Remember me
▼ Content

Venus is hotter than Mercury?!?



Page 19 of 26<<<1718192021>>>
09-03-2020 22:40
tmiddlesProfile picture★★★★★
(3266)
One Punch Man wrote:
If people are against the greenhouse gas theory increasing the temperature of Venus above its effective/equilibrium temperature then are there other possibilities that you would consider rather than simply saying "measurements aren't valid". The heat emanating from the interior as suggested by the Velikovskian theory? Pressure? The thick Venusian cloud-layer perhaps?

Yes I think this is what's great about Venus. It's not insane for someone to think measurements on Earth aren't accurate enough to claim even a dozen degrees of accuracy (I don't agree but it's not insane). Venus has such a massive disparity the simple fact that it's way hotter on the ground than it's emitting temperature is indisputable. So we can move on to why.

IBD seemed to suggest the ideal gas law explained it. That's simply saying that the pressure on Venus would explain the higher temperature but that doesn't work. The ideal gas law can't "create energy" can it?

James___ wrote:...Venus is hotter than Mercury because of CO2, right?
What I wonder about is what the different contributing factors are. When we talk about CO2 on Earth it's never a question of mass, or a heavier atmosphere (in fact the burning reaction makes for a lighter atmosphere). So how much of what's going on with Venus is due to the greenhouse properties of CO2 and how much just due to the mass.

James___ wrote:...heat being released by sulphur gas responsible for the heat observed coming from Venus?
https://earthsky.org/science-wire/have-venus-volcanoes-been-caught-in-the-act
Enjoy

Venus volcanoes wasn't something I was aware of that's pretty cool.

James___ wrote:...Sulphur burns at about 455º C. and when it does, the heat it releases is about 1,250º C.
tmiddles, can you say if this influences how hot Venus is?
Well if it does it seems the key to Venus being so hot is that it doesn't lose all the extra energy it has near the ground through it's emitting surface. If you set of a huge heat bomb on Earth it would presumably throw us out of equilibrium with the Sun temporarily as that heat was lost to space. Venus is keeping it's heat.

Into the Night wrote:No proof available in an open system.
What do you mean by that ITN? An "open system" is defined as one in which energy AND matter are exchanged beyond the system. Are you under the impression Venus took an intergalactic dump I'm not aware of? How is matter entering or leaving the planet of Venus?

I sure don't see a rebuttal from you on your assertions, your bold claims, about SBL and PL being thoroughly debunked by Venus.

"Good tests kill flawed theories; we remain alive to guess again." - Karl Popper
ITN/IBD Fraud exposed:  The 2nd LTD add on claiming radiance from cooler bodies can't be absorbed Max Planck debunks, they can't explain:net-thermal-radiation-you-in-a-room-as-a-reference & Proof: no data is valid for IBD or ITN
10-03-2020 01:14
tmiddlesProfile picture★★★★★
(3266)
IBD's response to this thread posted elsewhere:
IBdaMann wrote:
tmiddles wrote:So I will adapt here just for you IBD. "Planetary temperature" and "equilibrium temperature" are both retired temporarliy. I will use a new title of my own:

SOLAR ABSORPTION TEMPERATURE
which I define as the temperature which corresponds to the emission of radiance for the entire surface area of any object with that emission of energy being equal to the energy absorbed from the Sun.

Thank you ... this makes for a great lead-in to your unambiguous definition which will look like a mathematical equation.

Go ahead and lay it on me.

tmiddles wrote:
IBdaMann wrote:We need an unambiguous hypothesis and your declared "acceptable" margin of error before we can begin.

Nah let me help you out with that. It is:

Mean[TABA(Venus)] - SOLAR ABSORPTION TEMP = 508C +/- 300 degC

Great ... so once you provide that unambiguous "SOLAR ABSORPTION TEMP" definition we can insert it into this equation and get your complete unambiguous hypothesis and your target margin of error.

You are making excellent headway.

tmiddles wrote:The temperature provided by NASA and the CCCP is 462C. We put a +/-200 degree margin on that simply to set aside extreme skepticism I don't myself possess.


We have been over why this is worthless. You are on tap to provide the raw data and we cannot proceed until we have it.

tmiddles wrote: The albedo of Venus is estimated at 0.6

Where did you get that idea?

If you want to discuss Venus' emissivity then you must provide the raw data for that ... which no one has. Until humanity is able to measure something like that it remains an unknown value between 0.0 and 1.0.

The requirements haven't changed. The same ball is still in your court.




.
10-03-2020 01:34
James___
★★★★★
(2849)
tmiddles wrote:
One Punch Man wrote:
If people are against the greenhouse gas theory increasing the temperature of Venus above its effective/equilibrium temperature then are there other possibilities that you would consider rather than simply saying "measurements aren't valid". The heat emanating from the interior as suggested by the Velikovskian theory? Pressure? The thick Venusian cloud-layer perhaps?

Yes I think this is what's great about Venus. It's not insane for someone to think measurements on Earth aren't accurate enough to claim even a dozen degrees of accuracy (I don't agree but it's not insane). Venus has such a massive disparity the simple fact that it's way hotter on the ground than it's emitting temperature is indisputable. So we can move on to why.

IBD seemed to suggest the ideal gas law explained it. That's simply saying that the pressure on Venus would explain the higher temperature but that doesn't work. The ideal gas law can't "create energy" can it?

James___ wrote:...Venus is hotter than Mercury because of CO2, right?
What I wonder about is what the different contributing factors are. When we talk about CO2 on Earth it's never a question of mass, or a heavier atmosphere (in fact the burning reaction makes for a lighter atmosphere). So how much of what's going on with Venus is due to the greenhouse properties of CO2 and how much just due to the mass.

James___ wrote:...heat being released by sulphur gas responsible for the heat observed coming from Venus?
https://earthsky.org/science-wire/have-venus-volcanoes-been-caught-in-the-act
Enjoy

Venus volcanoes wasn't something I was aware of that's pretty cool.

James___ wrote:...Sulphur burns at about 455º C. and when it does, the heat it releases is about 1,250º C.
tmiddles, can you say if this influences how hot Venus is?
Well if it does it seems the key to Venus being so hot is that it doesn't lose all the extra energy it has near the ground through it's emitting surface. If you set of a huge heat bomb on Earth it would presumably throw us out of equilibrium with the Sun temporarily as that heat was lost to space. Venus is keeping it's heat.

Into the Night wrote:No proof available in an open system.
What do you mean by that ITN? An "open system" is defined as one in which energy AND matter are exchanged beyond the system. Are you under the impression Venus took an intergalactic dump I'm not aware of? How is matter entering or leaving the planet of Venus?

I sure don't see a rebuttal from you on your assertions, your bold claims, about SBL and PL being thoroughly debunked by Venus.

"Good tests kill flawed theories; we remain alive to guess again." - Karl Popper
ITN/IBD Fraud exposed:  The 2nd LTD add on claiming radiance from cooler bodies can't be absorbed Max Planck debunks, they can't explain:net-thermal-radiation-you-in-a-room-as-a-reference & Proof: no data is valid for IBD or ITN



You didn't answer any of my questions. You deflected. Typical Amerikkaner.
You sound like a Christian. You really do. I used to like going to church but had the wrong accent or a hearing loss.
Your post disgusts me. It really does. You are one sick individual. With me, neither ITN or IBDM has ever posted anything so repulsive. I can show your messages to me. I think then they'd understand.
The difference is that we don't make it personal or try to discredit the other person. I can tell ITN and IBDM I think they're both stupid but I will not send them personal messages contradicting my posts.
I would say that all Native Americans got what they deserved but then my posts suggest otherwise. And all personal messages will not disagree with what I post.
Don't try to make me look bad because all you can say is CO2. To say that volcanoes on Venus are cool is to ignore their implications and ramifications on its atmosphere. And dude, I might not like Americans all that much. Please think before you decide to play games. One American is like any other American. But I will not tolerate your bull shit.
10-03-2020 01:49
tmiddlesProfile picture★★★★★
(3266)
IBdaMann wrote:
tmiddles wrote:
SOLAR ABSORPTION TEMPERATURE[/b][/u] which I define ....
...will look like a mathematical equation.


