Earth - What is the real age?

[swansont]

I won't be insisting on anything really for as I have already said I prefer the Moon Capture Theory. But as you have noted there is definitely an inconsistency in the Giant Impact theory for they need the impact to occur "after a few tens of millions of years" but the Iron Catastrophe (IC) appears to occur much later, yet the heating that occurred during the Giant Impact would have immediately set the IC in motion at the very least.

There definitely appears to be something incompatible between the two ideas.

 

Why would heating "set the IC in motion"?

[Robittybob1]

 

Why would heating "set the IC in motion"?

There are definitely varying timelines for the Iron Catastrophe, this one mentions 50 million years.

http://www.pbs.org/wgbh/nova/education/programs/3111_origins.html

 

describes the theory of the Iron Catastrophe, thought to have occurred almost 50 million years after Earth's formation, when internal heat from trapped radioactive elements and external heat from surface collisions caused the planet's iron to melt, sink, and form Earth's core.

 

 

This alternative model has a better timeline but it still requires heating to set it in motion.

http://en.wikipedia.org/wiki/Rain-out_model

 

The rain-out model is a model of planetary science which describes the first stage of planetary differentiation and core formation. According to this model, a planetary body is assumed to be composed primarily of silicate minerals and NiFe (i.e. a mixture of nickel and iron). If temperatures within this body reach about 1500 K, the minerals and the metals will melt. This will produce an emulsion in which globules of liquid NiFe are dispersed in a magma of liquid silicates, the two being immiscible. Because the NiFe globules are denser than the silicates, they will sink under the influence of gravity to the centre of the planetary body–in effect, the globules of metal will "rain out" from the emulsion to the centre, forming a core.

According to the rain-out model, core formation was a relatively rapid process, taking a few dozen millennia to reach completion. This occurred at the end of a lengthy process in which the planets were assembled from colliding planetary embryos. Only the collisions of such large embryos could generate enough heat to melt entire bodies. Furthermore, it was only after all of the iron and nickel delivered by impacting bodies had arrived that core formation could proceed to completion.

However, this process of core formation was preceded by a long period of partial differentiation, in which some of the nickel and iron within the planetary embryos had begun to separate.

This model allows for a preformed core and further melting at the time of the GI.

Quite an extensive timeline is detailed in this webpage:

http://carbonomics.net/MCcarbon/Carbonomics/13c10/13c10c.html

 

 

1.2: The Earth Becomes Molten.

‘The Day the Earth was Born’ proposes that 25 million years after its formation, the Earth became a molten ball of rock with temperatures around 1200C, "Four and a half billion years ago the Earth was an undifferentiated sphere of rock. Volcanoes spewed vast amounts of lava onto the hot, seething surface. Our hostile planet was about to endure the first of several momentous events that would change it forever. Radioactive elements trapped when the planet was formed were heating it from inside. At the same time, Earth’s gravity was pulling in huge quantities of debris from space, a bombardment that generated excessive amounts of heat on the surface. The combined effect was catastrophic. At eight minutes past midnight on the Earth clock, (about 25 million years after the formation of the Earth i.e. 4,525,000,000 years ago) the entire planet was like a furnace."

1.3: The Iron Catastrophe: the Migration of the Elements.

In this molten state, iron melted and sank towards the centre of the Earth, "We think the earth, at some point, was totally molten, a big droplet of melt just floating in space. When you have a totally molten droplet like this, the heaviest elements, and that includes things like iron, would sink to the centre of this droplet and the lightest elements, things rich in carbon and water, for instance, the light elements, would float to the top and float there like algae on a lake." (Michael Zolensky, Nasa).

Edited April 2, 2015 by Robittybob1

[Acme]

...

This model allows for a preformed core and further melting at the time of the GI.

Quite an extensive timeline is detailed in this webpage:

http://carbonomics.net/MCcarbon/Carbonomics/13c10/13c10c.html

Let me quote from that last link:

The mc is an independent, non-profit making, research group which invented Carbonomics, the new eco-nomics of the Earth, a means of measuring humans' impact on the environment and what humans must do to stabilize the Earth's climate.

