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Even the moon HAS a GHE


Even the moon HAS a GHE25-11-2017 05:02
Leitwolf
★☆☆☆☆
(87)
Well it is easy to tell, what the average surface temperature (AST) of the moon should be. With an albedo of about 0.13 and the other well known parameters, we can determine: ((1368/4)*((1-0.13)/5.67E-8))^0.25 = 269K.

The pivotal question is however, what is the AST of the moon in practice?
Wikipedia tells the AST at the equator is about 220K, with extremes reaching from 100 to 390K. The total average may be as low as 205, or 200K.

These figures make any comparisson between theoretical and observed temperatures obsolete. 269 > 200, that is all we can say. So why is there such a strong deviation? Is there an anti-GHE on the moon?

The problem is pretty simple and obvious. Emissions obhere to the Stefan Boltzmann law, and will increase by the power of 4 according to temperature. With the sun in zenith, the moon reaches 390K. As compared to some 90K on the dark side, there will be (390/90)^4 = 350times as much emissions. So the moon emits allmost all of the energy it absorbs from its hot day side.

The question should be, at what even temperature the moon would emit as much energy as it does, given the very uneven temperature distribution. That would be a geometric average.

To give a simplified example. We have two equally sized objects (or surfaces) with temperatures of 100K and 400K respectively. The arithmetic average then would be (400+100)/2= 250K, obviously. The geometric average however is ((400^4+100^4)/2)^0.25 = 336,7K. So both objects with the original temperatures will emit just as much energy, as if both had 336.7K, which is substantially higher than 250K.

So that is exactly what we need to do with the AST of the moon. The temperature on the moon resembles almost perfectly the 4th root of a sinus curve. The sinus curve would represent the amount of solar radiation, the fourth root represents the Stefan Boltzmann law.

Here is an illustration:



As named before, the temperature rises to 390K if the sun is in the zenith. At a 45° angle, it will be sin(45°)^0.25 = 357,6K, which represents the 0° latitude curve at 15:00 or 9:00 in the graph. Just as much, at a latitude of 60°, there should be a maximum temperature of sin(90-60)^0.25*390 = 328K

With all that in mind, and a little help from Excel, it is not hard to calculate the actual geometric average, which is about 328K for the illuminated side.

As the dark side is very cold and hardly contributes to overall emissions, there is no need to do much of investigation into that part. We can simply assume it would have like 100K all over. So it is ((328^4+100^4)/2)^0.25 = 276.4K.

Now that is a figure we can actually compare to the theoretical 269K, named earlier. The moon is indeed substantially warmer than it should be. In other words, we have a GHE on the moon, but no greenhouse, or atmosphere.

There is even a more simple approach coming to the same result. Allowing for the albedo, the maximum temperature should be (1368*((1-0,13)/5,67E-8))^0,25 = 380.6K. Again the observed 390K around the equator are substantially higher. However we can get a much better approximation if we skip the albedo. (1368*(1/5,67E-8))^0,25 = 394K.

So the moon is almost as warm as a perfect black body. This is only logical, as emissivity tends to equal absorptivity. A fact widely ignored by the GH-theory, and an error falsly producing GHEs in the first place.

Yet there is another problem. The surface of the moon is bigger than that of a sphere with the same diameter. The moon is not s sphere, its surface is not smooth. Rather there are valleys and mountains, and even on a micro level, the surface is simply not smooth.

This is a problem we may ignore on earth, where we mainly measure air temperatures. In the absence of air, the excessive amount of surface will produce slightly lower temperatures than in theory. I have no clue how much more surface there is, as compared to an ideal sphere. But this factor will, by no doubt, move the effective temperature of the moon even closer to that of a perfect black body, which is about 279K of course.

So as we can determine a GHE even for the moon, where there definitely is no greenhouse, we learn about the major and fundamental flaw of the basic approach.
25-11-2017 05:11
Into the Night
★★★★★
(5558)
We do not know the albedo (emissivity) of the Moon. It is not possible to determine it without knowing accurately the temperature of the Moon in the first place.

It is not possible to determine the temperature of the Moon. There are not enough instruments there to even begin a statistical analysis that makes any sense.

It is the same with the Earth. It is the same for Venus. It is the same for any other planet.

Emissivity varies dramatically in the space of fractions of an inch both on the Moon and on Earth. We do not know it's average value.
25-11-2017 05:39
litesong
★★★★★
(2297)
"old sick silly sleepy sleepy sleezy slimy steenkin' filthy vile reprobate rooting (& rotting) racist pukey proud pig AGW denier liar whiner badnight" bluffed:...not possible to determine the temperature of the Moon....
Correction:
The ego of "old sick silly sleepy sleepy sleezy slimy steenkin' filthy vile reprobate rooting (& rotting) racist pukey proud pig AGW denier liar whiner badnight" admits that if it can't determine the temperature of the Moon, no one can.
Heat water with "badnight's" hand in it, to 117degF & see if "old sick silly sleepy sleepy sleezy slimy steenkin' filthy vile reprobate rooting (& rotting) racist pukey proud pig AGW denier liar whiner badnight" can determine if its skin is scalded.
Edited on 25-11-2017 06:37
27-11-2017 20:28
Wake
★★★★★
(3396)
Leitwolf wrote:
Well it is easy to tell, what the average surface temperature (AST) of the moon should be. With an albedo of about 0.13 and the other well known parameters, we can determine: ((1368/4)*((1-0.13)/5.67E-8))^0.25 = 269K.