No problem. Here you go:



Tp denoting the "SOLAR ABSORPTION TEMPERATURE" for any object with albedo denoted by "a", "R" being the radius of the sun, and D being the distance from the sun in AU.


IBdaMann wrote:
tmiddles wrote:
Mean[TABA(Venus)] - SOLAR ABSORPTION TEMP = 508C +/- 300 degC

Great ... so once you provide that unambiguous "SOLAR ABSORPTION TEMP" definition we can insert it into this equation and get your complete unambiguous hypothesis and your target margin of error.
Already done. "SOLAR ABSORPTION TEMPERATURE" is simply a clarified alternative to "PLANETRY TEMPERATURE" so that it's clear it is a value that exists even without a planet being there, that it is NOT the temperature of the planet but a reference based entirely on the energy received from the Sun. So the math was done previeously and here you go:
tmiddles wrote:
For Venus we have the distance from the sun is 0.72 AU
AU is 1.0 for Earth of course and 1.0 AU is 1.5*10^8KM
So Venus is 1.08*10^8KM from the Sun.
T = (temperature of the sun)*1.0(full emissivity)^1/4 * [(radius of sun)/2*distance from the sun]^1/2
Or 5770K*1.0*[(7*10^5)/[2*(1.08*10^8)]]^0.5
5770*(7/2160)^0.5=328K


IBdaMann wrote:
tmiddles wrote:The temperature provided by NASA and the CCCP is 462C. We put a +/-200 degree margin on that simply to set aside extreme skepticism I don't myself possess.

We have been over why this is worthless. You are on tap to provide the raw data and we cannot proceed until we have it.
First of all your requirements are just that, yours. There is no "RAW DATA AS DEFINED BY IBD" I am aware of and I don't care at all about YOUR requirement as you've been resoundingly debunked here: no data is valid for IBD. Because you have never defined it. However you have stated the following regarding the data provided by the CCCP and NASA about Venus:
IBdaMann wrote: I accept the measurements with a substantial margin of error.

So what margin? Go ahead that's your wording not mine.

Of course it's obvious you DESPERATELY want to avoid this discussion because you have already lost. But I'm asking anyway to give you the benefit of the doubt.

What did YOU mean when you said that?
IBdaMann wrote:
You suggested a 200deg margin of error and I said "OK" for discussion's sake.
So let's discuss.

We have NOT:
IBdaMann wrote:...been over why this is worthless.
You said you didn't think the Russian's were dishonest, I asked if you thought they were incompetent and you've said nothing.

IBdaMann wrote:If you want to discuss Venus' emissivity then you must provide the raw data for that ...
Not necessary. We already know it's not 1.0 or greater because that's impossible. My margin of error works this in so we are all good. Sorry to rob you of a hiding place but if you have an argument make it.

"Good tests kill flawed theories; we remain alive to guess again." - Karl Popper
ITN/IBD Fraud exposed:  The 2nd LTD add on claiming radiance from cooler bodies can't be absorbed Max Planck debunks, they can't explain:net-thermal-radiation-you-in-a-room-as-a-reference & Proof: no data is valid for IBD or ITN
Edited on 10-03-2020 02:14
10-03-2020 02:56
James___
★★★★★
(2849)
tmiddles wrote:
IBdaMann wrote:
tmiddles wrote:
SOLAR ABSORPTION TEMPERATURE[/b][/u] which I define ....
...will look like a mathematical equation.


No problem. Here you go:



Tp denoting the "SOLAR ABSORPTION TEMPERATURE" for any object with albedo denoted by "a", "R" being the radius of the sun, and D being the distance from the sun in AU.


IBdaMann wrote:
tmiddles wrote:
Mean[TABA(Venus)] - SOLAR ABSORPTION TEMP = 508C +/- 300 degC

Great ... so once you provide that unambiguous "SOLAR ABSORPTION TEMP" definition we can insert it into this equation and get your complete unambiguous hypothesis and your target margin of error.
Already done. "SOLAR ABSORPTION TEMPERATURE" is simply a clarified alternative to "PLANETRY TEMPERATURE" so that it's clear it is a value that exists even without a planet being there, that it is NOT the temperature of the planet but a reference based entirely on the energy received from the Sun. So the math was done previeously and here you go:
tmiddles wrote:
For Venus we have the distance from the sun is 0.72 AU
AU is 1.0 for Earth of course and 1.0 AU is 1.5*10^8KM
So Venus is 1.08*10^8KM from the Sun.
T = (temperature of the sun)*1.0(full emissivity)^1/4 * [(radius of sun)/2*distance from the sun]^1/2
Or 5770K*1.0*[(7*10^5)/[2*(1.08*10^8)]]^0.5
5770*(7/2160)^0.5=328K


IBdaMann wrote:
tmiddles wrote:The temperature provided by NASA and the CCCP is 462C. We put a +/-200 degree margin on that simply to set aside extreme skepticism I don't myself possess.

We have been over why this is worthless. You are on tap to provide the raw data and we cannot proceed until we have it.
First of all your requirements are just that, yours. There is no "RAW DATA AS DEFINED BY IBD" I am aware of and I don't care at all about YOUR requirement as you've been resoundingly debunked here: no data is valid for IBD. Because you have never defined it. However you have stated the following regarding the data provided by the CCCP and NASA about Venus:
IBdaMann wrote: I accept the measurements with a substantial margin of error.

So what margin? Go ahead that's your wording not mine.

Of course it's obvious you DESPERATELY want to avoid this discussion because you have already lost. But I'm asking anyway to give you the benefit of the doubt.

What did YOU mean when you said that?
IBdaMann wrote:
You suggested a 200deg margin of error and I said "OK" for discussion's sake.
So let's discuss.

We have NOT:
IBdaMann wrote:...been over why this is worthless.
You said you didn't think the Russian's were dishonest, I asked if you thought they were incompetent and you've said nothing.

IBdaMann wrote:If you want to discuss Venus' emissivity then you must provide the raw data for that ...
Not necessary. We already know it's not 1.0 or greater because that's impossible. My margin of error works this in so we are all good. Sorry to rob you of a hiding place but if you have an argument make it.

"Good tests kill flawed theories; we remain alive to guess again." - Karl Popper
ITN/IBD Fraud exposed:  The 2nd LTD add on claiming radiance from cooler bodies can't be absorbed Max Planck debunks, they can't explain:net-thermal-radiation-you-in-a-room-as-a-reference & Proof: no data is valid for IBD or ITN



I'll take it easy on you on this one, okay?
Is it the 4th root or the quarter root of the radius of the Sun dived by 2AU?
Just having some fun here. I'd like to think they'd separate it more if it was the quarter root of the radius of the Sun dived by 2AU.
Any idea between the difference in temperatures and the radius of the Sun? They are associated both differently and the same. Kind of helps to know.
This is funny though. I am being polite to you. Almost respectful. Please don't let either IBDM or ITN know. It's our secret, okay ?
10-03-2020 03:46
One Punch ManProfile picture★☆☆☆☆
(134)
James___ wrote:
This is for tmiddles and for everyone else to consider.