...

The Elephants sitting in the Corner of the Room

This website refuses to ignore the Elephants sitting in the corner of the room i.e. the overwhelming evidence of zionist world domination; zionists' initiation of the so-called war on terrorism which the zionists refer to as either world war III or world war IV; the benefits reaped from world war III/IV by the zionist state in palestine; zionists' dominance of pre bush administrations; zionists' dominance of the first bush administration; zionists' dominance of the second bush administration; the American Zionist Lobby: zionists' indoctrination of christians i.e. de-christianized Zionists (the so-called Christian Zionists): zionists' dominance of the american media; ...

Yeah, that's a scientifically authoritive source. Nice work.

 

Moreover, none of the material in the last pages is on the [misbegotten] topic of the Earth's age being grossly underestimated, a speculation that was shown to be in error long ago.

[Robittybob1]

...

Moreover, none of the material in the last pages is on the [misbegotten] topic of the Earth's age being grossly underestimated, a speculation that was shown to be in error long ago.

 

I have largely been trying to answer Swansont's persistent questions, and they have lead the thread along this path. We are trying to see the events that heated the Earth in the beginning so to me it is very relevant in the overall answer to the speculation.

[Acme]

I have largely been trying to answer Swansont's persistent questions, and they have lead the thread along this path. We are trying to see the events that heated the Earth in the beginning so to me it is very relevant in the overall answer to the speculation.

Mind you I am not attacking you personally, just attacking your idea that you have contributed anything with your 'persistence'. Since the Earth would have to have already formed to experience the iron catastrophe or a giant impact, or any post-formation events, then the particulars of those events have no bearing on when the Earth formed.

 

It has been made clear by actual scientific research that the Earth's formation, as for all the planets, planetoids, moons, etcetera, was/were coincident with the formation of the entire solar system. Further 'persistence' is unwarranted and unproductive.

[Robittybob1]

Mind you I am not attacking you personally, just attacking your idea that you have contributed anything with your 'persistence'. Since the Earth would have to have already formed to experience the iron catastrophe or a giant impact, or any post-formation events, then the particulars of those events have no bearing on when the Earth formed.

 

It has been made clear by actual scientific research that the Earth's formation, as for all the planets, planetoids, moons, etcetera, was/were coincident with the formation of the entire solar system. Further 'persistence' is unwarranted and unproductive.

It might be unproductive for you but I have learnt a lot. Give your insights please, but I get annoyed with your attitude. I prefer if you just contribute to the science.

Let me quote from that last link:

Yeah, that's a scientifically authoritive source. Nice work.

 

Moreover, none of the material in the last pages is on the [misbegotten] topic of the Earth's age being grossly underestimated, a speculation that was shown to be in error long ago.

I am not arguing that the Earth's age is underestimated.

[Greg H.]

 

This website refuses to ignore the Elephants sitting in the corner of the room i.e. the overwhelming evidence of zionist world domination; zionists' initiation of the so-called war on terrorism which the zionists refer to as either world war III or world war IV; the benefits reaped from world war III/IV by the zionist state in palestine; zionists' dominance of pre bush administrations; zionists' dominance of the first bush administration; zionists' dominance of the second bush administration; the American Zionist Lobby: zionists' indoctrination of christians i.e. de-christianized Zionists (the so-called Christian Zionists): zionists' dominance of the american media

Off topic, but, these people are bat shit crazy.

[David Levy]

Please see the following article about Kelvin Cooling Time for the Earth:

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime2.html#c1

It is stated that: "The Kelvin cooling time for the Earth is then about 30,000 years"

The formula is very clear. All the signs are also clear.

Full explanation is given by

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime.html#c1

However, the temperature calculation it's not fully clear to me:

I assume that T (final) = 300K, But what is the value of T (Hot)? Is it 1000K?

Why we can't see it in the calculation?

It should be a function of ( 1/T (final) >3 – 1/T(hot)>3)

Edited April 2, 2015 by David Levy

[John Cuthber]

Did you read the first page you cited?

