New Mars Forums

Hey kbd512,

It's extremely rare for someone with a strong opinion to change it via online debate and discussion, and the fact that you are willing to do so shows great strength of character and laudable intellectual honesty.  It's an example I will seek to follow.

I myself am no expert on climate modelling.  Terraformer says that GCR's have an effect on cloud formation.  I don't know anything about that. 

My initial, first-order guess would be that, outside of sunlight, no other source of energy input to Earth's thermal system is important.  But that's not a satisfying answer so I'd like to drill down into it.

I might categorize the energy inputs to Earth's thermal system as follows:

-Sunlight: 1366 W/m^2 parallel to the direction of the rays or 342 W/m^2 if averaged over the surface of the Earth

And:

-Solar Particle Radiation (Both the solar wind and higher energy solar flare particles)
-Solar Electromagnetic Radiation (X-rays and Gammas; I know sunlight is technically electromagnetic radiation but I am not referring to blackbody spectrum radiation)
-Cosmic Radiation (including both high energy particles, the cosmic microwave background, starlight, and other radiation from outside the solar system)
-Reflected sunlight from the Moon and planets
-Geothermal Energy (originating from radioactive decay inside the Earth)
-Human Generation (Nuclear energy contributes to Earth's thermal equilibrium, and the energy that went into fossil fuels was sequestered long enough ago that it also does, for all intents and purposes)

Let's run some numbers.

Solar Particle Radiation

Comes in two flavors: The Solar wind and solar flare particles.  The solar wind averages roughly 2 protons per cubic centimeter at a speed of roughly 400 km/s. 400 km/s corresponds to 80 GJ/kg (800 eV/amu).  2 protons per cubic centimeter at 400 km/s corresponds to 8e11 protons per second or 1.3e-15 kg/s.  Multiplying the two and then dividing by 4 to distribute the energy over Earth's surface, the total is 2.6e-5 W/m^2.  I don't imagine that this actually gets deposited in Earth's atmosphere directly because it is stopped by the magnetosphere.  It's entirely possible it gets deposited in Earth's outer core.  It's also possible that it radiates away into space.  Very hard to say.

Solar flares are much rarer, although I struggle to find information on exactly how rare.  While they release much higher energy particles and may release more of them, major storms seem to occur with a frequency of once every few years and don't last for very long.  Given the paucity of easily available information, it seems reasonable to round my estimate up to 3e-5 W/m^2, minus that fraction which does not enter Earth's thermal system which I would estimate (Based on no data) as 50%.

Solar Electromagnetic Radiation

Thanks to this explanation on Spaceweather.com, I feel that I have a pretty good handle on the magnitude of photon radiation from the Sun.  Anything above 1e-6 W/m^2 is a flare.  Considering that the Sun typically is not in a flare state and we have to divide by 4 to distribute the energy over Earth's surface, mean energies are presumably in the range of 1e-8 to 1e-7 W/m^2.  This energy will presumably become part of Earth's thermal system.

Cosmic Radiation

Consisting of Cosmic Ray Particles, the Microwave Background Radiation, and nonsolar X-rays and gamma rays.

Wikipedia happens to have some highly useful figures on this score.  From the article on Cosmic Rays:

For particles with energies this high, it is reasonable to assume speeds comparable to the speed of light. Summing the four given terms and rounding up you have roughly 2 eV/cm^2; At the speed of light this corresponds to 6e14 eV per cubic meter per second or 1e-4 W.  More than half of this comes from cosmic rays, which are substantially attenuated by the Sun's own magnetic field and particle radiation; I'm not sure how the galactic magnetic field behaves in a practical sense.  Taking just starlight and CMB you have about 3e-5 W/m^2 with cosmic rays adding another 1-5e-5 W/m^2.  This page, where NASA summarizes results from Voyager, suggest that the Sun manages to block a big part of the cosmic rays and therefore perhaps the lower end of that spectrum is reasonable.