1 mol is about 22.4 m^3 of space. With Venus' temperature and pressure, according to the Ideal Gas Law, it should have about 32,500 mols of gas in the 22.4 m^3 of space.
What this suggests is that gas molecules under pressure release heat content, ie., gases under pressure cool because they release their heat. Like steam condensing into water. If I remember correctly, 1 quart of water can make 16,000 quarts of steam unless it's under pressure which would require more energy for it to remain as steam.
After all, Venus is hotter than Mercury because of CO2, right? By all accounts CO2 should be close to being a liquid. Since it's atmosphere is a gas, please explain.

I was just perusing through some of my older comments and was reminded of a few things that I thought were rather interesting, in particular, Jupiter and Uranus. What is your opinion on Jupiter, James? Jupiter is dubbed "The Failed Star" and it's often accepted that if it had 70 times more mass than it currently does then it would be able to generate enough heat through gravitational compression for nuclear fusion to occur. Jupiter has a core temperature of 24,000K (18,800,000,000 W/m2) even though solar isolation is only 50 W/m2 and meanwhile Uranus has a tropospheric base that is radiating at 600 W/m2 (around 320K) despite it receiving only 3.7 W/m2 from the Sun.


Nathan-D
Edited on 10-03-2020 03:50
10-03-2020 04:07
James___
★★★★★
(2849)
One Punch Man wrote:
James___ wrote:
This is for tmiddles and for everyone else to consider.

1 mol is about 22.4 m^3 of space. With Venus' temperature and pressure, according to the Ideal Gas Law, it should have about 32,500 mols of gas in the 22.4 m^3 of space.
What this suggests is that gas molecules under pressure release heat content, ie., gases under pressure cool because they release their heat. Like steam condensing into water. If I remember correctly, 1 quart of water can make 16,000 quarts of steam unless it's under pressure which would require more energy for it to remain as steam.
After all, Venus is hotter than Mercury because of CO2, right? By all accounts CO2 should be close to being a liquid. Since it's atmosphere is a gas, please explain.

I was just perusing through some of my older comments and was reminded of a few things that I thought were rather interesting, in particular, Jupiter and Uranus. What is your opinion on Jupiter, James? Jupiter is dubbed "The Failed Star" and it's often accepted that if it had 70 times more mass than it currently does then it would be able to generate enough heat through gravitational compression for nuclear fusion to occur. Jupiter has a core temperature of 24,000K (18,800,000,000 W/m2) even though solar isolation is only 50 W/m2 and meanwhile Uranus has a tropospheric base that is radiating at 600 W/m2 (around 320K) despite it receiving only 3.7 W/m2 from the Sun.


I'll have to get back to you on this. Basically I'll need to consider what you said and go over what's known. If it takes more than a few days, don't be surprised. It's because I'll be giving it some thought. It's obvious you have so I'll bookmark this post of yours. I might give you an update on what I'm thinking. It'd be more to let you now that I am thinking about it. And you might be able to tell me what I might consider a little differently to help me out.
I am talking about in the next couple of days. It'll give me a chance to at least diagram the parameters. Then for the relationships that you referenced, we'd be on the same page. That's a start, right?
Edited on 10-03-2020 04:15
10-03-2020 04:55
James___
★★★★★
(2849)
Early update; 1/d^2 is the solar radiance equivalent for all planets. It's the inverse square formula for any radiated energy. For what One Man Punch asked about, this is the basis for his question.
Does the inverse square of the propagation of energy account for what he is questioning. I will be spending a couple of days going over the math. It's possible that he has noticed something that does not agree with current theory. With science, things do require being considered.
If anyone else is interested, the Earth is one AU, ie., the Sun's radius divided by 2 AU ( a part of the equation tmiddles showed), what an AU is. Jupiter is 5.2 AUs while Uranus is 19.181 AUs. It's where debates being. Common information made known. Myself, I think gravity is a little bit stranger and all Newton did was to open Pandora's Box.
As for gravity being a little different than what your textbooks say, there is an old experiment from the 1800's that might support me. If so, it's only because I am familiar with history


@One Man Punch, there is the inverse square of the Sun's gravity and if there is a relationship between that and the iridescence of a planet. That's the math I'll be checking out.
Something you might consider is if the Earth is only one point in that consideration and that we don't have a second point. As we consider it, gravity is linear. What if it was exponential?
Edited on 10-03-2020 05:02
10-03-2020 05:43
James___
★★★★★
(2849)
@One Man Punch, I won't leave you hanging on this one. There is evidence that the velocity of gravity is linear while it's force is exponential. This is if you get into the history of science. And if I am right, I do know how to demonstrate it. As for your specific question, it might take someone with a PhD years to figure out but I will consider it Hopefully you will understand that if gravity isn't a completely linear concept that it could take some serious money to resolve.
What I can do is maybe show where the current linear mathematics shows that you saw where something was wrong. From there, it is slow process because overturning conventional thought rarely happens.
10-03-2020 06:11
James___
★★★★★
(2849)
If you monkeys don't get it or in the case of ITN, there's a reason that atmospheric air pressure on Earth is 14.7 psi or 1.03 kg/cm^2. There is something that atmospheric air pressure is based on going back to the 1800's. Isn't history "KEWL"?
And I am willing to bet that none of you know what that experiment was about. It determined what atmospheric air pressure is. Basically 0.98kg created 1.03kg of force. That's not possible unless you consider the history behind it.
That being said, this is a debate forum and that has nothing to do with history or what history has demonstrated.
10-03-2020 10:18
Into the NightProfile picture★★★★★
(12789)
tmiddles wrote:...deleted Mantra 25k...20e2...25g....4b...37a...25c...7....7...29...29...30...29...25g...25c...7...29...


No arguments presented. RQAA.


The Parrot Killer
10-03-2020 16:13
IBdaMannProfile picture★★★★★
(7068)
tmiddles wrote:No problem. Here you go:


Tp denoting the "SOLAR ABSORPTION TEMPERATURE" for any object with albedo denoted by "a", "R" being the radius of the sun, and D being the distance from the sun in AU.

You inserted a graphic. Please type it out and explain all your terms. You explained only some of the terms.

tmiddles wrote:
IBdaMann wrote:You are on tap to provide the raw data and we cannot proceed until we have it.
First of all your requirements are just that, yours.

Great. Let me know when something changes.


.


A Spaghetti strainer with the faucet running, retains water- tmiddles

Clouds don't trap heat. Clouds block cold. - Spongy Iris

Printing dollars to pay debt doesn't increase the number of dollars. - keepit

If Venus were a black body it would have a much much lower temperature than what we found there.- tmiddles

Ah the "Valid Data" myth of ITN/IBD. - tmiddles

Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafn

You are completely misunderstanding their use of the word "accumulation"! - Climate Scientist.

The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank

:*sigh* Not the "raw data" crap. - Leafsdude

IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist
10-03-2020 18:52
James___
★★★★★
(2849)
One Punch Man wrote:
James___ wrote:
This is for tmiddles and for everyone else to consider.

1 mol is about 22.4 m^3 of space. With Venus' temperature and pressure, according to the Ideal Gas Law, it should have about 32,500 mols of gas in the 22.4 m^3 of space.
What this suggests is that gas molecules under pressure release heat content, ie., gases under pressure cool because they release their heat. Like steam condensing into water. If I remember correctly, 1 quart of water can make 16,000 quarts of steam unless it's under pressure which would require more energy for it to remain as steam.
After all, Venus is hotter than Mercury because of CO2, right? By all accounts CO2 should be close to being a liquid. Since it's atmosphere is a gas, please explain.

