increase atmospheric pressure and trap more heat in the atmosphere, leading to more water evaporation, if this happened to Earth, and all of Earth's water evaporated into the atmosphere, the pressure would be over ~358atm (given average ocean depth), because all the Earth's oceans are above you in the atmosphere - and Earth would very probably turn into a planet like Venus this way.

However we should acknowledge that climate change is still debated, and we probably haven't seen it's effects to a full extent, so our current estimations are in risk of being rather underestimates of the real thing. Assuming our early/current estimates are somewhat correct, this means the world warms about 2~3ºC per doubling of CO2, however in times like the Carboniferous period, where CO2 levels were 2x what they are today, global averages were about 20ºC, in fact 1,3x hotter when it should theoretically be ~18ºC, of course, at the time Earth would have a lower albedo due large rainforests and large oceans, but it stops making that much direct sense when you consider that since the Cretaceous period, CO2 levels been decreasing drastically, and even so, temperatures are equivalent to that of the late Devonian when the CO2 were 5~6x current levels.

It would take about 15~16 doublings to make an entire Earth atmosphere out of just CO², with the 3ºC per 2xCO2 rule that's a 42~45ºC increase to about 65ºC avg temp - of course if such a disaster happened we would have water cycle feedback loop set long before it reaches 40ºC.

Talking about water cycle, there is this paper published in Nature that considered an oceanic planet like Earth in different scenarios. And despite the water vapor absorption bands overlap CO2 emission, thus negating some of the climatic change, they still managed to set a moist-greenhouse effect (1,10x Earth's current level) to heat up the planet to ~67ºC.

They've found that exists this threshold between 300K and 330K (26ºC and 56ºC) where the climate enters a rather unstable regime, ie, minor increases in the radiative intake can lead to drastic decreases and increases in temperature, and above 330K exists this warm regime, where for example at 340K it would be needed a full drop in the radiative index back to current Earth levels for temperatures to stabilize back at ~292K, but over a period of 120 years, 80 of which, things could have gone wrong again while in the unstable zone.

Things can be relatively safe up the mark of 1,05x current radiative intake, whereas it would take over 90~100yrs for it to reach an equilibrium at ~62ºC.

However, a minor 1,03x radiative intake is still safe inside the cold regime zone, bellow 300K, despite the temperature going up to 298K (~25 ºC).

This aspect of the thing leads us to the...

VENUS ZONE PROBLEM

The Venus Zone is related to a theoretical area around a star where Earth-like planets would eventually turn into Venus-relative planets through various processes, but mainly by triggering a runaway greenhouse effect.

This could also be related to the finding of a Mercury Zone, which is the area around a star where planet's are unable to retain an atmosphere due solar wind blowing it away from the planet.

Catching back what we talked previously on that matter, I kicked that extra 10% radiative intake could indeed push Earth into a runaway greenhouse effect - the Nature's paper just confirmed that point of view, however, it is fair to admit it didn't really turned Earth into a Venusian planet, in any case, a paper published by James F Kasting and his team in 1993, concluded that a conservative estimate for a continuously habitable Habitable Zone over the course of 4,6Gyr would be squeezed between 0,93AU and 1,37AU, surely between 0,95AU and 1,15 AU, but could as well be larger with different atmospheric conditions than those of Earth.

For them, the Venus Zone starts at 0,84AU, with a radiative intake of 1,41x.

Water loss threshold is on that 1,10x RI mark at 0,95AU.

Greenhouse effect can be very helpful up to the 1,67AU mark, where the RI is 0,36x that of Earth's.

However, if greenhouse effect is not addressed, the 1st condensation of atmospheric CO2 happens at the 1,37AU mark, with RI 0,53x.

Greater changes in albedo, atmospheric water and CO2 are not needed at all if the planet orbits a low-mass star, like a K or M class star, as their output light is redhshifted - H2O and CO2 proved to be better suited to absorb IR light, so the more red-shifted the light from the star, the more of it (proportionally) they will absorb.

They also calculated different planet sizes.

The minimum distance for a Mars-like planet is 0,88AU for runaway greenhouse, and 0,98 for water loss.

The minimum distance for a planet with 2,55 Earth's gravity, is 0,81AU for greenhouse, and 0,91 for water loss.

Funfact: They also believe that 5% of double S-type systems (planet orbits 1 of 2 binary stars) could be habitable, whereas 50% of P-type systems (planet orbits a binary star system) could be habitable, in which case, has great implications for SETI.

ATMOSPHERIC MODELING

Atmospheric modeling is something I must admit that I wasn't so confident about when doing my stuff, if I could say I really knew anything about it, it would mainly be something similar to what we would normally see in a 5~6th grade natural sciences book - and later on specific one about the atmosphere structure, in the latter, only the very basic.

It isn't something we are usually taught about for real, and it isn't really something I got to study in the college's library because of the Coronavirus Pandemic (2020) so far.

All I have looked for so far, is a peek in climate simulation / greenhouse correction, scale height, and habitability.

However after further research, I had worked on a general spreadsheet for the past 3 days non-stop, so you can derive basic data about your star, planet, planet's climate and atmosphere, all in one place - it can only do 1 star and 1 planet at once, but it was built over these research topics I have been discussing for long here on this blog, and updated.

Don't be shy to report bugs, corrections, or suggest more items to it in future versions.

Star & Planet Calculator v.1.0.0

- M.O. Valent, 24/05/2020