Moonlight and Planetlight

I am going to disregard planetlight, because it seems plain to me that moonlight is orders of magnitude bigger than everything else combined.

Per Wikipedia, the apparent magnitude of the mean full moon is -12.74; This corresponds to 0.26 lux.  Assuming the luminous efficacy is 20 lumens per watt (this is a guess which is intended to also include thermal radiation from the Moon) the energy received from the Moon is 0.013 W/m^2.  Divide by 4 to account for Earth's surface area and 2 to account for the average phase of the Moon (which is a half Moon) and you have 2e-3 W/m^2.

Geothermal Energy

Estimated at around 87 mW/m^2 globally, 8.7e-2 W/m^2.

Human Generation

A reasonable first-order approximation is that virtually all human generation comes either from nuclear or fossil fuels.  Total human energy usage is about 18 TW.  Averaged over the surface of the planet this is 3.5e-2 W/m^2.

So, I have the following estimates, from biggest to smallest:

Geothermal Energy:       0.09              W/m^2
Human Generation:        0.04              W/m^2
Moonlight:                     0.002            W/m^2
Cosmic Radiation:          0.00004         W/m^2
Solar Particle Radiation:  0.00003         W/m^2
Solar Photon Radiation:  0.00000005    W/m^2

Of these, geothermal energy is most significant at ~0.03% of the incoming solar energy and solar photon radiation is the least significant at a level that is not meaningfully different from zero. Naturally all of these numbers are approximate but I believe that if the numbers I have provided are close to accurate they show that (absent some other effect which multiplies their significance relative to their magnitude) none has a major effect on Earth's heat balance.

One really interesting thing about this is that louis earlier made the claim that studies could not find evidence of harms caused by vaccines and is now making the claim that there is an obvious problem with an obvious answer.  Generally when things are obvious they are also easier to study.

Anyone reading or posting on this forum is by definition a person who interacts with a group of people that might be called (in the most positive possible way) "internet nerds".  Based on anecdotal evidence, autism and aspergers are much more common among that subset of the population.  I have to say that in my opinion our community and many others are better off for it.

So here's another lens through which to look at it: We have an imaginary relationship between vaccines and ASDs, vs a very real relationship between a lack of vaccines and all sorts of archaic diseases.

I strongly question the notion that cities will suspend their congestion charges for electric vehicles in the long term.

While air quality is one reason to try to limit the number of vehicles in a CBD, the main reason for congestion charges is exactly what it sounds like--congestion.

Consider the following image, which shows the amount of road space taken up by the same number of people on a bus, on a bike, and when driving alone:

Perhaps in the suburbs and in the country, space is not an issue.  But in the main downtown of a large city (the sort of area typically affected by a congestion charge) there simply is not enough space for everyone to get places by car.  Land has value, and nowhere more so than at the heart of a big city.  While road space is typically built and maintained as a free public service supported by taxes, demand in these circumstances exceeds supply and it is therefore necessary to establish a use fee as a disincentive.  There's a funny take on the above image, which I can't seem to find, that has three panels that are the same as the rightmost.  The first is labeled "gasoline car", the second "electric car", the third "autonomous car".  The point being, neither of the two solves the main problem.

As an example: A Manhattan apartment building within the CBD (traditionally defined as the entire island below 60th St) is typically five stories high, with a typical two bedroom apartment containing two people with an area of perhaps 700 sqft and a rent for that apartment of perhaps $3,000 per month.  If you assume 20% of the building is given over to stairs, storage, and entryway you will have 875 sqft of ground-land generating $15,000 per month or $180,000 per year.  Our hypothetical "typical" building was built in, let's say, 1925 and therefore construction was paid down long ago.

By comparison, one parking space is perhaps 8 ft wide by 15 ft long, and therefore 120 sqft (roughly 1/7 of that land on which 5 apartments or more might be built; in other words the value of that parking spot might be calculated to be $25,000 per year), and is free for anyone to park in (if you can find a spot, and with the exception of various posted limitations). 