I was just perusing through some of my older comments and was reminded of a few things that I thought were rather interesting, in particular, Jupiter and Uranus. What is your opinion on Jupiter, James? Jupiter is dubbed "The Failed Star" and it's often accepted that if it had 70 times more mass than it currently does then it would be able to generate enough heat through gravitational compression for nuclear fusion to occur. Jupiter has a core temperature of 24,000K (18,800,000,000 W/m2) even though solar isolation is only 50 W/m2 and meanwhile Uranus has a tropospheric base that is radiating at 600 W/m2 (around 320K) despite it receiving only 3.7 W/m2 from the Sun.


I'll give you an idea of what I'm thinking. The Earth is 1 AU from the Sun and Mars is somewhere around 1 1/2 AUs from the Sun. It's gravity is 3.41 m/s^2 which is about the inverse square of the Earth's.
What I'm wondering is how much effect the Sun's gravity (field) has on a planet's own gravity.
10-03-2020 19:38
Into the NightProfile picture★★★★★
(12789)
James___ wrote:
One Punch Man wrote:
James___ wrote:
This is for tmiddles and for everyone else to consider.

1 mol is about 22.4 m^3 of space. With Venus' temperature and pressure, according to the Ideal Gas Law, it should have about 32,500 mols of gas in the 22.4 m^3 of space.
What this suggests is that gas molecules under pressure release heat content, ie., gases under pressure cool because they release their heat. Like steam condensing into water. If I remember correctly, 1 quart of water can make 16,000 quarts of steam unless it's under pressure which would require more energy for it to remain as steam.
After all, Venus is hotter than Mercury because of CO2, right? By all accounts CO2 should be close to being a liquid. Since it's atmosphere is a gas, please explain.

I was just perusing through some of my older comments and was reminded of a few things that I thought were rather interesting, in particular, Jupiter and Uranus. What is your opinion on Jupiter, James? Jupiter is dubbed "The Failed Star" and it's often accepted that if it had 70 times more mass than it currently does then it would be able to generate enough heat through gravitational compression for nuclear fusion to occur. Jupiter has a core temperature of 24,000K (18,800,000,000 W/m2) even though solar isolation is only 50 W/m2 and meanwhile Uranus has a tropospheric base that is radiating at 600 W/m2 (around 320K) despite it receiving only 3.7 W/m2 from the Sun.


I'll give you an idea of what I'm thinking. The Earth is 1 AU from the Sun and Mars is somewhere around 1 1/2 AUs from the Sun. It's gravity is 3.41 m/s^2 which is about the inverse square of the Earth's.
What I'm wondering is how much effect the Sun's gravity (field) has on a planet's own gravity.

Maybe you should go study Galileo, Kepler, and Newton some more.


The Parrot Killer
10-03-2020 21:32
James___
★★★★★
(2849)
One Punch Man wrote:
James___ wrote:
This is for tmiddles and for everyone else to consider.

1 mol is about 22.4 m^3 of space. With Venus' temperature and pressure, according to the Ideal Gas Law, it should have about 32,500 mols of gas in the 22.4 m^3 of space.
What this suggests is that gas molecules under pressure release heat content, ie., gases under pressure cool because they release their heat. Like steam condensing into water. If I remember correctly, 1 quart of water can make 16,000 quarts of steam unless it's under pressure which would require more energy for it to remain as steam.
After all, Venus is hotter than Mercury because of CO2, right? By all accounts CO2 should be close to being a liquid. Since it's atmosphere is a gas, please explain.

I was just perusing through some of my older comments and was reminded of a few things that I thought were rather interesting, in particular, Jupiter and Uranus. What is your opinion on Jupiter, James? Jupiter is dubbed "The Failed Star" and it's often accepted that if it had 70 times more mass than it currently does then it would be able to generate enough heat through gravitational compression for nuclear fusion to occur. Jupiter has a core temperature of 24,000K (18,800,000,000 W/m2) even though solar isolation is only 50 W/m2 and meanwhile Uranus has a tropospheric base that is radiating at 600 W/m2 (around 320K) despite it receiving only 3.7 W/m2 from the Sun.



I have been pursuing an experiment that would be an attempt to show that gravity does work besides accelerating matter. An example of this is that a satellite stays in orbit because our planet's gravity transfers force to them.
As far as the gas giants goes, the Sun's gravitational field most likely has less of an effect on them. This would mean that the gravitational fields of the gas giants should be able to excite their atmospheres more efficiently as a result. It's possible that if they were closer to the Sun that they might burn of their atmospheres and then have a different atmosphere.
10-03-2020 23:12
tmiddlesProfile picture★★★★★
(3266)
IBdaMann wrote:
tmiddles wrote:No problem. Here you go:


Tp denoting the "SOLAR ABSORPTION TEMPERATURE" for any object with albedo denoted by "a", "R" being the radius of the sun, and D being the distance from the sun in AU.

You inserted a graphic. Please type it out and explain all your terms.
No a graphic is better. Ascii is too difficult to represent everything.

I explained all of the terms already.

The equation reads as follows:
The "SOLAR ABSORPTION TEMPERATURE" is equal to:
The temperature of the Sun multiplied by the 4th root (or to the 1/4 power) of 1 minus the albedo (or the emissivity) multiplied by the square root of the radius of the Sun divided by two times the distance to the sun.

Is anything unclear for you?
10-03-2020 23:50
IBdaMannProfile picture★★★★★
(7068)
tmiddles wrote: No a graphic is better. Ascii is too difficult to represent everything.

1) I would like your text to quote.
2) I have seen a fairly good sampling of your posts and given your typing proficiency it would be a piece of cake for you to type out the equation in text.

I explained all of the terms already.

tmiddles wrote: Is anything unclear for you?

Yes. For starters ... does this equation factor into your your equation?



If so, where is the emissivity term? I notice that emissivity is included in this equation:



.


A Spaghetti strainer with the faucet running, retains water- tmiddles

Clouds don't trap heat. Clouds block cold. - Spongy Iris

Printing dollars to pay debt doesn't increase the number of dollars. - keepit

If Venus were a black body it would have a much much lower temperature than what we found there.- tmiddles

Ah the "Valid Data" myth of ITN/IBD. - tmiddles

Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafn

You are completely misunderstanding their use of the word "accumulation"! - Climate Scientist.

The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank

:*sigh* Not the "raw data" crap. - Leafsdude

IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist
11-03-2020 00:58
James___
★★★★★
(2849)
This is the same as (1-a) to the 1/4 power. It is not multiplied. It's like the 4th root of 64 is 2.82842712475. Multiplied by itself 4 times then he can be expressed using a radical sign. With the attached image and link if the image doesn't attach, just showing another way to write a part of the expression being discussed.
https://photos.app.goo.gl/cgtc4gyS2edVNcja9
11-03-2020 01:31
tmiddlesProfile picture★★★★★
(3266)
IBdaMann wrote:
1) I would like your text to quote.
We'll I've written out what the equation is. So you can quote that.

IBdaMann wrote:....does this equation factor into your your equation?

Yes I am following this solution here:
http://burro.case.edu/Academics/Astr221/SolarSys/equiltemp.html
Again this is a rerun with you we've already been over it.

IBdaMann wrote:...where is the emissivity term? I notice that emissivity is included in this equation:


You don't see Tp in that equation? Look closer, it's the last variable shown taken to the 4th power.

And Tp, or the "SOLAR ABSORPTION TEMPERATURE", includes the (1-a) emissivity as already shown.

I'm assuming you don't have a solution of your own to this problem and aren't just stalling?

Was there a purpose to your questions? Do you doubt that the math has been done correctly and if so why?
11-03-2020 04:47
Into the NightProfile picture★★★★★
(12789)
tmiddles wrote:...deleted TMSa7...


No argument presented. RQAA.


