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The greenhouse effect denier Enceladus


The greenhouse effect denier Enceladus08-03-2017 20:55
Leitwolf
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I do not know why that is, but apparently there is little to talk about the probably most significant evidence falsifying climate change, or rather its foundation, the assumed "greenhouse effect" (GHE).

The story is simple and straight forward. The GHE is totally based on the formula that suggests earth should have a surface temperature of 255°K. That again is derived from a perfect black body (PB
, which would have in earths position 280°K. Then if you take albedo into account this results in (1 - 0.31) ^0.25 * 280 = 255.2 or so..

Enceladus, a moon of Saturn, has an extremely high albedo of 0.99, which is why it puts this approach to the test. Saturn is 9.58 AU from the sun, so a PBB should have 280 / 9.58^0.5 = 90.5°K. Now if we allow for the albedo, we get to (1 - 0.99)^0.25 *90.5 = 28.6°K.

So while Enceladus should be only 28.6°K cold in theory, it actually has like 75°K, or 2 1/2 times as much. There is a "greenhouse effect" of over 45°K on this moon. Of course, Enceladus does not hold any atmosphere, and GHE is definitely impossible. Rather this case prooves that the formula and the assumptions it is based on, will not give us the right results, ie. it is falsified.

As the existence of the GHE itself is falsified, it is rather pointless to discuss the conclusions (most notable greenhouse gases, climate change, etc..) based solely on it.
08-03-2017 21:30
Tim the plumber
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Given that the heat from the Sun at that distance is very weak and the extreme reflectivity of the body other factors such as it's own internal heat will be very big in comparison to earth's situation.

It may be heating from the tidal effects of Saturn or residual heat from impacts. The high albedo means that it will radiate very little heat.

Edited on 08-03-2017 21:30
08-03-2017 22:14
Leitwolf
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Well it is certainly true that emissions run low at these temperatures. Compared to earth it would only be like (75 / 288) ^4 = 0.46%.
But for surface temperatures inner heat is really playing a negligible role, and we are not talking about "heat" as such at all, so mainting these 0.46% just from geothermia would be an awefull lot. Then Enceladus has much colder temperatures at the poles, like only 33°K, which is a clear indication, that its temperature is driven by the sun, and nothing else.
08-03-2017 23:52
still learning
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(244)
Leitwolf wrote:
.....The GHE is totally based on the formula that suggests earth should have a surface temperature of 255°K......


Not so.

That result of 255 K caused folks to search for a reason why things are warmer than that.

Eventually the greenhouse effect was figured out.

Encaladus' warmth is apparently a mystery, true. Doesn't prove or disprove anything else, near as I can tell.
09-03-2017 00:07
Leitwolf
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(91)
Well it is only a mystery when you assume that it is perfectly reasonable to insinuate an absorptivity of 0.01 (1-0.99) while holding emissivity still at 1
)) It does not work for Enceladus, it does not work anywhere else. In fact there is a natural tendency for absorptivity and emissivity to be about equal. So the flaw in this formula, or rather the assumptions it is based on are quite obvious.

And well .. I guess in this respect Enceladus is just as irrelevant as a black swan to the theory that all swans were white..

And most importantly, and I brought this issue up in my other post, with regard to water surface emissivity seems to be lower than absorptivity. And that relation yields an "interesting" result, because (0.93 / 0.84) ^0.25 * 280 = 287°K (!!!).
26-04-2017 18:34
Wake
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(4026)
Leitwolf wrote:
Well it is certainly true that emissions run low at these temperatures. Compared to earth it would only be like (75 / 288) ^4 = 0.46%.
But for surface temperatures inner heat is really playing a negligible role, and we are not talking about "heat" as such at all, so mainting these 0.46% just from geothermia would be an awefull lot. Then Enceladus has much colder temperatures at the poles, like only 33°K, which is a clear indication, that its temperature is driven by the sun, and nothing else.


Accepting your statements of the heat levels on Enceladus we have to consider two things: the mass of the moon is FAR too small to have any internal heat driven by anything other than the tidal effects of Saturn.

However, these would be equal at the equator and the poles and your temperatures clearly show they are not.