The point here is that there really ought to be a congestion charge on EVs the same as on gasoline cars (or perhaps just slightly lower), because in the core sections of dense cities people really ought to be taking public transit.  It's simple geometry: Much like a forest, a car is mostly empty space.  If governments choose to exempt EVs as a sort of subsidy or incentive I suppose that's their prerogative but there's no law of urban planning that means it must be so.

Indeed, NYC is about to implement just such a charge, and my understanding is that there will be no exception for electric cars.

Quoted because I think this is a great synopsis of the state of our understanding.  Reflected light into shielded greenhouses may in fact be necessary.

On the topic of radiation shielding: I don't like it (I think the need for continuous refrigeration is something of a kludge) but it probably would work

Hey GW,

Seems to me there's an error in your math there.

57 kN/m^2 corresponds to 5.8 tonnes of atmosphere, not kilograms.  This corresponds, in turn, to 5.8 meters of water, not 0.58 centimeters.

While Earth's atmosphere probably would be adequate shielding to protect us from most cosmic radiation it does not have to serve that purpose because Earth has a strong magnetic field which protects us in its stead.

Kbd512,

I'd be interested in having a look at the studies you mentioned.  As louis pointed out they have a direct bearing on whether we can use natural light greenhouses or not.  My assumption thus far is that the radiation would not be a big issue for agriculture (agrifacturing?), but obviously if research paints a different picture I will have to update that.

On the topic of artificial magnetospheres, the first source notes (on page 7) that the system being described depends on a plasma environment to work.  Mars's atmosphere is thin, but not that thin.  I'm not sure that particular paper applies.

The second article mentions a system designed for Mars.  I believe we had a member of that Lake Matthew Team right here on the fora.  I don't totally understand how the system is supposed to work, but it seems that the basic idea is to use a large, superconducting loop (much larger than the area being protected) to generate a big magnetic field to push charged particles away.

The theory is sound, and I suppose I trust their math, but superconducting loops and carbon nanotubes are a substantial investment as far as infrastructure goes.  Is this better than burying things under dirt and ice?  Probably, I guess, if you're building on a big scale anyway.  It does also require continuous power though which is a substantial drawback.

How much mass is in a forest?  This paper based on research in Mexican forests, found an average Aboveground Biomass around 80 Mg/Ha with an average canopy height around 15 m.  That's roughly 8 kg/m^2, and with a height of 15 m roughly 0.5 kg/m^3.  That is to say, a forest has roughly double the amount of radiation protection offered by a comparable volume of non-forested air at 1/2 atmosphere (including the radiation protection coming from the air in the forest).

This is to say, not much.  Not much at all.

A key point to remember here is that line-of-sight is not a good metric for radiation protection as (unlike visible light) cosmic rays are barely affected by normal materials.  It's worth remembering that real forests are in fact overwhelmingly composed of air, which is what makes it possible to walk around and see through them.

If what you have in mind is a solid wall of thick-trunked trees you may not have a real-life forest in mind at all but rather an imaginary one.  In the real world trees need space for their root systems, especially big ones.

Given the height limit of the dome ceiling trees will also be of limited utility in protecting from radiation.  More cosmic radiation comes from the zenith than the horizon because the atmosphere provides more shielding horizontally than vertically (this is the same reason that sunsets are red).  Furthermore, thousands of kilograms per square meter are required for adequate shielding, corresponding to a couple meters of regolith.  An overhead structure used for shielding will necessarily be more than a metal roof over an area; it would have to be a pretty thick piece of solid material.

I say all this not to suggest that such a park is impossible or that radiation is an insurmountable problem.  I would, as I said, look to a thicker dome (actually the forces on a 200m pressurized dome would be immense anyway and would require substantial structures to hold in) rather than trees and roofs, as well as simply limiting exposure time.

