- Sky Dragon is up and running. We need people to sign up and add content. Introduce yourself at the General discussion forum. Speak up to welcome a few others, start a thread, or contribute to someone else's thread.
|Welcome to Sky Dragon.|
This site has been set up to allow convenient discussion of Earth's atmospheric greenhouse effect. The main topic for discussion here is the physics of how (and if) the atmosphere helps to determine Earth's current temperature. Other spin off topics are also supported.
Registration is not required to join in. Anyone can join in immediately. Anyone may also register as a user, and gain access to some additional features. Registration is simple, fast, and completely free.
|The giant space ship example; Can you calculate surface temperatures given a moon sized interstellar space ship with its own atmosphere?|
|Tweet Topic Started: Dec 4 2011, 03:29 AM (1,119 Views)|
|Chris Ho-Stuart||Dec 4 2011, 03:29 AM Post #1|
This example is due to "Gbaikie", in a comment at the Climate Etc blog. (Here is a link to the original comment.) I think this is a very well-conceived problem, so I am putting a copy here. I have rephrased a little; but using Gbaikie's original comment as the basis for this topic. Thanks, Gbaikie!
Suppose you wanted to travel to a different star system and your starship was going to be a moon size body.
The propulsion method is not important. You might, perhaps, use a lot of nuclear bombs for propulsion, so this moon size object might be a giant Orion nuclear starship. (See Project Orion (nuclear propulsion) at Wikipedia.) We will assume that somehow it can be made to go at perhaps 1/10th the speed of light, and so takes quite a while to get to another star system.
Suppose it is about the size of the largest known Kuiper belt object: Eris, about 2400 km in diameter. (Ref: largest Kuiper Belt objects.)
My question isn't where it's going or how nukes are needed to get to 1/10th light speed, but rather: how much heat would be needed to heat an atmosphere? The atmosphere could be rather low in pressure compared to Earth, but high enough pressure so people could use oxygen masks rather than pressure suits. Let us suppose a pressure similar to Earth at 8 km altitude (similar to the top of Mt Everst) , or 5 psi.
There will be no sunlight for most of trip and idea is to have air pressure about 5 psi and temperature about 10 C at the surface. The air will heated with fission nuclear furnaces; so heating the moon as one would a house, only on vastly larger scale.
Assume it has 1/10th earth gravity, and 1/3 the surface air pressure.
Given the temperature, air pressure, surface gravity and size... how much energy would be lost to the black universe?
|Chris Ho-Stuart||Dec 4 2011, 04:32 AM Post #2|
Here's why I think this is an excellent problem.
The giant starship gives a realistic (or at least conceptually sensible) completely defined problem, that naturally avoids many of the additional complications that would be involved on planet Earth.
There are some additional constraints that are needed to get an answer. We need to know the atmospheric composition. If other people have alternative methods to answer the problem, then they should indicate what additional assumptions are needed and propose sensible constraints.
To solve this problem we don't need to worry about any of the following complications.
I am proposing to give an estimate for power required to maintain the starship surface temperature, using the conventional atmospheric greenhouse effect as taught in conventional physics or meteorology courses. Critics of atmospheric greenhouse theory may propose an alternative theory, and show how they would answer the question posed for the topic.
I was struck at how effectively Gbaikie's problem managed to strip away various complications and give a clear and concrete case for which a reasonably direct calculation is feasible; but would which still requires a non-trivial application of physical theory for the impact of an atmosphere.
The easy solution is with a hypothetical idealized "grey" gas, with the same absorptivity across the spectrum.
A realistic but still pretty straightforward solution could use an atmosphere of only Oxygen and Nitrogen, plus just one real greenhouse gas, with CO2 being an obvious choice.
I think that about 2% CO2 by volume is the maximum credible concentration to allow passengers to live and work "outside" for extended periods without harm. (On Earth the concentration is about 0.04%.) So I am hoping to calculate answers for various compositions of the starship atmosphere, using 21% Oxygen, and Nitrogen with Carbon Dioxide for the rest. I'll use a range of concentrations, from 0 to 2%, for the Carbon Dioxide.