This then leaves us with the question: what is the bandwidth of the reflections of Enceladus?

And the plain test of the absorption theory you have is to measure the absolute temperature of the moon when it is in Saturn's shadow.
26-04-2017 18:39
Wake
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still learning wrote:
Leitwolf wrote:
.....The GHE is totally based on the formula that suggests earth should have a surface temperature of 255°K......


Not so.

That result of 255 K caused folks to search for a reason why things are warmer than that.

Eventually the greenhouse effect was figured out.

Encaladus' warmth is apparently a mystery, true. Doesn't prove or disprove anything else, near as I can tell.


Please don't give us more of this garbage. The Earth's temperature has nothing whatsoever to do with what is supposed to be the "greenhouse effect" but ONLY what the true greenhouse effect is - nothing more than a blanketing effect of the total atmosphere and no specific gases in it.
26-04-2017 22:40
Into the Night
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(8592)
Leitwolf wrote:
Well it is only a mystery when you assume that it is perfectly reasonable to insinuate an absorptivity of 0.01 (1-0.99) while holding emissivity still at 1
)) It does not work for Enceladus, it does not work anywhere else. In fact there is a natural tendency for absorptivity and emissivity to be about equal. So the flaw in this formula, or rather the assumptions it is based on are quite obvious.

And well .. I guess in this respect Enceladus is just as irrelevant as a black swan to the theory that all swans were white..

And most importantly, and I brought this issue up in my other post, with regard to water surface emissivity seems to be lower than absorptivity. And that relation yields an "interesting" result, because (0.93 / 0.84) ^0.25 * 280 = 287°K (!!!).


You don't know the emissivity of Earth. Emissivity changes dramatically every few inches.

It is not possible to calculate what the temperature of the Earth 'should' be.

It is not possible to measure the temperature of the Earth to any useful degree of accuracy.


The Parrot Killer
26-04-2017 23:03
Wake
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(4026)
Into the Night wrote:
Leitwolf wrote:
Well it is only a mystery when you assume that it is perfectly reasonable to insinuate an absorptivity of 0.01 (1-0.99) while holding emissivity still at 1
)) It does not work for Enceladus, it does not work anywhere else. In fact there is a natural tendency for absorptivity and emissivity to be about equal. So the flaw in this formula, or rather the assumptions it is based on are quite obvious.

And well .. I guess in this respect Enceladus is just as irrelevant as a black swan to the theory that all swans were white..

And most importantly, and I brought this issue up in my other post, with regard to water surface emissivity seems to be lower than absorptivity. And that relation yields an "interesting" result, because (0.93 / 0.84) ^0.25 * 280 = 287°K (!!!).


You don't know the emissivity of Earth. Emissivity changes dramatically every few inches.

It is not possible to calculate what the temperature of the Earth 'should' be.

It is not possible to measure the temperature of the Earth to any useful degree of accuracy.


Where are you getting this? Why do you quote the laws of thermodynamics and then make statements like this as if these laws do not exist?

We calculate the amount of energy that strikes and is absorbed by the Earth and so we can then use the laws to calculate what the Earth emits.

Are you playing some sort of game that we have to have any knowledge at all of the specific emissivity of every cm^2 in order to make any assumptions at all?
Edited on 26-04-2017 23:17
27-04-2017 01:35
Into the Night
★★★★★
(8592)
Wake wrote:
Into the Night wrote:
Leitwolf wrote:
Well it is only a mystery when you assume that it is perfectly reasonable to insinuate an absorptivity of 0.01 (1-0.99) while holding emissivity still at 1
)) It does not work for Enceladus, it does not work anywhere else. In fact there is a natural tendency for absorptivity and emissivity to be about equal. So the flaw in this formula, or rather the assumptions it is based on are quite obvious.

And well .. I guess in this respect Enceladus is just as irrelevant as a black swan to the theory that all swans were white..

And most importantly, and I brought this issue up in my other post, with regard to water surface emissivity seems to be lower than absorptivity. And that relation yields an "interesting" result, because (0.93 / 0.84) ^0.25 * 280 = 287°K (!!!).