Radiation at low altitudes on Mars is roughly 50 times higher than Earth.  If I were to design a radiation management strategy, it would be for habitats to be extremely well-shielded so as to have virtually no radiation within, and then for residents to minimize their time in higher-radiation zones to keep their total dosage down.

Given the mental health benefits of open/green space I would say a park like this one is a net win even with higher radiation exposure while you're there.  For what it's worth, if I were designing for maximum effect I would put the vegetation towards the center and make the outer wall an open vista, ideally an impressive crater or canyon.

Here's an entirely reasonable explanation for their actions:

SpaceX wants to land rockets on barges because it requires less fuel than returning to the launch site.

Therefore they try to land rockets on barges.

Landing rockets on barges is hard and their first few tries fail.

They try landing on land (which, as you say, is obvious) as an easier target to develop technology.

Once accomplished, they go back to barges.

The words you used when making the claim initially were as follows: "I would like to believe that SpaceX acted on my suggestion".  I would like to gently suggest to you that in using those words you gave away more than you intended.

On the topic, broadly, of whether Newmars has any influence on technical/policy discussions: I would say that as a default the answer is no but we may show up on a google search every once in a while.  To my knowledge we haven't had a member who was also in a place of power in the aerospace industry while on the forums, but Mr. Musk without a doubt is exposed to many similar sources--Zubrin's writings, the principles of physics and engineering, news articles about space, and the trends in science fiction.

The short answer to that question is no, vegetation is not enough to protect from radiation.  You can get a good level of radiation protection from a few thousand kg per square meter, and even a tropical rainforest canopy is almost definitely at least an order of magnitude less than that.

Hey Elderflower,

I also assumed that Cp*m=0. For a more rigorous model you would be completely right but for this one whether the gas is contained at constant volume or permitted to expand at constant pressure is (if you'll pardon the pun) immaterial

Hey kbd512,

I’m always glad to share knowledge when I have knowledge to share.

There are a number of ways I can respond to this, all of which are unsatisfying in some respects. 

The first is tautological: This model does not include a time component, therefore a model with a time component would be a different model.

The second is semantic: Because this model does not include a methodology for measuring or calculating λ from other known quantities (the composition and pressure of Earth’s atmosphere, for example) it isn’t really a model at all so much as a learning tool to help understand the greenhouse effect or to analyze the approximate effect of changing certain basic parameters.  Therefore trying to use it to model the actual Earth would be to mistake its purpose.

The third is logistical: This model has many flaws as a climate model.  Time invariance is one and spatial invariance is another, and treating the atmosphere as a single uniform layer is a third.  Of these, addressing time invariance is the hardest and also (possibly) the least significant because you need to think very carefully about how the heat capacity of the various layers of Earth’s thermal system act and interact through all three modes of heat transfer  (If you don’t, you’ll get a nighttime temperature estimate of 0 K).  It makes more sense, in my opinion, for a research effort to try to address the other two first because you’ll get a better return on your labor.

Hopefully by taking these three answers together you understand holistically what my response is.  We went into this acknowledging that the model is simple and limited and time invariance is certainly one of its flaws.  We have been working around this by using the “hack” of mean insolation levels, which is reasonable but imperfect.  By addressing this and other issues we could build a better and more reliable model.

Now, to go into your thought experiment.  I find it always helps to draw a picture, so here's mine:

Here are the assumptions for this model, which I believe accord with the scenario you have described:

• Yellow is CO2

• Blue is water

• CO2 and H2O are completely opaque to thermal radiation

• The gray exterior does not allow any heat energy to enter or leave, or alternatively it perfectly reflects that thermal radiation right back into the CO2

• Laser light with a power of 10 W enters through an infinitesimally small opening and is completely absorbed by the CO2

Your question is: What happens?

I want to start off by saying that there is no way to do an equilibrium calculation for this scenario because heat is entering and not leaving.  In effect, the parameters you have given are non-physical.  You have already recognized this and we can still analyze it, but to the extent that the results seem nonphysical it's because the inputs themselves also are.