The pivotal question is however, what is the AST of the moon in practice?
Wikipedia tells the AST at the equator is about 220K, with extremes reaching from 100 to 390K. The total average may be as low as 205, or 200K.

These figures make any comparisson between theoretical and observed temperatures obsolete. 269 > 200, that is all we can say. So why is there such a strong deviation? Is there an anti-GHE on the moon?

The problem is pretty simple and obvious. Emissions obhere to the Stefan Boltzmann law, and will increase by the power of 4 according to temperature. With the sun in zenith, the moon reaches 390K. As compared to some 90K on the dark side, there will be (390/90)^4 = 350times as much emissions. So the moon emits allmost all of the energy it absorbs from its hot day side.

The question should be, at what even temperature the moon would emit as much energy as it does, given the very uneven temperature distribution. That would be a geometric average.

To give a simplified example. We have two equally sized objects (or surfaces) with temperatures of 100K and 400K respectively. The arithmetic average then would be (400+100)/2= 250K, obviously. The geometric average however is ((400^4+100^4)/2)^0.25 = 336,7K. So both objects with the original temperatures will emit just as much energy, as if both had 336.7K, which is substantially higher than 250K.

So that is exactly what we need to do with the AST of the moon. The temperature on the moon resembles almost perfectly the 4th root of a sinus curve. The sinus curve would represent the amount of solar radiation, the fourth root represents the Stefan Boltzmann law.

Here is an illustration:



As named before, the temperature rises to 390K if the sun is in the zenith. At a 45° angle, it will be sin(45°)^0.25 = 357,6K, which represents the 0° latitude curve at 15:00 or 9:00 in the graph. Just as much, at a latitude of 60°, there should be a maximum temperature of sin(90-60)^0.25*390 = 328K

With all that in mind, and a little help from Excel, it is not hard to calculate the actual geometric average, which is about 328K for the illuminated side.

As the dark side is very cold and hardly contributes to overall emissions, there is no need to do much of investigation into that part. We can simply assume it would have like 100K all over. So it is ((328^4+100^4)/2)^0.25 = 276.4K.

Now that is a figure we can actually compare to the theoretical 269K, named earlier. The moon is indeed substantially warmer than it should be. In other words, we have a GHE on the moon, but no greenhouse, or atmosphere.

There is even a more simple approach coming to the same result. Allowing for the albedo, the maximum temperature should be (1368*((1-0,13)/5,67E-8))^0,25 = 380.6K. Again the observed 390K around the equator are substantially higher. However we can get a much better approximation if we skip the albedo. (1368*(1/5,67E-8))^0,25 = 394K.

So the moon is almost as warm as a perfect black body. This is only logical, as emissivity tends to equal absorptivity. A fact widely ignored by the GH-theory, and an error falsly producing GHEs in the first place.

Yet there is another problem. The surface of the moon is bigger than that of a sphere with the same diameter. The moon is not s sphere, its surface is not smooth. Rather there are valleys and mountains, and even on a micro level, the surface is simply not smooth.

This is a problem we may ignore on earth, where we mainly measure air temperatures. In the absence of air, the excessive amount of surface will produce slightly lower temperatures than in theory. I have no clue how much more surface there is, as compared to an ideal sphere. But this factor will, by no doubt, move the effective temperature of the moon even closer to that of a perfect black body, which is about 279K of course.

So as we can determine a GHE even for the moon, where there definitely is no greenhouse, we learn about the major and fundamental flaw of the basic approach.


That is a very nice determination. You could add that the reason that we don't have to worry about albedo of Earth is because the troposphere is about 25 km thick. This causes almost perfect mixing and hence a temperature in the stratosphere that allows (with the help of solar output) to radiate all of our received emissions back into deep space.

We know that we have 340.4 watts/meter^2 applied to the outer atmosphere and we know that what is lost is 339.8 making this 99.82% efficient or almost the perfect black body. The extremely small amount of energy left behind is used by plants as a growth medium.

GHE are entirely due to H2O in it's various phases in the atmosphere trapping heat or more accurately redirecting it. The maximum levels of H2O cannot increase from what it presently is simply because of the weight of H2O and the density of the atmosphere. CO2 in any case has such a narrow absorption frequency band that it has reached saturation somewhere around 250-280 ppm. At this point no increase in CO2 would cause any significant increase in absorption. (there would be a slight increase at any levels because of local increases in the absorption bands of CO2) This would be easy to test with the research grant to do so and you see that they are intent on not doing so.