You don't know the emissivity of Earth. Emissivity changes dramatically every few inches.

It is not possible to calculate what the temperature of the Earth 'should' be.

It is not possible to measure the temperature of the Earth to any useful degree of accuracy.


Where are you getting this? Why do you quote the laws of thermodynamics and then make statements like this as if these laws do not exist?

Nothing I have said violates or ignores any law of thermodynamics.

The problem is not a science problem. It is a math problem. Statistical analysis has a very rigid set of rules for how sampling is to occur before the analysis begins. For sampling something like a temperature, the population is not the available thermometers, but the possible temperature gradient. While we don't know what that is, we have observed gradients as high as 20 deg F in a single mile.

We do not have enough thermometers to produce a usable analysis.

Emissivity is even worse. It changes dramatically every few inches. The way you measure emissivity is by using contact thermometers and working backwards against the ideal references. We don't know the emissivity of Earth. We don't know the temperature of the Earth to any useful accuracy.

Wake wrote:
We calculate the amount of energy that strikes and is absorbed by the Earth and so we can then use the laws to calculate what the Earth emits.

You can't. You need to have a value for emissivity to do that. You don't know what it is.
Wake wrote:
Are you playing some sort of game that we have to have any knowledge at all of the specific emissivity of every cm^2 in order to make any assumptions at all?

Math doesn't 'assume'. Math is math. The margin of error is too great to calculate the emissivity or temperature with available instrumentation to an accuracy that makes any sense. The only assumptions that math makes are the founding axioms.

Now you can calculate how much wattage of sunlight hits a square (pick a unit) of Earth, but you don't know what happens to it. Some is reflected, some is absorbed and converted into thermal energy, some is absorbed and converted into chemical reactions, some is scattered, etc.

All your work calculating the wattage per square (pick a unit) really is a lot of work for very little.


The Parrot Killer
Edited on 27-04-2017 01:37
27-04-2017 02:02
Wake
★★★★★
(4026)
Into the Night wrote:
Wake wrote:
Into the Night wrote:
Leitwolf wrote:
Well it is only a mystery when you assume that it is perfectly reasonable to insinuate an absorptivity of 0.01 (1-0.99) while holding emissivity still at 1
)) It does not work for Enceladus, it does not work anywhere else. In fact there is a natural tendency for absorptivity and emissivity to be about equal. So the flaw in this formula, or rather the assumptions it is based on are quite obvious.

And well .. I guess in this respect Enceladus is just as irrelevant as a black swan to the theory that all swans were white..

And most importantly, and I brought this issue up in my other post, with regard to water surface emissivity seems to be lower than absorptivity. And that relation yields an "interesting" result, because (0.93 / 0.84) ^0.25 * 280 = 287°K (!!!).


You don't know the emissivity of Earth. Emissivity changes dramatically every few inches.

It is not possible to calculate what the temperature of the Earth 'should' be.

It is not possible to measure the temperature of the Earth to any useful degree of accuracy.


Where are you getting this? Why do you quote the laws of thermodynamics and then make statements like this as if these laws do not exist?

Nothing I have said violates or ignores any law of thermodynamics.

The problem is not a science problem. It is a math problem. Statistical analysis has a very rigid set of rules for how sampling is to occur before the analysis begins. For sampling something like a temperature, the population is not the available thermometers, but the possible temperature gradient. While we don't know what that is, we have observed gradients as high as 20 deg F in a single mile.

We do not have enough thermometers to produce a usable analysis.

Emissivity is even worse. It changes dramatically every few inches. The way you measure emissivity is by using contact thermometers and working backwards against the ideal references. We don't know the emissivity of Earth. We don't know the temperature of the Earth to any useful accuracy.

Wake wrote:
We calculate the amount of energy that strikes and is absorbed by the Earth and so we can then use the laws to calculate what the Earth emits.

You can't. You need to have a value for emissivity to do that. You don't know what it is.
Wake wrote:
Are you playing some sort of game that we have to have any knowledge at all of the specific emissivity of every cm^2 in order to make any assumptions at all?