I'm going to introduce a couple more assumptions so that I can put actual numbers on things:

First, that the mass of the water is 239 g.  I have chosen this because, at a heat capacity of 4.184 J/g-K, it will take 1000 J to raise its temperature 1 degree (it's conveniently also just about 1 standard US cup of water).  I will assume that the CO2 has a negligible thermal mass in relation to the water and in relation to the power of the laser.  I will assume that the water-CO2 interface has a surface area of Ai=100 cm^2 (0.01 square meters) and that the CO2-container interface has an area of Ac=1000 cm^2 (0.1 square meters).  I will assume that both the CO2 and the water are at uniform, different temperatures but begin at 0 C.  I will assume that heat transfer is only through radiation. 

How long will the water take to boil?  You might naively say: ΔU=Cp*m*ΔT (Eq. 1), where "U" is increase in energy (in this case as heat), Cp is the specific heat, m is the mass, and ΔT is the change in temperature. I have chosen m so that the product Cp*m=1000.  If the temperature difference is 100 K, total energy is 100,000 J and it will therefore take 10,000 s (a bit under 3 hours).

This answer is correct, and I will use the rest of this post to explain why and how.

The analysis in this post is based fundamentally on the first law of thermodynamics.  In words, the first law of thermodynamics is conservation of energy, but it has been formalized in a form that can be used directly for analysis.  This form is:

ΔU=ΔH-ΔW (Eq. 2)

Where H is enthalpy (heat entering or leaving, with the sign convention that heat entering is positive), W is mechanical work done (work having the technical meaning of FΔx, force times distance, or equivalent), and U is a quantity known as internal energy.  Alas, while I am quite comfortable talking about how heat moves I have a much harder time talking about what heat is.  Internal energy is a measure of the amount of energy stored in other sorts of forms, including molecular vibration (heat) but also chemical potential energy, etc. 

While Eq 2 does hold for the entire universe taken as a system, it's much more useful analytically to cut a slice out of the universe and analyze it independently.  There are two approaches to this, "Control Mass" and "Control Volume".  If you do your analysis correctly, each will give identical results to the other.  Typically, you will choose the one that is easiest to use for the problem you are looking at.

In the control mass approach, you look at a control mass--a particular set of atoms--and see how they change under the influence of whatever conditions they experience.  In the control volume approach you define a volume (defined by a surface through which heat and/or matter can enter or leave) and you look at the flux of various quantities through that surface.

This is a great control mass problem, which is good because the control mass approach tends to be more intuitive when you can apply it.

I will define two control masses: The water and the CO2.

Let's look at the water first:

ΔU=ΔH-W

Because the water is neither being compressed nor expanding by an appreciable amount, the work done is zero.  Therefore we have ΔU=ΔH

ΔH is equal to the heat entering (Ew) minus the heat leaving the water (Lw).  ΔU is as in Eq. 1.  Therefore we have the following equation that is valid for the water:

Cp*m*ΔT=Ew-Lw (Eq. 3a)

I will return to this later, but first I would like to derive a comparable equation for the CO2 gas.

ΔU=ΔH-W

It is normally unavoidably the case that gases do work when they are heated, either in increasing pressure at constant volume or by expanding at the same pressure (or a bit of both).  In a more realistic scenario I would want to include that in my model; in an equilibrium scenario it would be irrelevant because the temperature would not be changing.  In this scenario I am going to ignore it.  I don’t think this is the property of CO2 you’re trying to get at and it adds unnecessary complication.  For the purposes of this calculation it is as if CO2 is a solid with the optical properties you have specified.  Therefore W=0.