The Parrot Killer
11-03-2020 04:58
IBdaMannProfile picture★★★★★
(7068)
tmiddles wrote: Was there a purpose to your questions?

Were you planning on making a list of approved purposes for questions?

tmiddles wrote: Do you doubt that the math has been done correctly and if so why?

I need to look it over some more. I'm not buying your Tp. At first glance it appears your are squaring the surface area and altering the emissivity ... but I can't devote any more time to this right now.


.






.


A Spaghetti strainer with the faucet running, retains water- tmiddles

Clouds don't trap heat. Clouds block cold. - Spongy Iris

Printing dollars to pay debt doesn't increase the number of dollars. - keepit

If Venus were a black body it would have a much much lower temperature than what we found there.- tmiddles

Ah the "Valid Data" myth of ITN/IBD. - tmiddles

Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafn

You are completely misunderstanding their use of the word "accumulation"! - Climate Scientist.

The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank

:*sigh* Not the "raw data" crap. - Leafsdude

IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist
11-03-2020 05:13
tmiddlesProfile picture★★★★★
(3266)
IBdaMann wrote:
I need to look it over some more. I'm not buying your Tp. At first glance it appears your are squaring the surface area and altering the emissivity ... =


So to circle back with the overall "what you're not buying" it is that the ground level temperature on Venus, the "Mean[TABA(Venus)]", is greater than a temperature emitting in equilibrium with the energy being received from the Sun (what I've called the "SOLAR ABSORPTION TEMPERATURE"). Is that the case? NASA and CCCP numbers have it as over 500 degree greater but you don't buy that correct?

You believe NASA and the CCCP have gotten it so wrong they are off by more than 500 degrees if I'm understanding you. Not due to dishonesty but due to _____? What exactly?

To clarify my own understanding none of this is a hypothesis at the moment. The ground level temperature of Venus and the energy being received by the Sun are quantities that have been repeatedly measured directly.

Do you concede that your use of the 1st LTD and Placks Law/Stefan Boltzmann are in error if indeed the ground level of Venus is hotter than the "SOLAR ABSORPTION TEMPERATURE"/Equilibrium Temperature ?

ITN gave up long ago so you're free to do that. But I'm here to debate.
11-03-2020 06:33
Into the NightProfile picture★★★★★
(12789)
tmiddles wrote:
IBdaMann wrote:
I need to look it over some more. I'm not buying your Tp. At first glance it appears your are squaring the surface area and altering the emissivity ... =


So to circle back with the overall "what you're not buying" it is that the ground level temperature on Venus, the "Mean[TABA(Venus)]", is greater than a temperature emitting in equilibrium with the energy being received from the Sun (what I've called the "SOLAR ABSORPTION TEMPERATURE"). Is that the case? NASA and CCCP numbers have it as over 500 degree greater but you don't buy that correct?

You believe NASA and the CCCP have gotten it so wrong they are off by more than 500 degrees if I'm understanding you. Not due to dishonesty but due to _____? What exactly?

Neither NASA nor the CCCP has measured the temperature of Venus nor the emissivity of Venus. Mantras 25g...4d...20o...37a...29.
tmiddles wrote:
To clarify my own understanding none of this is a hypothesis at the moment.

Learn what 'hypothesis' means. A hypothesis isn't a theory, observation, or data.
tmiddles wrote:
The ground level temperature of Venus and the energy being received by the Sun are quantities that have been repeatedly measured directly.

Never have. Mantras 25g.
tmiddles wrote:
Do you concede that your use of the 1st LTD and Placks Law/Stefan Boltzmann are in error if indeed the ground level of Venus is hotter than the "SOLAR ABSORPTION TEMPERATURE"/Equilibrium Temperature ?

Meaningless buzzwords. Mantras 22...20a1...25k...20e1...20e1.
tmiddles wrote:
ITN gave up long ago so you're free to do that. But I'm here to debate.

Mantras 17...lie. You are not here to debate. You are here to preach. You are not debating at all.

No argument presented. RQAA.


The Parrot Killer
Edited on 11-03-2020 06:34
11-03-2020 07:40
tmiddlesProfile picture★★★★★
(3266)
IBdaMann wrote:....it appears your are squaring the surface area and altering the emissivity ... =
So disregarding emissivity for a moment (my margin of error has):

I will use "R" for the radius of the Sun and "r" for the radius of the object. And "T" will be the temperature of the Sun and "t" for the temperature the object would have emitting in equilibrium. "D" is the distance to the Sun.

The fraction of a sphere emitting or receiving radiance in one direction is a circle: πR^2

So the energy from the sun as per Stefan-Boltzman is: σT^4*πR^2 , while the absorbing surface of the object is: πr^2

The Sun's light disipates as the inverse square: 1/D^2

So the ENERGY IN received by the object from the sun is:
(πr^2)*(σT^4*πR^2)*(1/D^2)


The ENERGY OUT, the object emitting at temperature "t" is σt^4*4πr^2 (note that it's emitting as a sphere, 4πr^2)

If there is equilibrium then:

ENERGY OUT = ENERGY IN

σt^4*4πr^2=(πr^2)*(σT^4*πR^2)*(1/D^2)

Now we can simply cancel: σ,π,r^2

Leaving:
t^4*4=(T^4*R^2)*(1/D^2)

Now take the square root:

t^2*2=T^2*R*(1/D)

Divide both sides by 2

t^2=T^2*R(1/D)(1/2)

t^2=T^2(R/2D)

Take the square root again:

t=T(R/2D)^0.5

which is:


Now if you wanted to include emissivity for the object then it would have been (1-a)(πr^2), 1 minus the albedo times the circle area receving the Sun's light, and after taking the square root twice you'd have a (1-a)^0.25 in there.

My margin of error covers all possible values of emissivity up to 1.0 so it's irrelevant.
Edited on 11-03-2020 07:43
11-03-2020 07:59
tmiddlesProfile picture★★★★★
(3266)
Into the Night wrote:
Neither NASA nor the CCCP has measured the temperature of Venus nor the emissivity of Venus.
What about the chemical composition of the the atmosphere of Venus? ITN? Have they measured that? How about the atmospheric pressure of Venus? Have they measured that?

Oh wait a minute.... You were about to say RQAA but... what's this!!!!
ITN HAS ANSWERED ME!!!!!!!!!!


Into the Night wrote:In the case of Venus, that atmosphere has almost no hydrogen in it at all.

Into the Night wrote:The Venusian atmosphere is almost all CO2.

My head is spinning! The confidence with which ITN is laying out the "realities" of Venus. He just knows what he's talking about suddenly. Wait! There's more!:

Into the Night wrote:The surface pressure is 90 times the surface pressure on Earth.

Into the Night wrote:...the high temperatures of Venus.


Wow! Here is the link to ITN throwing down the deets like a pro!:
the DATA MINE

So ITN, I gotta ask, did someone hack your account when the Data Mine was written? You like, knew stuff, I hardly recognize your posts.

What happened man?

You do have a lot to answer for there....I will not hold my breath.

R to the Q to the A A

"Good tests kill flawed theories; we remain alive to guess again." - Karl Popper
ITN/IBD Fraud exposed:  The 2nd LTD add on claiming radiance from cooler bodies can't be absorbed Max Planck debunks, they can't explain:net-thermal-radiation-you-in-a-room-as-a-reference & Proof: no data is valid for IBD or ITN
Edited on 11-03-2020 08:32
11-03-2020 11:30
tmiddlesProfile picture★★★★★
(3266)
Oops, I goofed up the math a bit and have a left over pie.
I would never intentionally have left over pie.

tmiddles wrote:
σt^4*4πr^2=(πr^2)*(σT^4*πR^2)*(1/D^2)


So let's try a better approach:
So I like this link:
https://www.astro.princeton.edu/~strauss/FRS113/writeup3/

Again I will use "R" for the radius of the Sun and "r" for the radius of the object. And "T" will be the temperature of the Sun and "t" for the temperature the object would have emitting in equilibrium. "D" is the distance to the Sun.