Math doesn't 'assume'. Math is math. The margin of error is too great to calculate the emissivity or temperature with available instrumentation to an accuracy that makes any sense. The only assumptions that math makes are the founding axioms.

Now you can calculate how much wattage of sunlight hits a square (pick a unit) of Earth, but you don't know what happens to it. Some is reflected, some is absorbed and converted into thermal energy, some is absorbed and converted into chemical reactions, some is scattered, etc.

All your work calculating the wattage per square (pick a unit) really is a lot of work for very little.


It isn't statistical analysis and it doesn't matter what happens to it - whether it is reflect, absorbed, radiated, moved about by conduction and convention or waved about by hand - the energy from the Sun is lost totally by the Earth or the Earth would have been nothing but a charred ball before life ever began.

This amount of energy from the Sun is calculable without resorting to any statistical analysis and Leitwolf was demonstrating how it is done though a bit out of synch yet.

You have the diameter of the Sun and it's surface temperature; with that you can calculate the energy emitted.

Using the distance from the Sun to the (average distance) of the Earth's orbit you can expand the surface area and so reduce the energy per square unit.

At this point you know the radius of the Earth and hence its surface area exposed to the Sun's energy. You can then calculate the energy per square unit that the Earth is exposed to.

Corrections have to be made on Earth for the day to day difference in distance to the Sun.

Corrections have to be made for the loss of energy due to the reflections for the energy striking the Earth at an angle because the Earth is not a flat plane but a sphere. Corrections must also be made because the Sun is a sphere.

Corrections have to be made because though extremely tenuous the top two layers of atmosphere reflect enough energy that that actually affecting the Earth is lowered.

Leitwolf was attempting to make corrections due to cloud cover and the effects of that.

At what point in those calculations offer any requirement for statistical analysis?
27-04-2017 21:59
Leitwolf
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(91)
Wake wrote:
And the plain test of the absorption theory you have is to measure the absolute temperature of the moon when it is in Saturn's shadow.


Well, next I am there I might do so. It might prove tricky however, as Enceladus will spent a maximum of 2h40 in the shadow. And given it's low temperatures, you would not see much cooling within that period.

Into the night wrote:
It is not possible to calculate what the temperature of the Earth 'should' be.

It is not possible to measure the temperature of the Earth to any useful degree of accuracy.


Given so many impossibilities, we should all rather stay in bed, all day, all time. Me however, I love to do the impossible.
27-04-2017 22:10
litesong
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(2297)
Leitwolf wrote:
"old sick silly sleepy sleezy slimy steenkin' filthy vile reprobate rooting (& rotting) racist pukey proud pig AGW denier liar whiner wake-me-up" woofed: It is not possible to measure the temperature of the Earth to any useful degree of accuracy.

Given so many impossibilities, we should all rather stay in bed, all day, all time. Me however, I love to do the impossible.

Remember.... if "old sick silly sleepy sleezy slimy steenkin' filthy vile reprobate rooting (& rotting) racist pukey proud pig AGW denier liar whiner wake-me-up" can't do it, probably means others CAN & HAVE DONE IT.
27-04-2017 23:14
Into the Night
★★★★★
(8592)
Wake wrote:
Into the Night wrote:
Wake wrote:
Into the Night wrote:
Leitwolf wrote:
Well it is only a mystery when you assume that it is perfectly reasonable to insinuate an absorptivity of 0.01 (1-0.99) while holding emissivity still at 1
)) It does not work for Enceladus, it does not work anywhere else. In fact there is a natural tendency for absorptivity and emissivity to be about equal. So the flaw in this formula, or rather the assumptions it is based on are quite obvious.

And well .. I guess in this respect Enceladus is just as irrelevant as a black swan to the theory that all swans were white..

And most importantly, and I brought this issue up in my other post, with regard to water surface emissivity seems to be lower than absorptivity. And that relation yields an "interesting" result, because (0.93 / 0.84) ^0.25 * 280 = 287°K (!!!).


You don't know the emissivity of Earth. Emissivity changes dramatically every few inches.

It is not possible to calculate what the temperature of the Earth 'should' be.