My congratulations on a well considered calculation.
27-11-2017 20:57
Into the Night
★★★★★
(5558)
Wake wrote:
Leitwolf wrote:
Well it is easy to tell, what the average surface temperature (AST) of the moon should be. With an albedo of about 0.13 and the other well known parameters, we can determine: ((1368/4)*((1-0.13)/5.67E-8))^0.25 = 269K.

The pivotal question is however, what is the AST of the moon in practice?
Wikipedia tells the AST at the equator is about 220K, with extremes reaching from 100 to 390K. The total average may be as low as 205, or 200K.

These figures make any comparisson between theoretical and observed temperatures obsolete. 269 > 200, that is all we can say. So why is there such a strong deviation? Is there an anti-GHE on the moon?

The problem is pretty simple and obvious. Emissions obhere to the Stefan Boltzmann law, and will increase by the power of 4 according to temperature. With the sun in zenith, the moon reaches 390K. As compared to some 90K on the dark side, there will be (390/90)^4 = 350times as much emissions. So the moon emits allmost all of the energy it absorbs from its hot day side.

The question should be, at what even temperature the moon would emit as much energy as it does, given the very uneven temperature distribution. That would be a geometric average.

To give a simplified example. We have two equally sized objects (or surfaces) with temperatures of 100K and 400K respectively. The arithmetic average then would be (400+100)/2= 250K, obviously. The geometric average however is ((400^4+100^4)/2)^0.25 = 336,7K. So both objects with the original temperatures will emit just as much energy, as if both had 336.7K, which is substantially higher than 250K.

So that is exactly what we need to do with the AST of the moon. The temperature on the moon resembles almost perfectly the 4th root of a sinus curve. The sinus curve would represent the amount of solar radiation, the fourth root represents the Stefan Boltzmann law.

Here is an illustration:



As named before, the temperature rises to 390K if the sun is in the zenith. At a 45° angle, it will be sin(45°)^0.25 = 357,6K, which represents the 0° latitude curve at 15:00 or 9:00 in the graph. Just as much, at a latitude of 60°, there should be a maximum temperature of sin(90-60)^0.25*390 = 328K

With all that in mind, and a little help from Excel, it is not hard to calculate the actual geometric average, which is about 328K for the illuminated side.

As the dark side is very cold and hardly contributes to overall emissions, there is no need to do much of investigation into that part. We can simply assume it would have like 100K all over. So it is ((328^4+100^4)/2)^0.25 = 276.4K.

Now that is a figure we can actually compare to the theoretical 269K, named earlier. The moon is indeed substantially warmer than it should be. In other words, we have a GHE on the moon, but no greenhouse, or atmosphere.

There is even a more simple approach coming to the same result. Allowing for the albedo, the maximum temperature should be (1368*((1-0,13)/5,67E-8))^0,25 = 380.6K. Again the observed 390K around the equator are substantially higher. However we can get a much better approximation if we skip the albedo. (1368*(1/5,67E-8))^0,25 = 394K.

So the moon is almost as warm as a perfect black body. This is only logical, as emissivity tends to equal absorptivity. A fact widely ignored by the GH-theory, and an error falsly producing GHEs in the first place.

Yet there is another problem. The surface of the moon is bigger than that of a sphere with the same diameter. The moon is not s sphere, its surface is not smooth. Rather there are valleys and mountains, and even on a micro level, the surface is simply not smooth.

This is a problem we may ignore on earth, where we mainly measure air temperatures. In the absence of air, the excessive amount of surface will produce slightly lower temperatures than in theory. I have no clue how much more surface there is, as compared to an ideal sphere. But this factor will, by no doubt, move the effective temperature of the moon even closer to that of a perfect black body, which is about 279K of course.

So as we can determine a GHE even for the moon, where there definitely is no greenhouse, we learn about the major and fundamental flaw of the basic approach.


That is a very nice determination. You could add that the reason that we don't have to worry about albedo of Earth is because the troposphere is about 25 km thick.

The albedo of the Earth is not affected by the thickness of the troposphere at all. The troposphere is only about 9km thick (13km near the equator).
Wake wrote:
This causes almost perfect mixing

Thickness of the troposphere does not affect mixing. The atmosphere is not uniform, even in the troposphere. It if was, we wouldn't have storms.
Wake wrote:
and hence a temperature in the stratosphere that allows (with the help of solar output) to radiate all of our received emissions back into deep space.

Most radiance from the Earth comes from the surface, not the atmosphere. ALL of the atmosphere also radiates. There is no magick point in the atmosphere where radiance suddenly is possible.
Wake wrote:
We know that we have 340.4 watts/meter^2 applied to the outer atmosphere and we know that what is lost is 339.8 making this 99.82% efficient or almost the perfect black body. The extremely small amount of energy left behind is used by plants as a growth medium.

We don't know the albedo (emissivity) of Earth.
Wake wrote:
GHE are entirely due to H2O in it's various phases in the atmosphere trapping heat

There is no such thing as a 'greenhouse' gas. Not even H2O can do it. It is not possible to trap heat.
Wake wrote:
or more accurately redirecting it.