I'm not ready to give answers yet, but I love the problem!
People who think it is sufficient to use the lapse rate, or the specific heat in the atmosphere, should try this example also. I think they will find they really do need to consider additional properties of the atmosphere; in particular the optical depth at various frequencies in the infra red.
|gbaikie||Dec 4 2011, 08:23 AM Post #3|
I think this would be good dialogue about greenhouse effect.
I think it could also lead to possibly extending the definition of habitable zones.
There is considerable effort these days finding worlds in nearby stars which could support life.
At the moment we just finding the bigger planets- mainly due current lack of telescope resolution, but within a decade or so that could be remedied.
"Fifty new alien worlds, including 16 "super Earths," have been found—the largest extrasolar planet haul announced at one time, astronomers say.
The discoveries bring the total number of known extrasolar planets, or exoplanets, to 645.
"The harvest of discoveries ... has exceeded all expectations, and includes an exceptionally rich population of super-Earths and Neptune-type planets hosted by stars very similar to our sun," study leader Michel Mayor, an astronomer at the University of Geneva in Switzerland, said in a statement."
|Chris Ho-Stuart||Dec 4 2011, 10:30 AM Post #4|
Welcome aboard, gbaikie! You have the distinction of being the first person to join and post a comment here, apart from myself who set up the board.
I agree. New planets being found around other stars introduce a host of strange new worlds for scientists to study, many of them unlike anything in our own solar system. Everyone involved is keen to glean as much information as possible. Being able to go visit another star would be the ultimate buzz.
Astronomers can make all kinds of useful inferences about these distant planets from tiny scraps of information. In a few cases they've even been able to make inferences about atmospheres on the planets at a distant star! Your exercise in this thread is similarly a case of trying to make sensible inferences from very limited data, using universal principles of physics that (apparently) apply throughout the observable universe.
|Chris Ho-Stuart||Dec 4 2011, 10:48 AM Post #5|
At Climate Etc, gbaikie provided some comparisons and calculations to get started with the starship. I'll repeat some of them here, in tabular form.
The following numbers define what is given for the ship, and can be a basis for comparison with Earth, or the moon, or other bodies.
Values for bodies in the solar system are available from NASA at the Planetary Fact Sheet.
Calculating values invariably works better with SI units, so that's what I will be using. Some points to bear in mind on units:
Temperature. Many thermodynamic calculations need to use an absolute temperature scale, where the coldest possible temperature is given as 0. Absolute temperature is measured in Kelvins. The temperature in degrees Celsius is equal to the temperature in Kelvin, plus 273.16; adding 273 is adequate for rough conversions.
Pressure. The SI unit for pressure is the Pascal (Pa), which is the same as one Newton per square meter. The force on one kilogram in Earth's gravity is 9.8 Newtons. Earth's atmospheric pressure at sea level is very close to 100,000 Pa. The unit "hectoPascal" (hPa) is sometimes used; a hectoPascal is 100 Pascals, and this is equivalent to 1 millibar, and a "bar" is an unit representing an early estimate of pressure at Earth's surface. The "standard atmosphere" pressure is now defined to be 101325 Pa. One psi is equal to 6895 Pa.
The pressure at Earth's surface in Pascals is the mass of the atmospheric column above a square meter of surface, times the gravitational acceleration.
The gravitational acceleration g at the surface of a uniform sphere of mass M and radius R is equal to GM/R2, where G = 6.673 e-11 is the gravitational constant. Hence M = gR2/G. The volume V of a sphere of radius R is (4/3).pi.R3, so the density will be M/V = 0.75*g/G.R.pi.
The dry adiabatic lapse rate (DALR) in an atmosphere is -g/Cp (unit K/km), where Cp is the specific heat at constant pressure (in units J/g/K). For SI purists, note that these are not quite SI units; Cp is typically given as Joule per gram per Kelvin (not per kilogram), and the lapse rate in Kelvin per kilometer (not per meter). I'll use these units also, where it shows up.
On Earth, where the atmosphere is mainly Nitrogen and Oxygen, Cp is very close to 1. On Mars, the atmosphere is mainly Carbon Dioxide, with Cp being about 0.83, so there the DALR is about -3.71/0.83, or the DALR is thus about 9.8.