It is not possible to measure the temperature of the Earth to any useful degree of accuracy.


Where are you getting this? Why do you quote the laws of thermodynamics and then make statements like this as if these laws do not exist?

Nothing I have said violates or ignores any law of thermodynamics.

The problem is not a science problem. It is a math problem. Statistical analysis has a very rigid set of rules for how sampling is to occur before the analysis begins. For sampling something like a temperature, the population is not the available thermometers, but the possible temperature gradient. While we don't know what that is, we have observed gradients as high as 20 deg F in a single mile.

We do not have enough thermometers to produce a usable analysis.

Emissivity is even worse. It changes dramatically every few inches. The way you measure emissivity is by using contact thermometers and working backwards against the ideal references. We don't know the emissivity of Earth. We don't know the temperature of the Earth to any useful accuracy.

Wake wrote:
We calculate the amount of energy that strikes and is absorbed by the Earth and so we can then use the laws to calculate what the Earth emits.

You can't. You need to have a value for emissivity to do that. You don't know what it is.
Wake wrote:
Are you playing some sort of game that we have to have any knowledge at all of the specific emissivity of every cm^2 in order to make any assumptions at all?

Math doesn't 'assume'. Math is math. The margin of error is too great to calculate the emissivity or temperature with available instrumentation to an accuracy that makes any sense. The only assumptions that math makes are the founding axioms.

Now you can calculate how much wattage of sunlight hits a square (pick a unit) of Earth, but you don't know what happens to it. Some is reflected, some is absorbed and converted into thermal energy, some is absorbed and converted into chemical reactions, some is scattered, etc.

All your work calculating the wattage per square (pick a unit) really is a lot of work for very little.


It isn't statistical analysis and it doesn't matter what happens to it - whether it is reflect, absorbed, radiated, moved about by conduction and convention or waved about by hand - the energy from the Sun is lost totally by the Earth or the Earth would have been nothing but a charred ball before life ever began.

This amount of energy from the Sun is calculable without resorting to any statistical analysis and Leitwolf was demonstrating how it is done though a bit out of synch yet.

You have the diameter of the Sun and it's surface temperature; with that you can calculate the energy emitted.
You don't know the temperature of the Sun. It is only estimated by applying Wien's law backward and comparing against known reference points here on Earth.
Wake wrote:
At what point in those calculations offer any requirement for statistical analysis?

Wattage striking the Earth isn't the issue. Absorption is, and only that absorption that results in conversion to thermal energy. Some is reflected, some is absorbed. You don't have an emissivity value for Earth. Emissivity changes dramatically every few inches. You don't have anything like sufficient instrumentation to measure it.


The Parrot Killer
Edited on 27-04-2017 23:16
30-04-2017 01:42
LifeIsThermal
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(39)
Leitwolf wrote:
I do not know why that is, but apparently there is little to talk about the probably most significant evidence falsifying climate change, or rather its foundation, the assumed "greenhouse effect" (GHE).

The story is simple and straight forward. The GHE is totally based on the formula that suggests earth should have a surface temperature of 255°K. That again is derived from a perfect black body (PB
, which would have in earths position 280°K. Then if you take albedo into account this results in (1 - 0.31) ^0.25 * 280 = 255.2 or so..

Enceladus, a moon of Saturn, has an extremely high albedo of 0.99, which is why it puts this approach to the test. Saturn is 9.58 AU from the sun, so a PBB should have 280 / 9.58^0.5 = 90.5°K. Now if we allow for the albedo, we get to (1 - 0.99)^0.25 *90.5 = 28.6°K.

So while Enceladus should be only 28.6°K cold in theory, it actually has like 75°K, or 2 1/2 times as much. There is a "greenhouse effect" of over 45°K on this moon. Of course, Enceladus does not hold any atmosphere, and GHE is definitely impossible. Rather this case prooves that the formula and the assumptions it is based on, will not give us the right results, ie. it is falsified.

As the existence of the GHE itself is falsified, it is rather pointless to discuss the conclusions (most notable greenhouse gases, climate change, etc..) based solely on it.