So the object/planet (a sphere) is getting a flat circle of exposure from the sun: πr^2



The sun sends out radiance in an expanding sphere. So the sphere of radiance that reaches the planet, at distance D is 4πD^2.

So if you know the total energy output of the Sun, the fraction that the planet gets is:
πr^2/4πD^2=r^2/2D^2

Now the total output of the Sun is:
σT^4*4πR^2

So the total ENERGY IN is:
σT^4*4πR^2*r^2/2D^2

The ENERGY OUT is:
σt^4*4πr^2

If there is an equilibrium, ENERGY IN = ENERGY OUT

σT^4*4πR^2*r^2/2D^2=σt^4*4πr^2

We divide both sides by: σ, π, 4, and r^2

T^4*R^2/2D^2=t^4

Take the square root

T^2*R/2D=t^2

Take the square toot again

t=T(R/2D)^0.5

The "SOLAR ABSORPTION TEMPERATURE"/"Equilibrium Temperature" (for an emissivity of 1.0) is the temperature of the Sun times the square root of: the radius of the Sun divided by the distance to the sun squared.

"Good tests kill flawed theories; we remain alive to guess again." - Karl Popper
ITN/IBD Fraud exposed:  The 2nd LTD add on claiming radiance from cooler bodies can't be absorbed Max Planck debunks, they can't explain:net-thermal-radiation-you-in-a-room-as-a-reference & Proof: no data is valid for IBD or ITN
Edited on 11-03-2020 11:36
11-03-2020 14:27
tmiddlesProfile picture★★★★★
(3266)
I meant it is the temperature of the Sun times the square root of: the radius of the Sun divided by the twice the distance to the sun.

11-03-2020 16:31
James___
★★★★★
(2849)
Square root of Radius of the Sun/2D.
Is interesting though. The corona of the Sun is a few million degrees Celsius to the Sun's surface of 6,000° C. but emits 1 w/m^2.
We know the Earth's Van Allen radiation belts cools the Earth. Could the Sun's corona lower its emissivity?
Also is there any relationship between the Sun's surface temperature and that of satellites or planets surface temperature that has little or no atmosphere?
An example is the Earth is 213.5 of the Sun's radius from it. Every time the radius doubles, heat would be 1/2. This is if the Sun's surface temperature is relative to the heat it's radiating.
Or we could just say 2,718,671.5 mi /213 = 12,763
That would give a temperature of 2.127º C. at the Earth's orbit. Or about 35.8º F.

>> If Earth had no atmosphere and we had to rely upon the Sun's energy alone, Earth would be a frigid place. Its mean global temperature would be about 0°F.
https://www.climate.gov/news-features/understanding-climate/climate-change-incoming-sunlight <<
Edited on 11-03-2020 17:02
11-03-2020 17:33
James___
★★★★★
(2849)
The numbers I gave for radiant heat from the Sun should be about 2° C. higher.
11-03-2020 20:16
James___
★★★★★
(2849)
I thought I'd mention something misleading when going by º C. or º F.
As I mentioned, the Sun's surface temperature is about 6,000º C. While that = 6273.15 kelvins.
I'm using 300 because it's a round number.
The Earth is about 112 diameters of the Sun away from it. The basic calculation for kelvins is;
6,300/112 = 56.25º kelvin which = -216.9º C.

For Celsius it's;
6,000/112 = 53.57º C.

Notice how not considering temperature in kelvins changes things. When the IPCC considers how CO2 levels influence our atmosphere, they are considering it from about 287º kelvin while ignoring that part of the equation.
Basically, does CO2 influence increase above 285º kelvin? Or would its warming influence be proportional to when it can be a gas at 194.65º kelvin?
Edited on 11-03-2020 20:17
11-03-2020 20:47
Into the NightProfile picture★★★★★
(12789)
tmiddles wrote:
[quote]IBdaMann wrote:....it appears your are squaring the surface area and altering the emissivity ... =
So disregarding emissivity for a moment (my margin of error has):

I will use "R" for the radius of the Sun and "r" for the radius of the object. And "T" will be the temperature of the Sun and "t" for the temperature the object would have emitting in equilibrium. "D" is the distance to the Sun.
Math error. Define 'emitting in equilibrium'. 't' remains undefined. Temperature doesn't emit anything. Unit error.
tmiddles wrote:
The fraction of a sphere emitting or receiving radiance in one direction is a circle: πR^2

So the energy from the sun as per Stefan-Boltzman is: σT^4*πR^2 , while the absorbing surface of the object is: πr^2

The Sun's light disipates as the inverse square: 1/D^2

So the ENERGY IN received by the object from the sun is:
(πr^2)*(σT^4*πR^2)*(1/D^2)

Math error. Failure to include arc section of emission by the Sun.
tmiddles wrote:
The ENERGY OUT, the object emitting at temperature "t" is σt^4*4πr^2 (note that it's emitting as a sphere, 4πr^2)

Math error. The shape of the emitting object is immaterial to the power emitted. You are introducing a new term into the Stefan-Boltzmann law.
Math error. You removed the emissivity constant from the Stefan-Boltzmann law.
tmiddles wrote:
If there is equilibrium then:

ENERGY OUT = ENERGY IN

σt^4*4πr^2=(πr^2)*(σT^4*πR^2)*(1/D^2)

Math error. 't' is undefined. Introduced extraneous term into the Stefan-Boltzmann law.
tmiddles wrote:
Now we can simply cancel: σ,π,r^2

Leaving:
t^4*4=(T^4*R^2)*(1/D^2)

Math error. You canceled out the right side twice.
tmiddles wrote:
Now take the square root:

t^2*2=T^2*R*(1/D)

Divide both sides by 2

t^2=T^2*R(1/D)(1/2)

t^2=T^2(R/2D)

Math error. You failed to divide T^2 by 2.
tmiddles wrote:
Take the square root again:

t=T(R/2D)^0.5

which is:


Not the same equation.
tmiddles wrote:
Now if you wanted to include emissivity for the object then it would have been (1-a)(πr^2), 1 minus the albedo times the circle area receving the Sun's light, and after taking the square root twice you'd have a (1-a)^0.25 in there.

Math error. Emissivity is not affected by the area of the object. Math error. Assumption of uniform emissivity of the object.
tmiddles wrote:
My margin of error covers all possible values of emissivity up to 1.0 so it's irrelevant.

Math error. Margin of error is a calculated value. You can't just pick it out of the air. Failure to declare variance.

Your 'proof' contains numerous math errors and invalid assumptions. The 'proof' is invalid, therefore the resulting equations you managed to come up with are also invalid.


The Parrot Killer
Edited on 11-03-2020 20:50
11-03-2020 21:33
IBdaMannProfile picture★★★★★
(7068)
tmiddles wrote: So to circle back with the overall "what you're not buying" it is that the ground level temperature on Venus, the "Mean[TABA(Venus)]", is greater than a temperature emitting in equilibrium with the energy being received from the Sun (what I've called the "SOLAR ABSORPTION TEMPERATURE"). Is that the case?

Frankly, I am having trouble parsing your words.

For any given planet that has an atmopshere, the average temperature at the bottom of the atmosphere will be higher than the planet's average black-body science temperature.

1) I am trying to figure out if the equations you are using are the same as the blackbody science laws, i.e. Planck's, Stefan-Boltzmann, etc... We could save time if you would just use those. If your equations differ then I will have a problem with them to that extent.