It is not possible to heat a warmer surface using a colder gas. You are violating the 2nd law of thermodynamics. You can't make hot coffee with an ice cube.
Wake wrote:
The maximum levels of H2O cannot increase from what it presently is simply because of the weight of H2O and the density of the atmosphere.

H2O has a molecular weight of 18. Oxygen (O2) has a molecular weight of 32. Nitrogen (N2) has a molecular weight of 28. Carbon dioxide has a molecular weight of 44.

H2O is has a lighter molecular weight than the bulk of the atmosphere. Something is wrong with your model. What is it?

Wake wrote:
CO2 in any case has such a narrow absorption frequency band that it has reached saturation somewhere around 250-280 ppm. At this point no increase in CO2 would cause any significant increase in absorption. (there would be a slight increase at any levels because of local increases in the absorption bands of CO2) This would be easy to test with the research grant to do so and you see that they are intent on not doing so.

CO2, by absorbing some frequencies of infrared light, help to COOL the surface, not warm it. It is just another way for the surface to heat the atmosphere.
Wake wrote:
My congratulations on a well considered calculation.

Well considered? The only thing this calculation does is show the Stefan-Boltzmann law in action, regardless of the actual emissivity. The conclusion that he jumps to is without substance.


The Parrot Killer
28-11-2017 22:34
litesong
★★★★★
(2297)
"old sick silly sleepy sleezy slimy steenkin' filthy vile reprobate rooting (& rotting) racist pukey proud pig AGW denier liar whiner & many time (plus 1) threatener wake-me-up" wiffed:... entirely due to...
The character of "old sick silly sleepy sleezy slimy steenkin' filthy vile reprobate rooting (& rotting) racist pukey proud pig AGW denier liar whiner & many time (plus 1) threatener wake-me-up" is almost entirely due to old sick silly sleepy sleezy slimy steenkin' filthy vile reprobate rooting (& rotting) racist pukey proud pig AGW denier liar whinerisms & many time (plus 1) threats.
29-11-2017 00:52
Wake
★★★★★
(3396)
Into the Night wrote:
Wake wrote:
Leitwolf wrote:
Well it is easy to tell, what the average surface temperature (AST) of the moon should be. With an albedo of about 0.13 and the other well known parameters, we can determine: ((1368/4)*((1-0.13)/5.67E-8))^0.25 = 269K.

The pivotal question is however, what is the AST of the moon in practice?
Wikipedia tells the AST at the equator is about 220K, with extremes reaching from 100 to 390K. The total average may be as low as 205, or 200K.

These figures make any comparisson between theoretical and observed temperatures obsolete. 269 > 200, that is all we can say. So why is there such a strong deviation? Is there an anti-GHE on the moon?

The problem is pretty simple and obvious. Emissions obhere to the Stefan Boltzmann law, and will increase by the power of 4 according to temperature. With the sun in zenith, the moon reaches 390K. As compared to some 90K on the dark side, there will be (390/90)^4 = 350times as much emissions. So the moon emits allmost all of the energy it absorbs from its hot day side.

The question should be, at what even temperature the moon would emit as much energy as it does, given the very uneven temperature distribution. That would be a geometric average.

To give a simplified example. We have two equally sized objects (or surfaces) with temperatures of 100K and 400K respectively. The arithmetic average then would be (400+100)/2= 250K, obviously. The geometric average however is ((400^4+100^4)/2)^0.25 = 336,7K. So both objects with the original temperatures will emit just as much energy, as if both had 336.7K, which is substantially higher than 250K.

So that is exactly what we need to do with the AST of the moon. The temperature on the moon resembles almost perfectly the 4th root of a sinus curve. The sinus curve would represent the amount of solar radiation, the fourth root represents the Stefan Boltzmann law.

Here is an illustration:



As named before, the temperature rises to 390K if the sun is in the zenith. At a 45° angle, it will be sin(45°)^0.25 = 357,6K, which represents the 0° latitude curve at 15:00 or 9:00 in the graph. Just as much, at a latitude of 60°, there should be a maximum temperature of sin(90-60)^0.25*390 = 328K

With all that in mind, and a little help from Excel, it is not hard to calculate the actual geometric average, which is about 328K for the illuminated side.

As the dark side is very cold and hardly contributes to overall emissions, there is no need to do much of investigation into that part. We can simply assume it would have like 100K all over. So it is ((328^4+100^4)/2)^0.25 = 276.4K.

Now that is a figure we can actually compare to the theoretical 269K, named earlier. The moon is indeed substantially warmer than it should be. In other words, we have a GHE on the moon, but no greenhouse, or atmosphere.

There is even a more simple approach coming to the same result. Allowing for the albedo, the maximum temperature should be (1368*((1-0,13)/5,67E-8))^0,25 = 380.6K. Again the observed 390K around the equator are substantially higher. However we can get a much better approximation if we skip the albedo. (1368*(1/5,67E-8))^0,25 = 394K.