For an atmosphere consisting mainly of Nitrogen and Oxygen, the specific heat (at constant pressure) Cp is pretty close to 1 (unit J/g/K); and the DALR is about 4.5. In the starship, Cp should be similar to Earth, since you want an atmosphere that is mostly Nitrogen and Oxygen, both of which have Cp close to 1. Hence the DALR will be about 1 as well.
Using these formulae, we can find some basic properties for Earth, Mars, our Moon, and the starship.
(late edit to fix the DALR on Mars; I previously had -3.7 which failed to account for the reduced Cp value.)
Gbaikie, this matches your estimates so far, apart from lapse rate. You were correct to suppose a weaker lapse rate on the starship than on Earth or Mars; but it's weaker even than you had guessed.
Edited by Chris Ho-Stuart, Dec 4 2011, 09:59 PM.
|Chris Ho-Stuart||Dec 4 2011, 11:52 AM Post #6|
The easiest way to answer the original question is to simply use a Nitrogen/Oxygen mix in the starship atmosphere. Nitrogen and Oxygen are almost totally transparent to infrared radiation, which means that there is no greenhouse effect in the starship atmosphere. The thermal radiation emitted from the surface will pass straight out into space, pretty much the same as if there was no atmosphere at all.
The starship surface is at a temperature of 10 C, or 283 K. The Stefan Boltzmann relation gives a very good approximation for the thermal radiation from most conceivable surfaces. This relation is
Q = sigma.T4
For T = 283, this gives Q = 364.
Now the surface area of the starship is 4.pi.R2, with R = 1.2e6 m. So surface area is 1.8e13 m2 (18 million sq km).
Combining, the total thermal energy being emitted out to space is 364 * 1.8e13 = 6.58e15 Watts, or
(Value in GigaWatts corrected with a late edit.)
That's the power that will need to be supplied by the onboard heating system. It's several hundred times greater than all power generated world wide today. Wikipedia tells me (at Electricy Generation) that the total power output from electricity generation over a year (2008) is now up to 20,261 TWh. A year has 8760 hours, so that works out to 2313 GigaWatts.
The starship engineers would love to reduce that... how much benefit could they get from an atmospheric greenhouse effect? That's what I have to consider next.
Edited by Chris Ho-Stuart, Dec 4 2011, 09:57 PM.
|gbaikie||Dec 4 2011, 07:40 PM Post #7|
"You were correct to suppose a weaker lapse rate on the starship than on Earth or Mars; but it's weaker even than you had guessed."
So lapse rate for 1/10th gravity is 1 C per 1000 meters?
And so follows gravity exactly.
What factors would make it not, precisely follow gravity?
Or is that definitional, and any variation would called something else.
This ref http://daphne.palomar.edu/jthorngren/adiabatic_processes.htm
Defines dry adiabatic lapse rate as 10 C per 1000 meter on earth which is number
you using. It also says:
"Make sure you notice that we are talking about moving air (rising or subsiding), not still air. The change in temperature of still air (that is, air that is not rising or subsiding) follows the Environmental Lapse Rate, which varies considerably, but averages about 6.5 deg C/1000 meters (3.6 deg/1000 feet)."
The average lapse rate will determine height of the atmosphere, if it's 1 C per 1000 meter, the atmosphere will be higher than I thought it would be.
In general terms having this much pressure on such relatively low mass body will require a high atmosphere.
Having cooler atmosphere will lower it's height.
In general I would tend think this atmosphere would warmer on average than Mars [the top surface of ground temperature during daylight on Mars may reach around 25 C- but it's air temperature drops very significantly in first meter of elevation].
I would think this atmosphere would be warmer [percentage of bulk of air], but with low lapse rate coupled with inversion layers- related to any greenhouse gases- e.g. H2O- and tendency to have still air, my expectation could be wrong- the bulk of air could as cool or cooler than Mars.
It seems if it rains on this planet, it will require more generated heat- raining is disruptive: dumps heat when vapor liquifies into droplets, causing upper air to warm. But having humidity in the air should cause inversions layers, allowing cooler air above it.