But the greenhouse effect is falsified by the facts. Dry ice is cold. Water cools surfaces. Air cools surfaces. Submerge the whole thing in the ultimate heat sink of space, vacuum at 3 Kelvin, the only observed infinity. It swallows all heat without increasing its temperature even a little tiny bit. Infinity.

A greenhouse without radiant barriers, just freezingly cold damp air with the potent heat absorber dry ice in the infinitely cold vacuum. Come on, who was stupid enough to come up with this shit? And why did anyone listen?
Edited on 30-04-2017 01:43
30-04-2017 11:58
Tim the plumber
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(1295)
Leitwolf wrote:
Well it is certainly true that emissions run low at these temperatures. Compared to earth it would only be like (75 / 288) ^4 = 0.46%.
But for surface temperatures inner heat is really playing a negligible role, and we are not talking about "heat" as such at all, so mainting these 0.46% just from geothermia would be an awefull lot. Then Enceladus has much colder temperatures at the poles, like only 33°K, which is a clear indication, that its temperature is driven by the sun, and nothing else.


Not at all.

The tidal effects and magnetic effects may well be concentrated at the equator of the moon.

The power of the sunshine is very very low and almost all of it (99%) is reflected so there can be no justification for such a wide temperature difference from sunshine. That bit is easy to work out.
30-04-2017 23:50
Leitwolf
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(91)
Tim the plumber wrote:
Not at all.

The tidal effects and magnetic effects may well be concentrated at the equator of the moon.

The power of the sunshine is very very low and almost all of it (99%) is reflected so there can be no justification for such a wide temperature difference from sunshine. That bit is easy to work out.


Your argument is blue eyed. First, there is not a single indication that tidal forces and furthermore geothermia is responsible for Enceladus' surface temperature. The only reason you "invent" this idea is to hold on to a theory that has been falsified on many other occasions.
The case of Enceladus serves to outline the absurdity of putting absorptivity to 0.01, while holding emissivity to 1. As stated before, both A and E are loosly correlated, and unless we have specific information on which of the two is bigger, we should be assume they are about equal. In other words completely ignore the albedo, and assume the behaviour of a perfect black body. In this way we get much better approximations of surface temperatures for almost all objects in the solar system.

Indeed and in reality, we have surface types, where E is smaller than A, which means they become warmer than a PBB. Notoriously important examples would be water, or sand. Both play a major role in earth's climate.
Of course I brought up the case water, which is pivotal for earths temperature, here:
http://www.climate-debate.com/forum/emissivity-of-the-ocean-d6-e1342.php
01-05-2017 01:43
Surface Detail
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(1673)
Leitwolf wrote:
Tim the plumber wrote:
Not at all.

The tidal effects and magnetic effects may well be concentrated at the equator of the moon.

The power of the sunshine is very very low and almost all of it (99%) is reflected so there can be no justification for such a wide temperature difference from sunshine. That bit is easy to work out.


Your argument is blue eyed. First, there is not a single indication that tidal forces and furthermore geothermia is responsible for Enceladus' surface temperature. The only reason you "invent" this idea is to hold on to a theory that has been falsified on many other occasions.
The case of Enceladus serves to outline the absurdity of putting absorptivity to 0.01, while holding emissivity to 1. As stated before, both A and E are loosly correlated, and unless we have specific information on which of the two is bigger, we should be assume they are about equal. In other words completely ignore the albedo, and assume the behaviour of a perfect black body. In this way we get much better approximations of surface temperatures for almost all objects in the solar system.

Indeed and in reality, we have surface types, where E is smaller than A, which means they become warmer than a PBB. Notoriously important examples would be water, or sand. Both play a major role in earth's climate.
Of course I brought up the case water, which is pivotal for earths temperature, here:
http://www.climate-debate.com/forum/emissivity-of-the-ocean-d6-e1342.php

Are you German? The correct translation of blauäugig (literally blue-eyed) would be actually be naive in English. There's no association between blue eyes and nativity in English
On the whole, though, your English is excellent.