So the moon is almost as warm as a perfect black body. This is only logical, as emissivity tends to equal absorptivity. A fact widely ignored by the GH-theory, and an error falsly producing GHEs in the first place.

Yet there is another problem. The surface of the moon is bigger than that of a sphere with the same diameter. The moon is not s sphere, its surface is not smooth. Rather there are valleys and mountains, and even on a micro level, the surface is simply not smooth.

This is a problem we may ignore on earth, where we mainly measure air temperatures. In the absence of air, the excessive amount of surface will produce slightly lower temperatures than in theory. I have no clue how much more surface there is, as compared to an ideal sphere. But this factor will, by no doubt, move the effective temperature of the moon even closer to that of a perfect black body, which is about 279K of course.

So as we can determine a GHE even for the moon, where there definitely is no greenhouse, we learn about the major and fundamental flaw of the basic approach.


That is a very nice determination. You could add that the reason that we don't have to worry about albedo of Earth is because the troposphere is about 25 km thick.

The albedo of the Earth is not affected by the thickness of the troposphere at all. The troposphere is only about 9km thick (13km near the equator).
Wake wrote:
This causes almost perfect mixing

Thickness of the troposphere does not affect mixing. The atmosphere is not uniform, even in the troposphere. It if was, we wouldn't have storms.
Wake wrote:
and hence a temperature in the stratosphere that allows (with the help of solar output) to radiate all of our received emissions back into deep space.

Most radiance from the Earth comes from the surface, not the atmosphere. ALL of the atmosphere also radiates. There is no magick point in the atmosphere where radiance suddenly is possible.
Wake wrote:
We know that we have 340.4 watts/meter^2 applied to the outer atmosphere and we know that what is lost is 339.8 making this 99.82% efficient or almost the perfect black body. The extremely small amount of energy left behind is used by plants as a growth medium.

We don't know the albedo (emissivity) of Earth.
Wake wrote:
GHE are entirely due to H2O in it's various phases in the atmosphere trapping heat

There is no such thing as a 'greenhouse' gas. Not even H2O can do it. It is not possible to trap heat.
Wake wrote:
or more accurately redirecting it.

It is not possible to heat a warmer surface using a colder gas. You are violating the 2nd law of thermodynamics. You can't make hot coffee with an ice cube.
Wake wrote:
The maximum levels of H2O cannot increase from what it presently is simply because of the weight of H2O and the density of the atmosphere.

H2O has a molecular weight of 18. Oxygen (O2) has a molecular weight of 32. Nitrogen (N2) has a molecular weight of 28. Carbon dioxide has a molecular weight of 44.

H2O is has a lighter molecular weight than the bulk of the atmosphere. Something is wrong with your model. What is it?

Wake wrote:
CO2 in any case has such a narrow absorption frequency band that it has reached saturation somewhere around 250-280 ppm. At this point no increase in CO2 would cause any significant increase in absorption. (there would be a slight increase at any levels because of local increases in the absorption bands of CO2) This would be easy to test with the research grant to do so and you see that they are intent on not doing so.

CO2, by absorbing some frequencies of infrared light, help to COOL the surface, not warm it. It is just another way for the surface to heat the atmosphere.
Wake wrote:
My congratulations on a well considered calculation.

Well considered? The only thing this calculation does is show the Stefan-Boltzmann law in action, regardless of the actual emissivity. The conclusion that he jumps to is without substance.


You continue to demonstrate the intellectual capacity of a frankfurter smothered in mustard. The difference in the hottest and coldest temperatures EVER recorded on Earth is less than 1% of average surface temperature measured in Kelvin. But with your total ignorance that probably went above your head.

The depth of the troposphere in mid latitudes in the summer is between 20 and 25 km. You really have to learn that there is a difference between a mile and a kilometer. But I doubt you're capable.

Demonstrating again your incredible inability to think you do not understand that the top of the tropopause is about 210 Kelvin along it's entire surface due to the mixing of the air in the lower troposphere and the expansion of the rising air or reduction in pressure into the tropopause.

Your total ignorance also seems to miss the fact that WEATHER occurs in general half way up in the troposphere though extreme weather events can happen into the bottom of the stratosphere. These are unusual events. And they only occur in the middle latitudes.

And your stupidity in not understanding what clouds are and why they are typically around middle altitudes when they carry liquid water and clouds present in the stratosphere are nothing but ice would make just about any grade school child wonder if you ever went to school at all.

Can you offer us any explanation of why you shoot your mouth off in a continuous array of ignorance that would embarrass a sloth? Why do you continue to spray your total lack of intelligence around? You do not understand the Stefan-Boltzmann equation and insist that you do. You most assuredly do not understand the laws of thermodynamics and it is my impression that you are incapable of doing so.

Tell us what a calorimeter is. You're the one that threw that into the ring. If it's more than just a word that sounds "sciency" to you then why are you not telling us what it is. Perhaps you should look up "calorie"?
29-11-2017 02:24
Into the Night
★★★★★
(5558)
Wake wrote:
Into the Night wrote:
Wake wrote:
Leitwolf wrote:
Well it is easy to tell, what the average surface temperature (AST) of the moon should be. With an albedo of about 0.13 and the other well known parameters, we can determine: ((1368/4)*((1-0.13)/5.67E-8))^0.25 = 269K.

