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How can Uranus have storms hot enough to melt steel? A runaway greenhouse effect?

Written by hockeyschtick.blogspot.co.uk

Here’s a great question for CAGW enthusiasts:
 
psi 1According to recent reports, “giant storms on gassy Uranus have astronomers scratching their heads,” because they are extremely large [~5,760 miles wide] and especially because they are extremely hot [2,800°F = melting point of steel].

How does that happen on a planet 30 times further from the Sun than Earth, and without any volcanic activity or SUVs?

From a Mannian/Hansenian runaway-greenhouse radiative-forcing effect? 

Let’s see if that’s possible.
 

According to infrared images taken with the Keck telescope, these storms radiate at peak wavelengths of 1.6 – 2.2 microns. Using Wien’s Displacement Law, we can calculate the temperature of a blackbody radiating at these peak wavelengths as:

1.6 microns → 1,538°C, or 2,800°F or 1,811°K [steel melts at 2,600 – 2,800°F]

2.2 microns → 1,044°C or 1,911°F or 1,317°K

This is amazing considering that Uranus only receives a trifling 3.71 W/m2 energy from the Sun, which per the Stefan-Boltzmann Law is equivalent to a blackbody radiating at a temperature of -183°C, 90°K, or -298°F.
 
How do greenhouse gases violate the 1st law of thermodynamics to amplify 3.71 W/m2 incomingradiation from the Sun to 610,143 W/m2 radiated by a blackbody at 1,538C with peak emission at 1.6 microns, an amplification of outgoing radiation to space over incoming radiation from the Sun by afactor of 83,596 times?
 
Furthermore, how do greenhouse gases raise the outgoing radiating temperature by 1811°K – 90°K = 1721°K, a temperature increase of more than 20 times? [90°K is the Uranus equilibrium temperature with the Sun]
It’s clearly impossible, unless you program a GIGO climate model to say so. 
 
 

“the base of the troposphere on the planet Uranus is 320K, considerably hotter than on Earth [288K], despite being nearly 30 times further from the Sun. The base of the troposphere on Uranus is 320K at 100 bars pressure, despite the planet only receiving 3.71 W/m2 energy from the Sun. By the Stefan-Boltzmann Law, a 320K blackbody radiates 584.6 W/m2. This is 157.5 times the energy received from the Sun, due to the atmospheric temperature gradient produced within a planetary gravity field. The temperature at the base of the troposphere is determined by the ideal gas law PV=nRT, where pressure from gravity and atmospheric mass raise the temperature at the base of the troposphere from the equilibrium temperature with the Sun of Uranus of 89.94K to 320K, regardless of the atmospheric mixture of greenhouse gases.”

Atmospheric mass/pressure/gravity establish the lapse rate/tropospheric temperature profile of all of the planets in our solar system with thick atmospheres, as demonstrated by a paper in Nature by Robinson and Catling. Convection dominates over radiative forcing in the troposphere of each of these planets as also demonstrated by Robinson and Catling. 

Note also as stated by the report below, 

“These storms on an usually quiet Uranus have astronomers scratching their heads, De Pater said. The three other gassy giant planets — Jupiter, Saturn and Neptune — all seem to have strong internal heat sources, and that energy could help stir up storms in the atmosphere. But Uranus doesn’t appear to have one — which means that the Sun must be mostly responsible for generating such disturbances in the planet’s atmosphere, astronomers had thought.”

“But seven years after the equinox, there is not enough sun to explain these massive storms, De Pater pointed out. So it means that the inner workings of Uranus, which remain hidden from view, are more complex than scientists expected.”

“Well, at least it tells us that the theories have to be adjusted,” De Pater said. “They are not representative of reality.”

Read more at: hockeyschtick.blogspot.co.uk

 

Comments (4)

  • Avatar

    Rosco

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    A textbook on Atmospheric Radiation I read begins with –

    “The most important process responsible for energy transfer in the atmosphere is
    electromagnetic radiation.”

    When one considers how much the Sun heats the Earth’s surfaces it is difficult to argue with that.

    Earth’s IR on the other hand is insignificant compared to the solar radiation despite Universities claiming it has equal heating power.

  • Avatar

    Jerry L Krause

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    Sorry, I see I submitted my second comment wrongly. So one needs to read from bottom to top

    Jerry

  • Avatar

    Jerry L Krause

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    Hi Whoever,

    “According to infrared images taken with the Keck telescope, these storms radiate at peak wavelengths of 1.6 – 2.2 microns. Using Wien’s Displacement Law, we can calculate the temperature of a blackbody radiating at these peak wavelengths as:
    1.6 microns → 1,538°C, or 2,800°F or 1,811°K [steel melts at 2,600 – 2,800°F]
    2.2 microns → 1,044°C or 1,911°F or 1,317°K
    This is amazing considering that Uranus only receives a trifling 3.71 W/m2 energy from the Sun, which per the Stefan-Boltzmann Law is equivalent to a blackbody radiating at a temperature of -183°C, 90°K, or -298°F.”

    In my previous comment I alluded to the sometimes considered observation that the Uranus atmosphere has a cloud deck below which the planet is cannot be viewed by the visible solar radiation incident upon it. However, the presence of cloud is visible even though the incident solar radiation is a tiny fraction of that incident upon the earth.