Edited by gbaikie, Dec 4 2011, 07:55 PM.
|gbaikie||Dec 4 2011, 08:22 PM Post #8|
"Combining, the total thermal energy being emitted out to space is 364 * 1.8e13 = 6.58e15 Watts, or 658,000 GigaWatts!"
Hmm my guess was 100,000 1 GW nuclear reactors. The 1 GW refers to electrical power rather the total thermal energy generated [waste heat]. And since efficiency generally less than 50% I assume 1 GW nuclear plant would give 2 [or more] GW of heat.
So 200,000 GW. My guess, your calculation fairly close.
But before going on to calculate other factors.
Before humans are living on this planet, one could have very cold atmosphere and still have gaseous atmosphere.
Suppose we start with atmosphere which was 150 K [and surface 150 k]
And there was these nuclear reactors putting out 658,000 GigaWatts of heat.
How much power is needed to increase the atmosphere by 1 K if using these reactors
making 658,000 GigaWatts of heat?
Or long does it take to warm from the average temperature of 150 K to 151 K?
Btw: love the ability to edit. Too bad Judith doesn't have this function.
I am always finding errors, after clicking the post button.
Edited by gbaikie, Dec 4 2011, 08:59 PM.
|Chris Ho-Stuart||Dec 4 2011, 09:52 PM Post #9|
Differences in Cp. The DALR is equal to g/Cp. An atmosphere of mainly N2 and O2 has Cp very close to 1, and so DALR = g.
But on Mars, the atmosphere is mostly CO2, with Cp of 0.83. On Mars, therefore, g = 3.7 and DALR = 4.5
You can see this in the table of various properties I gave. (Oops. I goofed again. I actually had the right number in my spread sheet, but transcribed incorrectly. I have gone back to fix up the table in my above comment. Edit post, how I love you!)
Any condensible component of the atmosphere adds a whole new dimension to the problem. I'm going to stick with dry atmospheres, so that we can use the DALR as a guide to the environmental lapse rate. It's going to be quite hard enough to solve without adding latent heat effects.
Hmm... you estimated 20 W/m2; I gave 364 Wm/2, so there should be a factor of almost 20 difference....
Aha! I goofed! I said "6.58e15 Watts, or 658,000 GigaWatts!" That should be 6,580,000 GigaWatts. In other words, my post converted to GigaWatts incorrectly.
I'll update my post accordingly. With this correction, our guesses are indeed substantially different. But note that I omitted any impact from the atmosphere at all. With a greenhouse effect, my estimates are going to come back down again -- though not by a factor of 20, I think.
That's a different question to the spaceship; it relates to the magnitude of Earth's effective heat sink. This is dominated by the ocean. More seriously, the question must consider that as temperatures increase on a planet, it radiates more energy. If you have a very simple planet, like the starship, we can ignore "climate sensitivity" worries and just use the "Planck response".
This is how it works.
For a surface at 150K, the thermal radiation is only 28.7 W/m^2. An additional 658,000 GW, spread out over Earth's surface area of 5.1e14 m2, means an extra 1.3 W/m2, so it enable the surface to radiate 30 W/m^2. That corresponds to a temperature of 152 K.
So you get an extra 2K of temperature. How long it takes depends on the heat capacity, which on Earth will be dominated by the ocean. On the starship, the heat capacity is dominated by the atmosphere. But once you have raised that additional 2K, you have to keep those reactors going full bore just to maintain the new equilibrium. Switch them off, and the surface will cool back down to where it started.
But an additional 6.58e15 W (which I should have used!) spread over the starship surface area of 1.8e13 m2 boosts the temperature to 288 K, an additional 138 degrees. That's 15C.
In other words, the same power output gives you 10C, or boosts from -123C to 15C.
Yes indeed... and the preview feature is a great help as well. We can do quite a bit more with a bulletin board structure. I've set the edit timeout to 24 hours. If you need an edit after that, just ask. You can ask using the "report post" feature, if you like. Similarly you can report any posts that you think ought to have some kind of moderator intervention of any kind. So far there are no rules. That will change only if problems start to arise!