The pivotal question is however, what is the AST of the moon in practice?
Wikipedia tells the AST at the equator is about 220K, with extremes reaching from 100 to 390K. The total average may be as low as 205, or 200K.

These figures make any comparisson between theoretical and observed temperatures obsolete. 269 > 200, that is all we can say. So why is there such a strong deviation? Is there an anti-GHE on the moon?

The problem is pretty simple and obvious. Emissions obhere to the Stefan Boltzmann law, and will increase by the power of 4 according to temperature. With the sun in zenith, the moon reaches 390K. As compared to some 90K on the dark side, there will be (390/90)^4 = 350times as much emissions. So the moon emits allmost all of the energy it absorbs from its hot day side.

The question should be, at what even temperature the moon would emit as much energy as it does, given the very uneven temperature distribution. That would be a geometric average.

To give a simplified example. We have two equally sized objects (or surfaces) with temperatures of 100K and 400K respectively. The arithmetic average then would be (400+100)/2= 250K, obviously. The geometric average however is ((400^4+100^4)/2)^0.25 = 336,7K. So both objects with the original temperatures will emit just as much energy, as if both had 336.7K, which is substantially higher than 250K.

So that is exactly what we need to do with the AST of the moon. The temperature on the moon resembles almost perfectly the 4th root of a sinus curve. The sinus curve would represent the amount of solar radiation, the fourth root represents the Stefan Boltzmann law.

Here is an illustration:



As named before, the temperature rises to 390K if the sun is in the zenith. At a 45° angle, it will be sin(45°)^0.25 = 357,6K, which represents the 0° latitude curve at 15:00 or 9:00 in the graph. Just as much, at a latitude of 60°, there should be a maximum temperature of sin(90-60)^0.25*390 = 328K

With all that in mind, and a little help from Excel, it is not hard to calculate the actual geometric average, which is about 328K for the illuminated side.

As the dark side is very cold and hardly contributes to overall emissions, there is no need to do much of investigation into that part. We can simply assume it would have like 100K all over. So it is ((328^4+100^4)/2)^0.25 = 276.4K.

Now that is a figure we can actually compare to the theoretical 269K, named earlier. The moon is indeed substantially warmer than it should be. In other words, we have a GHE on the moon, but no greenhouse, or atmosphere.

There is even a more simple approach coming to the same result. Allowing for the albedo, the maximum temperature should be (1368*((1-0,13)/5,67E-8))^0,25 = 380.6K. Again the observed 390K around the equator are substantially higher. However we can get a much better approximation if we skip the albedo. (1368*(1/5,67E-8))^0,25 = 394K.

So the moon is almost as warm as a perfect black body. This is only logical, as emissivity tends to equal absorptivity. A fact widely ignored by the GH-theory, and an error falsly producing GHEs in the first place.

Yet there is another problem. The surface of the moon is bigger than that of a sphere with the same diameter. The moon is not s sphere, its surface is not smooth. Rather there are valleys and mountains, and even on a micro level, the surface is simply not smooth.

This is a problem we may ignore on earth, where we mainly measure air temperatures. In the absence of air, the excessive amount of surface will produce slightly lower temperatures than in theory. I have no clue how much more surface there is, as compared to an ideal sphere. But this factor will, by no doubt, move the effective temperature of the moon even closer to that of a perfect black body, which is about 279K of course.

So as we can determine a GHE even for the moon, where there definitely is no greenhouse, we learn about the major and fundamental flaw of the basic approach.


That is a very nice determination. You could add that the reason that we don't have to worry about albedo of Earth is because the troposphere is about 25 km thick.

The albedo of the Earth is not affected by the thickness of the troposphere at all. The troposphere is only about 9km thick (13km near the equator).
Wake wrote:
This causes almost perfect mixing

Thickness of the troposphere does not affect mixing. The atmosphere is not uniform, even in the troposphere. It if was, we wouldn't have storms.
Wake wrote:
and hence a temperature in the stratosphere that allows (with the help of solar output) to radiate all of our received emissions back into deep space.

Most radiance from the Earth comes from the surface, not the atmosphere. ALL of the atmosphere also radiates. There is no magick point in the atmosphere where radiance suddenly is possible.
Wake wrote:
We know that we have 340.4 watts/meter^2 applied to the outer atmosphere and we know that what is lost is 339.8 making this 99.82% efficient or almost the perfect black body. The extremely small amount of energy left behind is used by plants as a growth medium.

We don't know the albedo (emissivity) of Earth.
Wake wrote:
GHE are entirely due to H2O in it's various phases in the atmosphere trapping heat

There is no such thing as a 'greenhouse' gas. Not even H2O can do it. It is not possible to trap heat.
Wake wrote:
or more accurately redirecting it.