2) Then we have to address your value for emissivity. You made it up. I'm not sure, however, if it matters much in this case so we can table that for the present. We can address it later if we need to.

3) You have not provided any raw data ... because you don't have any ... and so of course it's all my fault. This is going to be a problem if you make it over the other hurdles.

4) You opened by declaring a HUGE margin of error. I am certain that you don't know what you are doing, and that you don't realize that you are effectively ensuring that you are wasting your time. I am certain that you believe that your initial target margin of error applies to your data values and that you don't realize that it applies to your conclusions. You should be seeking a small target margin of error and working towards that; that way people will be inclined to agree with your conclusions and with your argument. If others find your margin of error to be too big they will dismiss your conclusions and your argument. If your target margin of error is intentionally huge, what's the point? You are constructing an argument that will either say absolutely nothing or that absolutely no one will accept.

5) Your argument is that the average temperature at the bottom of the atmosphere is at least a certain amount above the planet's average temperature, but you refuse to include the Ideal Gas Law to account for the temperature effects of the atmopshere. Good luck.

Would you like to address these points now or later?


tmiddles wrote: NASA and CCCP numbers have it as over 500 degree greater but you don't buy that correct?

Without the raw data, it is not the case that NASA and CCCP have anything to say on the matter.

tmiddles wrote: To clarify my own understanding none of this is a hypothesis at the moment. The ground level temperature of Venus and the energy being received by the Sun are quantities that have been repeatedly measured directly.

We need to focus on the definite article "the" in this case.

You simply don't have "the" Venus ground-level temperature. Period. Remember, you don't have any raw data and omniscience doesn't count.

You don't have Venus' average planetary temperature to any usable accuracy. You don't know Venus' emissivity. Nobody does. I'm still trying to figure out how you plan on performing a subtraction operation on two unknown values, and what would you be able to say about the margin of error of that operation? Would you claim that the answer = unknown +/- unknown?

I'm not sure this is a sufficiently compelling argument to be pursuing.



.


A Spaghetti strainer with the faucet running, retains water- tmiddles

Clouds don't trap heat. Clouds block cold. - Spongy Iris

Printing dollars to pay debt doesn't increase the number of dollars. - keepit

If Venus were a black body it would have a much much lower temperature than what we found there.- tmiddles

Ah the "Valid Data" myth of ITN/IBD. - tmiddles

Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafn

You are completely misunderstanding their use of the word "accumulation"! - Climate Scientist.

The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank

:*sigh* Not the "raw data" crap. - Leafsdude

IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist
11-03-2020 21:39
Into the NightProfile picture★★★★★
(12789)
tmiddles wrote:...deleted Mantras 25c...29...25c...29...16c...4c...29...4c...29...29...


No argument presented. RQAA.


The Parrot Killer
11-03-2020 21:42
Into the NightProfile picture★★★★★
(12789)
tmiddles wrote:
Oops, I goofed up the math a bit and have a left over pie.
I would never intentionally have left over pie.

tmiddles wrote:
σt^4*4πr^2=(πr^2)*(σT^4*πR^2)*(1/D^2)


So let's try a better approach:
So I like this link:
https://www.astro.princeton.edu/~strauss/FRS113/writeup3/

Again I will use "R" for the radius of the Sun and "r" for the radius of the object. And "T" will be the temperature of the Sun and "t" for the temperature the object would have emitting in equilibrium. "D" is the distance to the Sun.

So the object/planet (a sphere) is getting a flat circle of exposure from the sun: πr^2



The sun sends out radiance in an expanding sphere. So the sphere of radiance that reaches the planet, at distance D is 4πD^2.

So if you know the total energy output of the Sun, the fraction that the planet gets is:
πr^2/4πD^2=r^2/2D^2

Now the total output of the Sun is:
σT^4*4πR^2

So the total ENERGY IN is:
σT^4*4πR^2*r^2/2D^2

The ENERGY OUT is:
σt^4*4πr^2

If there is an equilibrium, ENERGY IN = ENERGY OUT

σT^4*4πR^2*r^2/2D^2=σt^4*4πr^2

We divide both sides by: σ, π, 4, and r^2

T^4*R^2/2D^2=t^4

Take the square root

T^2*R/2D=t^2

Take the square toot again

t=T(R/2D)^0.5

The "SOLAR ABSORPTION TEMPERATURE"/"Equilibrium Temperature" (for an emissivity of 1.0) is the temperature of the Sun times the square root of: the radius of the Sun divided by the distance to the sun squared.


Similar math errors as last time. You really gotta stop making these, dude.


The Parrot Killer
11-03-2020 22:02
tmiddlesProfile picture★★★★★
(3266)
Into the Night wrote:
tmiddles wrote:
So let's try a better approach:

Similar math errors as last time. You really gotta stop making these, dude.

I looked at your comments on my first proof (which was a waste of time since I replaced it) and YOUR numerous errors are shown. You only got one thing right actually which is the error I already caught of the extra pie.

Into the Night wrote:
tmiddles wrote:
I will use "R" for the radius ... "D" is the distance to the Sun.
....Define 'emitting in equilibrium'. 't' remains undefined. Temperature doesn't emit anything.
t is the temperature of the object emitting in equilibrium. Temperature determines emission as per SB. "emtting in equilibrium" means that the emission matches the absorption. The "Energy in" equal the "Energy out". That clear enough?

Into the Night wrote:Math error. Failure to include arc section of emission by the Sun.
This way of doing it basically treats the emission of the Sun as a giant sphere that has reached the object. So it works fine. IF there was an error you should be able to point out the RIGHT way to do it.

Into the Night wrote:
tmiddles wrote:..."t" is σt^4*4πr^2 (note that it's emitting as a sphere, 4πr^2)

Math error. The shape of the emitting object is immaterial to the power emitted. You are introducing a new term into the Stefan-Boltzmann law.
All I'm doing is putting Area into the equation where it belongs. That's just the formula for the are of a sphere. I was pointing out it was the sphere formula as we'd just used the circle area formula instead earlier.

Into the Night wrote:
tmiddles wrote:
σt^4*4πr^2=(πr^2)*(σT^4*πR^2)*(1/D^2)
Now we can simply cancel: σ,π,r^2
Leaving:
t^4*4=(T^4*R^2)*(1/D^2)

Math error. You canceled out the right side twice.

I already corrected this with my second proof which is better. Comment on that. There are two pie's on the right correct.
Look again:
σt^4*4πr^2=(πr^2)*(σT^4*πR^2)*(1/D^2)
t^4*4=(T^4*R^2)*(1/D^2)

Into the Night wrote:
tmiddles wrote:
...
t^2*2=T^2*R*(1/D)
Divide both sides by 2
t^2=T^2*R(1/D)(1/2)

Math error. You failed to divide T^2 by 2.
No it's right, see the 2 on the left divided out and the "* 1/2" now on the right. *1/2 is dividing by 2.

Into the Night wrote:Not the same equation.
I'm referencing this threads earlier discussion. Try to follow along.

Into the Night wrote: Emissivity is ....
Irrelevant for the purposes of this discussion. My margin of error includes all possibilities. Do you see a problem with that?

Into the Night wrote: Margin of error is a calculated value.
OK. So what you're saying is we can't talk about it? What a surprise!

Into the Night wrote:Your 'proof' contains numerous math errors
I have now pointed out all of your errors in assuming so.