Edited by Chris Ho-Stuart, Dec 4 2011, 10:09 PM.
|gbaikie||Dec 5 2011, 12:28 AM Post #10|
I view it as way to check your answer. I not concerned about effective heat sink. Effective heat sink has nothing to do with spaceship as there is no night and day.
Though you do have figure out the atmosphere's effective heat sink to determine the speed in which the atmosphere will warm.
I will illustrate what I mean. Suppose we cover the planet in 2 meters water and freeze it so it's ice at 2 K.
With this done, I can add the 150 K air. And heat the air with these 6,580,000 GigaWatts reactors.
The ice I am not trying to directly heat, and I could basically ignore it. Because the air will take "forever" to warm it.
But the time it takes to warm the air will inform you about how much energy is needed.
In other words to take an extreme example, suppose these reactors heat the air and increase the temperature by 1 K an hour. Which would mean that in 10 hours [roughly] it would get 10 K warmer- or say 20 K per day. So roughly in one day atmosphere gains 20 K and ice might warm as much as 1 K [any increase in temperate of ice is a loss but the effort is focused on heating the air though such a cold surface could add some interesting dynamics. Also heating the atmosphere so rapidly would have "interesting dynamics"- it would be very turbulent- the warm air goes to top. Heating that quickly should give fairly uniform temperature [not desired-huge losses].
Now you could dampen the above mentioned losses. Insulate the ice and heat large quantities of air and mix within "furnace" so the exhaust in exits at near surface temperature and low air velocity- which is harder technically to do, but isn't the issue.
Or instead doing the frozen water thing. One could start with entire atmosphere in frozen state- have nuclear furnace circulate air thru huge ducts. Air exits reactor at 600 K or something and returns at 100 K to be reheated. Anyways- I am getting bogged down in whatever technology is used.
That doesn't seem right- you seem to be saying 2 K every second.
So, we got how much total atmosphere.
Your chart says: 2.16e22 and 5.4e17. I think it's the latter.
So 5.4e17 kg.
And need specif heat per kg of air at 150 K
".715 Joules per gram per degree Kelvin "
Ok, it's about 1 kJ/kg.K
So 5.4e17 KJ per K
Or 5.4e20 watts/joules And have 6,580,000 GigaWatts
Which is 6.58 x 10^15 watts
So about 100,000 seconds and one day is 86400 seconds
So 1 or two days per 1 K increase
You said "For a surface at 150K, the thermal radiation is only 28.7 W/m^2. An additional 658,000 GW"
Do mean requires 658,000 GW to maintain 150 K?
If so that allows 6,580,000 - 658,000 which is 5,922,000 GW to add heat when at 150 K, meaning less than 1 1/2 days to increase one K, and 15 days to get 10 K increase.
Now, if we continue to ignore the ground. What we need is average temperature of air- it could be near 150 K, though should be warmer.
Air at surface is 10 C [282 K]. If temperature drops by 1 K per 1000 meters at 100,000 meters
it's 182 K.
At what elevation is half the atmosphere?
Take average temperature for one half and take average temperature of other half- and average
Or could simply count up to 100,000 meter [it has to be much higher than this]- and say average temperature is 282 minus 50 K which 232 K.
Then to continue with rough approximation, we could use heat pumps to cool the ground to 232 K
and don't have ignore the ground anymore.
The ground could like the cold concrete floor in a basement which can have warmer air above it.
Though without heat pumps drawing away heat from the ground this planet has been in blackness of universe for billions of years and don't have much internal heat and would stay cold for long time.
Or the 10 C air would in net, lose heat warming the ground and never get very far in this regard for centuries.
Edited by Chris Ho-Stuart, Dec 5 2011, 01:43 AM.
|Chris Ho-Stuart||Dec 5 2011, 02:22 AM Post #11|
Gbaikie, I have added quote tags to your post just above, which is the usual way to quote from a previous post. You can edit your own post again to have a look at how it works.
When you enter the text, it looks like this:
While you are entering your post, there is a row of buttons above the box where you enter text. These can add formatting to your post; it will appear as "BBCode tags", which are in square brackets. One of the formatting buttons is marked "Quote", which added anonymous quote tags (no name supplied).