I stipulated in my assumptions that the thermal mass of the CO2 is negligible.  What this means is that the product Cp*m ≈ 0.  Therefore:

ΔU=Cp*m*ΔT=0*ΔT=0

And we have:

ΔH=0=Ec-Lc (Eq. 4a)

Where Ec is the energy entering the CO2 and Lc is the energy leaving the CO2.

Going back to equation 3: The energy leaving the water is thermal emission from its surface, and the energy received is the thermal emission from the surface of the CO2.  Note that emittance is equal to 1 in all cases and therefore has been ignored.  We have:

Cp*m*ΔT=σAiTc^4-σAiTw^4 (Eq. 3b)

or, filling in:

1000ΔT=.01σ*(Tc^4-Tw^4) (Eq. 3b)

Now, for the CO2 it’s a bit more complicated.  Energy leaves via thermal radiation towards the wall and towards the water.  It’s received by thermal radiation back from the wall, thermal radiation from the water, and from the laser.  Plugging in:

0=10+AiσTw^4+AcσTc^4-AcσTc^4-AiσTc^4 (Eq. 4b)

The first term is the laser power, the second emission from the water, the third reflection from the wall, the fourth emission towards the wall, and the fifth emission towards the water.  Terms 3 and 4 cancel each other out; removing them and plugging in:

0=10+.01σTw^4-.01σTc^4 (Eq. 4b).

Rearranging:

10=.01σ*(Tc^4-Tw^4) (Eq. 4b)

We see the right sides of Eq. 3b and Eq 4b are equal to each other; and solving for ΔT we find:

ΔT=0.01 K/s (Eq. 5)

Which corresponds to the initial “naive” result.

What actually happens?  Initially, the water and CO2 are at equilibrium at 0 C.  The water radiates 3.15 W at the CO2; The CO2 radiates 3.15 W at the water and 31.5 W at the wall which is reflected back onto itself.  All told, 34.65 W are falling on the CO2 but because it also emits 34.65 W the temperature does not change. 

Then the laser turns on.  Because the CO2 has no thermal inertia it cannot hold any of the heat and heats up rapidly.  Soon, it has reached a temperature of 117 C while the water is still at 0 C.  At 117 C (390 K) it radiates much more: 13.15 W at the water and 131.5 W at the wall.  The water, which is still at 0 C, absorbs 13.15 W and emits 3.15 W and therefore absorbs a net energy of 10 watts.

This continues for 10,000 seconds (per our calculation) until the water is at 100 C.  The CO2 is now at a temperature of 166 C.  At 166 C it emits still more: 20.98 W at the water and 209.8 W at the wall.  The water, whose emissions have also gone up, absorbs 20.98 W and emits 10.98 W and therefore absorbs a net energy of 10 W.  Because of the fourth-power law, the CO2’s temperature has gone up by 49 C while the water has gone up by 100 C.

Once the laser is turned off, the CO2 rapidly cools down to 100 C.  The water is still at 100 C and radiates 10.98 W at the CO2; The CO2 radiates 10.98 W at the water and 109.8 W at the wall which is reflected back onto itself.  All told, 120.78 W are falling on the CO2 but because it also emits 120.78 W the temperature does not change. 

I’ve gone on for almost 2000 words in this post and, though I showed how I would analyze the problem you presented, I’m not sure how much I’ve done to clarify things.

If I had to extract a few key points, it would be these:

1. Blackbody radiation is everywhere, literally all around us all the time, and is an omnipresent feature of matter in the universe

2. Most often nearby objects are near thermal equilibrium and the thermal radiation going in one direction cancels out the radiation going in the opposite direction

3. Just because there is a significant amount of power bouncing around in the form of microwaves does not mean that you could actually use it for anything.  Consider the equation for carnot efficiency, which I believe is derived directly from the second law of thermodynamics: Efficiency=1-Tc/Th.  You generate energy from thermal gradients and not from temperature itself. Meanwhile thermal radiation is close to omnidirectional, meaning that no possible system of mirrors or lenses could concentrate it to a higher power.