Which, I'm afraid, is more than can be said for your science. Absorptivity and emissivity are indeed equal (as Kirchhoff's radiation law states), but, critically, emissivity (and hence absorptivity) is a function of wavelength. In the case of Enceladus, we have a very low emissivity in the visible part of the spectrum, but a much higher emissivity in the IR part of the spectrum. This difference in emissivities in the IR and the visible means that Enceladus should have a temperature much lower than its black body temperature. However, as Tim has mentioned, tidal effects could well be warming Enceladus. Indeed, images taken by the Cassini spacecraft showing jets or geysers on Enceladus support this explanation.
01-05-2017 02:25
Leitwolf
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(91)
Are you German?


No, but Germans are not the only ones to speak German.

You do not see how your argument is working against you. True, Enceladus has these "warm" water plumes on the south pole, which are said to reach up to 157K. The problem with that: otherwise temperatures are spread quite evenly over the surface, with typical cold polar regions.

If geothermia was generally responsible for E. temperature, it would not be evenly spread. Rather you have tectonic structures resulting in warm and cold spots. Also, the argument brought forward, that geothermia would most of all affect the equatorial regions, is falsified by these plumes.

So you are trying to flick something which does not work, by something else, which works even less. It pointless beyond absurdity.
01-05-2017 02:51
Surface Detail
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(1673)
Leitwolf wrote:
Are you German?


No, but Germans are not the only ones to speak German.

You do not see how your argument is working against you. True, Enceladus has these "warm" water plumes on the south pole, which are said to reach up to 157K. The problem with that: otherwise temperatures are spread quite evenly over the surface, with typical cold polar regions.

If geothermia was generally responsible for E. temperature, it would not be evenly spread. Rather you have tectonic structures resulting in warm and cold spots. Also, the argument brought forward, that geothermia would most of all affect the equatorial regions, is falsified by these plumes.

So you are trying to flick something which does not work, by something else, which works even less. It pointless beyond absurdity.

Austrian or Swiss, then, I guess.

Whatever the cause of the geysers, it is clear that something must be warming the moon from the inside and hence making the outside of the moon warmer than it would otherwise be. The source of the internal heat could be tidal (there are indeed warm and cold areas), chemical, radioactive or some combination of these. Whatever it is, it has absolutely nothing to say about the greenhouse effect, given that Enceladus has no atmosphere.

Also, English flick doesn't mean German flicken. It's a false friend. The phrase you're looking for is cobble together, patch up, or something similar.
01-05-2017 12:31
Tim the plumber
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(1295)
Leitwolf wrote:
Tim the plumber wrote:
Not at all.

The tidal effects and magnetic effects may well be concentrated at the equator of the moon.

The power of the sunshine is very very low and almost all of it (99%) is reflected so there can be no justification for such a wide temperature difference from sunshine. That bit is easy to work out.


Your argument is blue eyed. First, there is not a single indication that tidal forces and furthermore geothermia is responsible for Enceladus' surface temperature. The only reason you "invent" this idea is to hold on to a theory that has been falsified on many other occasions.
The case of Enceladus serves to outline the absurdity of putting absorptivity to 0.01, while holding emissivity to 1. As stated before, both A and E are loosly correlated, and unless we have specific information on which of the two is bigger, we should be assume they are about equal. In other words completely ignore the albedo, and assume the behaviour of a perfect black body. In this way we get much better approximations of surface temperatures for almost all objects in the solar system.

Indeed and in reality, we have surface types, where E is smaller than A, which means they become warmer than a PBB. Notoriously important examples would be water, or sand. Both play a major role in earth's climate.
Of course I brought up the case water, which is pivotal for earths temperature, here:
http://www.climate-debate.com/forum/emissivity-of-the-ocean-d6-e1342.php


I thought I was pointing out that we should be able to ignore a bit the effect of the sun because of the high albedo. That the high albedo is the reason the temperature can be out of line with the expected temp of a black body with no other heat source at that distance from the sun.

There may be other explanations. It may have had a recent asteroid strike. Who knows?

Edited on 01-05-2017 12:33
01-05-2017 17:41
Leitwolf
★☆☆☆☆
(91)
..or possibly a lot of aliens smoking so much weed that it heats the moon. Who knows..




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