You can also get quote boxes like this added automatically before you start to enter a reply. Click the little check box at the bottom of the post to which you are replying, which you see at the bottom right of the comment, and then use either the reply button, or the quote button next to the check box. Try it; see what happens.
This will start of your new post with the previous post already included in quote tags. You can remove the text you don't actually want to be quoting, and you can have multiple quotes (which is what I just added for you) simply by adding the tags to end a quoted passage " [/Quote] " and another tag to start quoting again " [Quote] ". Experiment... you can always fix problems with the edit facility.
As for your questions, they all relate to the different between how long it takes to heat up by a certain amount; and how far you can heat up.
When you add a new supply of energy to the surface, it does not keep warming indefinitely. It only warms up until the thermal radiation emitted is once again equal to total energy being supplied. I have ignored how long it takes to warm up, and considered only how far it will warm up.
Not really. The time it takes to warm the air informs you about how much heat capacity there is.
In the case of Nitrogen/Oxygen mix, really the only way you can heat the air is by putting the air in physical contact with a hot plate of some kind. That's exactly the same as putting in physical contact with a layer of ice.
By the way... water becomes ice at a temperature of 273 K. This is an absolute temperature scale. So water at 2 K is going to be so cold that it will freeze the air itself!
The final temperature of anything is the temperature at which it is no longer gaining thermal energy at all.
I recommend you don't worry about how long it takes to heat up and cool down. Perhaps we could make a new thread on thermodynamics in general for such questions. The question originally posed is about the power required to maintain a certain temperature in the starship. That is adding energy to make up for energy being lost as the starship radiates thermally to space.
|gbaikie||Dec 5 2011, 06:33 AM Post #12|
yes I know- warmer it become more heat loss.
But with just 10 K increase it wouldn't that significant.
Yeah, would it cause nitrogen liquidity or snow.
Also pretty hard to cool it to 2 K. Whereas 150 K would be quite easy, 150 being temperature water ice stops evaporating in vacuum.
yeah that would be good.
Now, it seems the most significant element is average atmosphere temperature- rather than ground temperature or surface air temperature.
And next second important element would significant air mass "bordering/closest" to vacuum of space. And this second element I have no clue how to quantify.
So if second important element is not significant or could reduced somehow, then isn't first element to main factor in determining heat loss from atmosphere?
Edited by gbaikie, Dec 5 2011, 06:38 AM.
|Chris Ho-Stuart||Dec 5 2011, 10:37 AM Post #13|
As the example of the Nitrogen/Oxygen atmosphere shows, we are going to need some measure of how effectively the atmosphere blocks infrared radiation. I'll be getting to that soon. With a transparent atmosphere (as you get with a mix of only O2 and N2) the heat loss from the atmosphere is zero; all the loss is from the surface directly. The atmosphere doesn't make any difference in this case.
When there is a non-trivial atmospheric absorption (absorptivity is a function of frequency and temperature) its gets more difficult, and this is the guts of the original problem you have posed. Basically, we need a way to describe the absorptivity of the atmosphere. If the atmosphere is N2, O2 and CO2 all well mixed, then we'll need the concentration of CO2, and some absorptivity data for CO2.
|gbaikie||Dec 5 2011, 06:52 PM Post #14|
So from view of using the least amount energy [as 6,580,000 GigaWatts mind-boggling amount of energy] and in terms costs- making that many power plant costs zillions but nor is it cheap getting 5.4e17 kg of atmosphere or other elements.
It seem to me making the ground remains cold would be very easy comparatively. "Making cold" in space is far easier than it is on earth.
As I mentioned elsewhere one simply needs to not heat an area of the surface and result will be to collapse the atmosphere in that area.
Having a Niagara falls [to put it mildly] of the atmosphere, adds it's own complication, but the benefit is it gives is unlimited source "portable" cold. If you need rivers of liquid nitrogen that is doable.
The point is a model to understand climate, but I think being able to change variables provides different view and I think the benefit in terms of a starship of having some portion of planet have collapsed atmosphere would dictate having it.