As a very brief follow-up to the previous post, for λ=0 the model simplifies to the UTSA model I discussed earlier.

In response to kbd512's post, #153:

If you want to split hairs, my recollection (I could be wrong!) of what I have said is that I have no interest talking about cold fusion.  Regardless, you are right that I wouldn't bother talking to someone pushing a perpetual motion machine.  The common thread between the two is that it's generally not worth my time to engage with the (often conspiracy-minded) people who will reject scientific consensus in favor of their own unsupported (and usually easily disprovable) nonsense.

It is for precisely this reason that I typically avoid engaging with climate change denialists.  I have relaxed that rule in this case because I respect your intelligence.  I can't say I've gleaned any benefit from doing so, but here we are.

Your question, as I understand it, is this: How can it be the case, if absorbed sunlight averages 240 W/m^2 over the surface of the Earth, that the Earth's emissions can be double that at 480 W/m^2?

Your question "Is this not a violation of the conservation of energy?" seems particularly strange in context, because the analysis on that page is based entirely on the conservation of energy.

I notice you made an error in defining your variables.  You said "G - greenhouse effect".  But this is not right.  G in this case is for "Ground"--the G-term refers to thermal emission by the ground.  The A-term is in fact the one more closely associated with the greenhouse effect.

Likewise the term λ is a bit more complicated than just the emissivity/emittance of the atmosphere.  An interesting fact about emittance and absorbance is that (for the same angle of incidence if the material is solid) they are the same value at the same wavelength. 

On the whole, I believe the page you linked does an excellent job of explaining a simple equilibrium model for the greenhouse effect. I'm not sure that I can explain it better or differently but I will try.  Please excuse the poor quality of my graphics.

In this first image, I look at the energy balance of the ground, the atmosphere, and free space.  "Free Space" can be considered to be a control volume that encloses the whole universe except for the Earth and its atmosphere.  For each of the three control volumes (ground, atmosphere, and free space) the energy input equals the energy output.

Then in the following image I created a sort of flow chart with equations.  In the webpage you linked to, S equals the energy absorbed by the Earth's surface; They ignored albedo effects by reducing S to the correct value.  I have followed that convention below.

Another way of explaining the answer to your question is to consider two parallel mirrors with a beam of light bouncing back and forth between.  The beam itself may have a power of 1 watt; but if it bounces back and forth 100 times between the mirrors you will find it to be the case that a control surface has 100 W of light passing through it in each direction.  It's not a violation of conservation an energy, just a consequence of comparing apples to oranges and looking at the wrong control surfaces and volumes.  The model presented in that webpage is mathematically analogous to a mirror which is partially reflective and partially transmissive (with a stipulation that it can be no more than 50% reflective because emissions go in both directions).

Hope this helps

Hey GW,

Thank you for your kind words.

I think it would be going too far to say that I am an expert on heat transfer.  However, it was something of a focus of mine in undergrad, including a couple of classes that got into thermal modelling, and something which I have occasionally had the chance to apply professionally.

As far as air friction heating goes, it would be irresponsible for me to go so far as to offer an opinion.  I have had the chance to work with supersonic gases on occasion in the form of a light gas gun (up to roughly 1.5 times the gases' speeds of sound) but in general I've had empirical correlations available to describe their behavior.

Talking about science is really hard.  Science journalists usually use metaphors, rather than numbers.  I understand why they do it (neither they nor their readers can really understand the physics behind most things) but if you would be able to understand it their explanations are usually pretty frustrating.

For anyone who may be interested:

I want to get into a longer description of my simplified temperature model for those who are interested, both in the interests of transparency and because I think it's sort of cool.  You can skip past the math and explanations if you'd like: I have a link to a google sheets calculator that will do the math for you at the end of this post.