It is not possible to heat a warmer surface using a colder gas. You are violating the 2nd law of thermodynamics. You can't make hot coffee with an ice cube.
Wake wrote:
The maximum levels of H2O cannot increase from what it presently is simply because of the weight of H2O and the density of the atmosphere.

H2O has a molecular weight of 18. Oxygen (O2) has a molecular weight of 32. Nitrogen (N2) has a molecular weight of 28. Carbon dioxide has a molecular weight of 44.

H2O is has a lighter molecular weight than the bulk of the atmosphere. Something is wrong with your model. What is it?

Wake wrote:
CO2 in any case has such a narrow absorption frequency band that it has reached saturation somewhere around 250-280 ppm. At this point no increase in CO2 would cause any significant increase in absorption. (there would be a slight increase at any levels because of local increases in the absorption bands of CO2) This would be easy to test with the research grant to do so and you see that they are intent on not doing so.

CO2, by absorbing some frequencies of infrared light, help to COOL the surface, not warm it. It is just another way for the surface to heat the atmosphere.
Wake wrote:
My congratulations on a well considered calculation.

Well considered? The only thing this calculation does is show the Stefan-Boltzmann law in action, regardless of the actual emissivity. The conclusion that he jumps to is without substance.


...deleted Mantras 2...2...1...
The difference in the hottest and coldest temperatures EVER recorded on Earth is less than 1% of average surface temperature measured in Kelvin.

So? The choice of scale makes no difference.
Wake wrote:
...deleted Mantras 1...
The depth of the troposphere in mid latitudes in the summer is between 20 and 25 km. You really have to learn that there is a difference between a mile and a kilometer.

Nope. The troposphere is about 9km high (not depth). Otherwise about 6 miles. This altitude is typical around the 45th parallel. Go look it up.
Wake wrote:
...deleted Mantras 1...2...
the top of the tropopause is about 210 Kelvin along it's entire surface due to the mixing of the air in the lower troposphere and the expansion of the rising air or reduction in pressure into the tropopause.

The temperature of the tropopause is not uniform. It is not cold because of lack of pressure. It is cold because an endothermic chemical reaction is taking place there. It is the coldest point in our atmosphere. Temperature begins to INCREASE above the tropopause, where the air is even thinner.
Wake wrote:
...deleted Mantras 2...WEATHER occurs in general half way up in the troposphere though extreme weather events can happen into the bottom of the stratosphere.

WRONG. Weather occurs all through the troposphere and into the stratosphere. It can occur at ground level as well (and does).
Wake wrote:
These are unusual events. And they only occur in the middle latitudes.

Weather in the stratosphere is not unusual. It occurs at all latitudes.
Wake wrote:
...deleted Mantras 2...1...
what clouds are and why they are typically around middle altitudes when they carry liquid water

Clouds don't 'carry' water. They ARE water.
Wake wrote:
and clouds present in the stratosphere are nothing but ice

...and your point?? You seem to be making up all kinds of stuff I supposedly said.
Wake wrote:
...deleted Mantras 1...2...1...1...2...2...1...
the Stefan-Boltzmann equation and insist that you do. ...the laws of thermodynamics

The Stefan Boltzmann law is:

radiance = SBconstant * emissivity * temperature ^ 4.

What about this do you not understand?

Heat flows from hot to cold. What about that do you not understand?

Wake wrote:
Tell us what a calorimeter is. You're the one that threw that into the ring.

Go look it up. You obviously never attended a physics course. You build one in a lab there.
Wake wrote:
If it's more than just a word that sounds "sciency" to you then why are you not telling us what it is. Perhaps you should look up "calorie"?

Which one are you interested in?


The Parrot Killer
29-11-2017 15:31
Wake
★★★★★
(3396)
Into the Night wrote:
Wake wrote:
If it's more than just a word that sounds "sciency" to you then why are you not telling us what it is. Perhaps you should look up "calorie"?

Which one are you interested in?


This is getting better and better all the time. You opened your mouth and stuck your foot in it and now will not remove it.
29-11-2017 16:40
litesong
★★★★★
(2297)
"old sick silly sleepy sleezy slimy slimebag steenkin' filthy vile reprobate rooting (& rotting) racist pukey proud pig AGW denier liar whiner badnight" signals: Clouds don't 'carry' water. They ARE water.

https://upload.wikimedia.org/wikipedia/commons/a/a3/Cumulus_clouds_panorama.jpg
"I've looked at clouds from both sides now."---Joni Mitchell
29-11-2017 21:32
Into the Night
★★★★★
(5558)
Wake wrote:
Into the Night wrote:
Wake wrote:
If it's more than just a word that sounds "sciency" to you then why are you not telling us what it is. Perhaps you should look up "calorie"?

Which one are you interested in?


This is getting better and better all the time. You opened your mouth and stuck your foot in it and now will not remove it.


Apparently you know nothing about calories either.


The Parrot Killer
29-11-2017 22:16
Wake
★★★★★
(3396)
Into the Night wrote:
Wake wrote:
Into the Night wrote:
Wake wrote:
If it's more than just a word that sounds "sciency" to you then why are you not telling us what it is. Perhaps you should look up "calorie"?