Keeping the condition the same- 10 C surface temperature, global atmosphere, 1/10th gravity, and 1/3rd earth's atmospheric pressure is desired.
But also asking how you "save in costs", if say you have -20 C instead of 10 C; What happens if surface ground is much colder than 10 C; What if there was 1/5th instead of 1/3rd atmosphere is
I believe [perhaps as a crazy religion] that a purpose of science is to lower costs- science makes technology [and technology makes science]. Engineers are all about lowering costs. Exploration challenges science and technology [and by itself adds knowledge]. That democracy, science, exploration, freedom, trade, and happiness are an ecology- are interconnected and inalienable.
Therefore whatever is cheaper or more economical shouldn't be ignored- or is a tradition that shouldn't be dropped. And I am space cadet and an idea of some huge spaceship is too fascinating.
Anyways, your statement: "With a transparent atmosphere (as you get with a mix of only O2 and N2) the heat loss from the atmosphere is zero."
Is interesting. One aspect of having a large atmosphere was the idea of reducing loss of heat from the surface. It would great to have full atmosphere, but that seemed to make too complicated on 1/10th gravity world [I think the sheer distance/height of atmosphere would bring into play the gravity difference as factor. Even with this dwarf planet I am uncertain the molecule speed will not exceed it's gravity- and more atmosphere makes it more uncertain.].
And be easier if smaller world, but wanted enough gravity to get as much atmosphere as possible. Btw, this world's atmosphere is unlikely to survive near the sun [earth distance]- or at least it adds a lot of more complications and unknowns- like when would atmosphere be blown away by the sun within a year or decades/centuries
So it's your contention that there little or no difference if one had 1/5th earth pressure
as compared to 1/3?
One could also perhaps make it 1/2 earth atmospheric pressure. Half an Earth pressure could give atmospheric height perhaps as high as Neptune [though Neptune is massive and has much thicker atmosphere]. And the models of Neptune atmosphere are not very well defined.
Oh, I just thought of something, what does dust do in terms of heat. Low gravity world- lots of dust is fairly easy. It would interfere with beautiful view of the universe, but one could have as much dust as forest fire or volcano fairly easily.
Edited by gbaikie, Dec 5 2011, 06:57 PM.
|Chris Ho-Stuart||Dec 7 2011, 04:33 AM Post #15|
IF the atmosphere is transparent.... then yes. The pressure is irrelevant in this case.
Remember, the atmosphere is not itself a source of energy. It can only "insulate" somehow; moderate the flow of heat from the surface to space. If the atmosphere is transparent, the the thermal radiation from the surface just goes straight out to space without loss. That is what being transparent means.
There are other flows of energy -- convection, conduction, etc, but all that can do is bring the temperature of the atmosphere into equilibrium with the temperature of the surface. The atmosphere has no way of passing any energy on out in to space because it is transparent. It neither absorbs nor emits thermal radiation. So when the atmosphere is transparent, the convection in the atmosphere will be very sluggish. There can be no large sustained flow of energy into the atmosphere because the atmosphere can't get rid of it again.
You must have some way for the atmosphere to actually absorb some of that thermal radiation from the surface before it can have any impact at all on the surface temperature.
This is a really fundamental point, and it explains why I am going to have to add some CO2 (or any other greenhouse gas) into the atmosphere and calculate the transparency in order to make any difference to the calculation of 6.58e15 Watts.
If this point is at all confusing, then what I am going to do next will be hard to follow. It's important to grasp why I have to do this to answer your original question; or at least have that clearly identified as a point of dispute or question.
Cheers -- Chris
PS. I'm being a bit slow, because I am trying to fix up a few things in the operation of the board itself. Judy has offered to us a guest thread at Climate Etc, and I have suggested your problem as a useful aspect of such a thread. But I'd also like the board here to be in a good state to accept an influx. Might take me a week. When we get to that point, I'd be happy to have input from you or others on how the initial blog post is put together!
|1 user reading this topic (1 Guest and 0 Anonymous)|
|Go to Next Page|
|« Previous Topic · Physical theory for climate · Next Topic »|