While I don't have the time or resources that it would take to construct a generalized, precise climate model of the planet (or any other planet), I can construct the much simpler model I described above.  I have been calling this model the UTSA Model: Uniform Temperature Sphere Approximation.  The basic assumptions that go into this model are as follows:

• A planet is a perfect sphere (Accurate to within 5% usually)

• A planet will either absorb or reflect away all of the sun's radiation that falls on it (basically accurate)

• A planet does not have any atmosphere that interferes either with the absorption (e.g. the Ozone layer or the reflective clouds of Venus) or emission (the greenhouse effect) of radiation (Not accurate; atmospheric effects are pretty much completely disregarded in this model)

• A planet has a uniform surface temperature (highly inaccurate)

• A planet radiates as a perfect blackbody, as modified by its emittance value (basically accurate once atmospheric effects are disregarded as I talked about in the previous post)

• A planet is in thermal equilibrium at all times, with energy received from the Sun being equal to energy lost to space (accurate to within 1%)

You can write the equation for the energy balance as follows:

P_in=P_out

(Power in is equal to power out; units are Watts)

P_in is the energy from the Sun falling on the planet multiplied by that portion of the energy which is absorbed.  The equation for this is:

P_in=(1-α)*I*A

Where:
α is the reflectance, also known as the albedo (unitless)
I is the irradiance, also known as the Solar Constant.  At 1 AU its value is 1366 W/m^2
A is the cross-sectional area of the planet in m^2; For a spherical planet this is A=π*r^2

Substituting for area:

P_in=(1-α)*I*π*r^2

Now, P_out is governed by blackbody radiation:

P_out=ϵ*σ*A*T^4

(From my previous post)

ϵ is emittance, a unitless constant between 0 and 1
σ is the Stefan-Boltzmann constant, 5.670e-8 W-m^-2-K^-4
A is the emitting area, in this case the entire planetary surface; area of a sphere is A=4*π*r^2
T is the absolute temperature in Kelvin

Substituting for area:

P_out=4*ϵ*σ*π*r^2*T^4

Setting the final equations for P_in and P_out equal to each other:

(1-α)*I*π*r^2=4*ϵ*σ*π*r^2*T^4

Simplifying:

(1-α)*I=4*ϵ*σ*T^4

Solving for T:

T=( (1-α)I/(4ϵσ) )^(1/4)

I is proportional to the inverse of the square of a planet's distance from the Sun.  In other words:

I=C/d^2

Where C is the Solar Constant at Earth (1366 W/m^2) and d is a planet's distance from the Sun in AU.
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UTSA Calculator Here

Instructions for use:

The calculator allows you to vary three parameters: Albedo (reflectance), Emittance, and Orbital Radius:

Albedo: Values can generally be looked up on Wikipedia; Earth is 0.3, Mars is 0.2, the Moon is 0.07.  For a rocky planet or asteroid with no cloud cover and no values available I recommend a value of 0.25

Emittance: Real values for real planets are probably roughly 0.95 but I recommend using a value of 1 because I don't believe it's possible to measure the emittance of a planet in any meaningful sense.

Orbital Radius: Can be looked up on Wikipedia or wherever; Orbital Radius in AU.  Earth and Moon are 1, Venus is 0.72, Mars is 1.52, etc.

This is a very simple calculator.  I have started it off with an orbital radius of 1.52, an albedo of 0.2, and an emittance of 1 to give the user an idea of how to use it.  Please reenter these values before you exit the calculator because they do not reset automatically.
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The only reason I included emittance as a value which can be modified is so you can calculate the "effective emittance" of each planet or body.  This is a non-physical value.  It's a number that can be used as a shorthand for atmospheric effects and nonuniform temperature and whatever else.  It is a fudge factor and an average, not a physical description of emittance behavior.  You can find it by entering the correct values for albedo and orbital radius, and then changing the albedo until the USTA temperature matches the observed temperature.  For the Earth and the Moon, the "effective emittance" is 0.61 and 1.7 respectively.

PS: Apologies to the Texans among us who hear UTSA and think "University of Texas, San Antonio"