    Now an observed fact is the solar radiation has a portion of ‘visible’ radiation and a portion of invisible IR radiation of wavelengths 1.6-2.2 microns. Albeit that the ‘intensity’ of this IR radiation is slight relative to that of the shorter wavelengths. But the intensity of the observed radiation is not reported, what is reported is the wavelengths 1.6 ¬– 2.2 microns from which the temperatures are calculated.

    Now, this seems a wonderful opportunity to review something which Richard Feynman taught freshman physic students at Caltech during the academic year 1961-1962. Which seems not to be commonly considered.

    “We have just explained that every atom scatters light, and of course the water vapor will scatter light, too. This mystery is why, when the water is condensed into clouds, does it scatter such a tremendously greater amount of light.?”

    I omit the theoretical reasoning presented by Feynman because I do not pretend to be able to comprehend it. But after this he continued: “That is to say, the scattering of water in lumps of N molecules each is N times more tense than the scattering of the single atoms. So as the water agglomerates the scattering increases. Does it increase ad infinitum? No! When does this analysis begin to fail? How many atoms can we put together before we cannot drive this argument any further? Answer: If the water drop gets so big that from one end to the other is a wavelength or so, then the atoms are no longer all in phase because they are too far apart. So as we keep increasing the size of the droplets we get more and more scattering, until such a time that a drop gets about the size of a wavelength, and then the scattering does not increase anywhere nearly as rapidly as the drop gets bigger. Furthermore, the blue disappears, because for long wavelengths the drops can be bigger, before this limit is reached, than they can be for short wavelengths. Although the short wave lengths scatter more per atom than the long waves, there is a bigger enhancement for the red end of the spectrum than for the blue end when all the drops are bigger than the wavelength, so the color is shifted from the blue toward the red.”

    There is more to the lecture and you might read everything about this topic that Feynman taught in The Feynman Lectures on Physics Vol I pp 32-8,9.

    So according to what he taught, I understand that if the cloud droplets had diameters of at least 2.2 microns, these (possibly cold) cloud tops should strongly scatter incident solar radiation of the wavelengths observed back toward space. Certainly this is something we might consider.

    Have a good day, Jerry

  • Avatar

    Jerry L Krause

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    Hi whoever,

    Doug Cotton, of whom I suspect you are aware, commonly refers to that atmosphere’s of Venus and Uranus for reasons similar to what you what you have just written. The temperatures of their atmospheres appear to be abnormally warm at the base of their tropospheres. But I have never discovered that he ever directly consider their atmosphere’s have another feature in common.

    “But seven years after the equinox, there is not enough sun to explain these massive storms, De Pater pointed out. So it means that the inner workings of Uranus, which remain hidden from view, are more complex than scientists expected.”

    This feature is alluded to in this quote from your article, but certainly not specifically identified for a reader not familiar with this planet’s atmosphere. And Cotton does not consider that this feature could have anything to the atmosphere’s temperature at the base of the troposphere. Nor does he, or you, call attention to a fundamental difference between these two atmospheres. In fact neither Cotton nor you completely describe the basic physical features of these two planets.

    So, for readers of your article, who might not be familiar with all of Uranus’s unique features, I review some of them and compare them with the earth’s so without describing each a reader by comparison might know what Uranus’ unique features are.

    Earth (E), Uranus (U) Semimajor Axis (A.U.) E=1, U=19.18; Sidereal Period E=365.26 days, U=84.01 years; Mass E=1, U=14.6; Density (water=1) E=5.52, U=1.3; Period of Rotation E=24 hrs, U=24 hrs; Inclination of Equator to Ecliptic E=23.5o, U=98o (Physical Science, Robert T. Dixon) Just because this textbook was convenient.

    The difference density leads to the conclusion that Uranus is a mass of gaseous matter and has no solid matter. The difference of the inclination of equator to ecliptic means at times during Uranus’s 84 year sidereal period one of its poles basically faces the sun for years as it rotates about axis with a period of 24 hours. The consequence of this is the same atmosphere nearly faces the sun during the entire 24 hour period. Then at other times of the 84 year period the equator of Uranus faces the sun with the result the atmosphere is illuminated by the sun about 12 hour (an equinox condition) and it is not illuminated by the sun about 12 hours as is the general case for the earth. Except, because of inclination of the earth’s equator to the ecliptic, we know the one of the earth’s poles is in total darkness for a half-year while the other is constantly illuminated by the sun for a half-year.

    I have not written the previous paragraph to confuse a reader but alert the reader that there is so much dynamics involved in this gaseous planet that it seems somewhat ridiculous to state: “So it means that the inner workings of Uranus, which remain hidden from view, are more complex than scientists expected.” Especially so when it is never reviewed why the inner working of Uranus are hidden from view. For this is what Doug Cotton fails to consider as a possible factor in understanding what you have generally reviewed. Given the basic information about Uranus, I believe an interesting scientific problem would be to describe the dynamic changes which this planet, as generally described, should undergoe as it orbits the sun with a period of 84 years.

    Have a good day, Jerry

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