Which one are you interested in?


This is getting better and better all the time. You opened your mouth and stuck your foot in it and now will not remove it.


Apparently you know nothing about calories either.


You still haven't told us what you think a calorimeter is nor how you can measure energy from afar with that but not from the Earth.
29-11-2017 22:50
Into the Night
★★★★★
(5558)
Wake wrote:
Into the Night wrote:
Wake wrote:
Into the Night wrote:
Wake wrote:
If it's more than just a word that sounds "sciency" to you then why are you not telling us what it is. Perhaps you should look up "calorie"?

Which one are you interested in?


This is getting better and better all the time. You opened your mouth and stuck your foot in it and now will not remove it.


Apparently you know nothing about calories either.


You still haven't told us what you think a calorimeter is nor how you can measure energy from afar with that but not from the Earth.


The Sun is a source of energy. The Earth is not.


The Parrot Killer
30-11-2017 15:43
Wake
★★★★★
(3396)
Into the Night wrote:
Wake wrote:
Into the Night wrote:
Wake wrote:
Into the Night wrote:
Wake wrote:
If it's more than just a word that sounds "sciency" to you then why are you not telling us what it is. Perhaps you should look up "calorie"?

Which one are you interested in?


This is getting better and better all the time. You opened your mouth and stuck your foot in it and now will not remove it.


Apparently you know nothing about calories either.


You still haven't told us what you think a calorimeter is nor how you can measure energy from afar with that but not from the Earth.


The Sun is a source of energy. The Earth is not.


So now not only can you not tell us what you think a calorimeter is but despite crying that the universe is controlled with the Stefan-Boltzmann equation now you tell us that it isn't.
30-11-2017 21:29
Into the Night
★★★★★
(5558)
Wake wrote:
Into the Night wrote:
Wake wrote:
Into the Night wrote:
Wake wrote:
Into the Night wrote:
Wake wrote:
If it's more than just a word that sounds "sciency" to you then why are you not telling us what it is. Perhaps you should look up "calorie"?

Which one are you interested in?


This is getting better and better all the time. You opened your mouth and stuck your foot in it and now will not remove it.


Apparently you know nothing about calories either.


You still haven't told us what you think a calorimeter is nor how you can measure energy from afar with that but not from the Earth.


The Sun is a source of energy. The Earth is not.


So now not only can you not tell us what you think a calorimeter is but despite crying that the universe is controlled with the Stefan-Boltzmann equation now you tell us that it isn't.


Never did say the universe was controlled solely by the Stefan-Boltzmann law. You are just making up shit again liar.

I also already told you what a calorimeter does, liar. Twice.


The Parrot Killer
Edited on 30-11-2017 21:30
01-12-2017 00:50
Wake
★★★★★
(3396)
Into the Night wrote:
Wake wrote:
Into the Night wrote:
Wake wrote:
Into the Night wrote:
Wake wrote:
Into the Night wrote:
Wake wrote:
If it's more than just a word that sounds "sciency" to you then why are you not telling us what it is. Perhaps you should look up "calorie"?

Which one are you interested in?


This is getting better and better all the time. You opened your mouth and stuck your foot in it and now will not remove it.


Apparently you know nothing about calories either.


You still haven't told us what you think a calorimeter is nor how you can measure energy from afar with that but not from the Earth.


The Sun is a source of energy. The Earth is not.


So now not only can you not tell us what you think a calorimeter is but despite crying that the universe is controlled with the Stefan-Boltzmann equation now you tell us that it isn't.


Never did say the universe was controlled solely by the Stefan-Boltzmann law. You are just making up shit again liar.

I also already told you what a calorimeter does, liar. Twice.


I'm waiting for that reference to where you explained what a calorimeter is and how they work. And I'm also now expecting you to explain how suddenly you don't know what the Stefan-Boltzmann equation is.
01-12-2017 01:00
Into the Night
★★★★★
(5558)
Wake wrote:
Into the Night wrote:
Wake wrote:
Into the Night wrote:
Wake wrote:
Into the Night wrote:
Wake wrote:
Into the Night wrote:
Wake wrote:
If it's more than just a word that sounds "sciency" to you then why are you not telling us what it is. Perhaps you should look up "calorie"?

Which one are you interested in?


This is getting better and better all the time. You opened your mouth and stuck your foot in it and now will not remove it.


Apparently you know nothing about calories either.


You still haven't told us what you think a calorimeter is nor how you can measure energy from afar with that but not from the Earth.


The Sun is a source of energy. The Earth is not.


So now not only can you not tell us what you think a calorimeter is but despite crying that the universe is controlled with the Stefan-Boltzmann equation now you tell us that it isn't.


Never did say the universe was controlled solely by the Stefan-Boltzmann law. You are just making up shit again liar.

I also already told you what a calorimeter does, liar. Twice.


...deleted answered questions...mantra 4...


You're just making shit up, liar.


The Parrot Killer




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