"Good tests kill flawed theories; we remain alive to guess again." - Karl Popper
ITN/IBD Fraud exposed:  The 2nd LTD add on claiming radiance from cooler bodies can't be absorbed Max Planck debunks, they can't explain:net-thermal-radiation-you-in-a-room-as-a-reference & Proof: no data is valid for IBD or ITN
Edited on 11-03-2020 22:05
11-03-2020 22:50
tmiddlesProfile picture★★★★★
(3266)
IBdaMann wrote:
For any given planet that has an atmosphere, the average temperature at the bottom of the atmosphere will be higher than the planet's average black-body science temperature.
This is you're saying this correct? Have you addressed this before? Perhaps I've been remiss in that somehow that's escaped our conversations here.

IBdaMann wrote:
1) I am trying to figure out if the equations you are using are the same as the blackbody science laws, i.e. Planck's, Stefan-Boltzmann, etc...
Have you looked at the derivation for t=T(R/2D)^0.5 above? It's perfectly conventional use of SB as far as I know. Corrections welcomed.

IBdaMann wrote:...we have to address your value for emissivity.
I'm assuming, for the sake of argument, we know absolutely nothing about the emissivity. It could be anything so we must include all possible values in our margin of error. Do you have a problem with that?

IBdaMann wrote:...You have not provided any raw data ...
For the sake of argument let's just ask this question: The CCCP and NASA have provided a value without a margin of error. Can we apply a margin of error to that value ourselves without "raw data"? Yes of course we are talking about making up a number for the sake of argument. You were willing to do that earlier.

Of course I have no clue what you mean by raw data as there are no examples you will acknowledge on this board. You and ITN were in agreement on this subject in the DATA MINE and you're both thoroughly debunked on it in my sig.

IBdaMann wrote:
You opened by declaring a HUGE margin of error.
Yes I did and I stand by it for the purposes of the debate. +/- 300 degrees when you factor in emissivity. That is +/- 300 degrees on the value provided by the "experts" from the CCCP and NASA, no IBD approved "Raw Data" at all, so a really big margin for this debate (because you're in it). I personally do not doubt their numbers.

IBdaMann wrote:...you believe that your initial target margin of error applies to your data values and that you don't realize that it applies to your conclusions.
Did you have a comment on my actual calculation?:
Mean[TABA(Venus)] - SOLAR ABSORPTION TEMP = 508C +/- 300 degC

What you've said makes no sense to me. Having a margin of error on input data will naturally effect the calculation but you do not apply the margin twice.

IBdaMann wrote:....you refuse to include the Ideal Gas Law to account for the temperature effects of the atmosphere. Good luck.
My calculated value, with the huge margin of error, successfully debunks most of your arguments over the past 5 years. I don't have to include any comment on the ideal gas law to debunk your errors because you didn't.

IBdaMann wrote:Would you like to address these points now or later?
All 5 are addressed above. Let me know if I left anything out.

IBdaMann wrote:...it is not the case that NASA and CCCP have anything to say on the matter.
That does not make sense as I read it. Care to elaborate?

IBdaMann wrote:You don't have Venus' average planetary temperature to any usable accuracy.
Really? Not about chemical composition? Temp? Pressure? Nothing? We know nothing at all?

Into the Night wrote:In the case of Venus, that atmosphere has almost no hydrogen in it at all.
Into the Night wrote:The Venusian atmosphere is almost all CO2.
Into the Night wrote:The surface pressure is 90 times the surface pressure on Earth.
Into the Night wrote:...the high temperatures of Venus.
from the DATA MINE
IBdaMann wrote:...the extreme heat of Venus, ...
link

Now I've gotta let ITN say it here:
Into the Night wrote: ....You have been EVADING a sensible conversation by ...demanding data to be formatted in a particular way...Start having a discussion about the meaning of the data you have!
From his thread, the Data Mine

I'm saying that the bottom of the atmosphere of Venus being 200+ degrees hotter than the equilibrium temperature for that distance from the sun is a reliable conclusion we can have confidence in. I can only assume you are evading even entertaining a discussion based on that be cause you know as well as I do that it destroys most of your arguments over the past 5 years.

"Good tests kill flawed theories; we remain alive to guess again." - Karl Popper
ITN/IBD Fraud exposed:  The 2nd LTD add on claiming radiance from cooler bodies can't be absorbed Max Planck debunks, they can't explain:net-thermal-radiation-you-in-a-room-as-a-reference & Proof: no data is valid for IBD or ITN
11-03-2020 23:30
James___
★★★★★
(2849)
tmiddles wrote:
Have you looked at the derivation for t=T(R/2D)^0.5 above? It's perfectly conventional use of SB as far as I know.



I have a question for you. The emissivity of the Earth is based on solar IR which as I've shown, by the time it gets here is about 60 kelvins. The Earth's emissivity would be based on that. Yet they say the solar constant is over 1,300 w/m^2 when not that much energy is coming from the Sun.
For the Sun to have an emissivity of about 6,300º kelvin, what is the source of energy? You don't have to answer this. While the Earth has an energy budget based on incoming solar IR, the Sun doesn't.

https://earthobservatory.nasa.gov/features/EnergyBalance
Edited on 11-03-2020 23:32
12-03-2020 01:21
James___
★★★★★
(2849)
As strange as it may seem people, there might be a relationship between the Sun's corona and it's surface temperature which possibly could have a relationship between the Earth's Van Allen radiation belts and its emissivity. But that's getting into astrophysics and this is a climate debate forum.
The Sun's relationship with its corona is 300:1 while the Earth's relationship to the Van Allen radiation belts is 700:1.
Could the Sun's gravitational field and the Earth's own gravitational field allow for the difference? You know, the Sun's field makes our radiation belts even hotter. Or just the difference in gravity of those 2 celestial bodies?
Edited on 12-03-2020 01:28
12-03-2020 02:30
Into the NightProfile picture★★★★★
(12789)
tmiddles wrote:...deleted Mantras 29...lie...25...20b2...20b (addition of area term * inverse square law term)...29...25c...20b2...25g...25c...16c...30...2...29...7...paradox q...25g...37a...37d...25c...37a...37d...20k...20o...22 (solar absorption temperature)...25c...25c...25...7...7...6...lie...29...2...29...25c...29...25c...29...25c...29...9a...16c...4c...29...4c...20a1...7...


No argument presented. RQAA.


The Parrot Killer
12-03-2020 02:54
tmiddlesProfile picture★★★★★
(3266)
Into the Night wrote:
[quote]tmiddles wrote:...(addition of area term * inverse square law term)...
It's not clear AT ALL what your commentary is but the math is now perfect.

There is nothing at all that is being summed or added, there is no "addition":

σT^4*4πR^2*r^2/2D^2=σt^4*4πr^2

The formula to find the energy output with Stefan-Boltzman requires the area of the emitter. In both the case of the Sun and the object that's a sphere. So of course you use 4πR^2 in:
σT^4*4πR^2
that is simply:
σT^4*A

Every one of the Math errors you thought you found (with the exception of the extra π I'd already caught) was actually your error. I see no rebuttal or admission of that.
Page 19 of 26<<<1718192021>>>





Join the debate Venus is hotter than Mercury?!?:

Remember me

Related content
ThreadsRepliesLast post
Why can't you say Venus is hotter than Mercury because Venus got CO2?12919-12-2019 17:10
I don't believe CO2 makes air hotter because I don't see any experimental proof509-10-2019 03:15
The only straw the Church of AGW can grasp is Venus8826-09-2019 05:49
The only straw the Church of AGW can grasp is Venus and Mercury418-09-2019 22:37
There is no evidence water vapor makes things hotter018-09-2019 21:34
▲ Top of page
Public Poll
Who is leading the renewable energy race?

US

EU

China

Japan

India

Brazil

Other

Don't know


Thanks for supporting Climate-Debate.com.
Copyright © 2009-2020 Climate-Debate.com | About | Contact