"It is reasonable to assume that the term involving the early hot temperature of the Earth can be neglected. If the hot temperature is more than 3.16 times the final temperature, then the contribution of the high temperature term is less than 1%. For example,"

[David Levy]

O.K.

 

So what is the level of T(hot)?

I would like to understand the calculation

In the formula, it is expected to see the following function:

(1/T (final) >3 – 1/T(hot)>3)

But it isn't there. Why?

[Robittybob1]

Did you read the first page you cited?

"It is reasonable to assume that the term involving the early hot temperature of the Earth can be neglected. If the hot temperature is more than 3.16 times the final temperature, then the contribution of the high temperature term is less than 1%. For example,"

So what that means is that Kelvin calculated that the very hot temperatures and hence high radiation early on had little effect on the time it took for a metal sphere to cool. So any starting temperature above 3.16 times the final temp measured in Kelvin had less than a 1% effect.

Initially that is hard to believe or accept but it is clear the time taken to cool once the surface nears the final temperature is so much longer than the initial rapid cooling, I came to understand this.

It is only considering the effect on the surface temperatures.

Edited April 2, 2015 by Robittybob1

[swansont]

I would like to point out that this is the same concept we discussed on pages 1 and 2. The radiation is T⁴, so you very rapidly radiate away energy when it's very hot — consequently it takes very little time. The bulk of the time is spent cooling after this initial drop, which means you can pick any T_hot as long as it (in this case) is bigger than ~1000 K and not appreciably affect the answer.

 

Check for yourself and see.

 

1/300³ = 3.7 x 10^-8

1/1000³ = 1 x 10^-9

1/2000³ = 1.25 x 10^-10

 

Above ~1000K you are essentially subtracting zero, to within a percent or two.

[imatfaal]

You can even got to the bottom of the radiative cooling time page and plug in final temp 300K, number of particles 1.1e55 etc and then plug in different hot temperatures and see the difference in the time (ie a few years in over 10,000)

[mp][/mp]

 

And can we have a shout across to the thread that claims that maths is not essential - until you get to grips with the T^4 power in Boltzman and the integration of the 1/t^4 that gives us inverse cubes then the answers are completely insane.

 

Over a period of 12,000 years something which is 100,000K will cool to about 2.5K higher than something that started at 1000K.

 

And is that funky stuff with splitting the derivative dT/dt and dividing though by dt really legit? Or do we accept it cos it works?

And David

 

The way to show an exponent is with the caret ^ It is above the six on my keyboard. Alternatives include using natural language (ie cubed, inverse square, raised to the sixth power etc) or using our superscript option (x³ x^-2 x⁶ ). What you have done is use the "greater than symbol" > this is a mathematical inequality symbol which denotes that the left hand side is bigger than the right hand side ie 8 > 4 read as "Eight is Greater than Four"

 

Or you can learn latex so that it looks pretty

[latex]

 

\frac{1}{T_{Final}^3}-\frac{1}{T_{Hot}^3}

[/latex]

[David Levy]

Thanks

Now it is clear, however it seems that there is a real paradox in the final formula for Cooling time - T(cooling).

Let's start with Stefan-Boltzmann law .

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime.html#c1

"Here P is the power emitted from the area, and E is the energy contained by the object. For very hot objects, the role of the ambient temperature can be neglected. If the hot temperature is more than 3.16 times the ambient, then the contribution of ambient terms is less than 1%. For example, for 300K ambient on the earth, an object of temperature higher than 1000K can be treated like a pure radiator into space."

Hence, based on Stefan-Boltzmann law :

P = dE/dt = function of: T(hot)⁴ – T(ambient)⁴

Hence, based on Stefan-Boltzmann law the : T(ambient)⁴ is neglected with regards to T(hot)⁴

If T(hot) = 1000K and T(ambient) = 300K we will neglect the T(ambient).

Therefore, it is assumed (by error of less than 1%) that:

P = dE/dt = function of: T(hot) ⁴

Hence, the power is a direct function of the high temperature T(hot), without any influence of T(ambient).

However, in order to calculate the Cooling Time it is needed to set an integral from T(hot) to T(final).

The result of this integral is that:

Cooling time is function of: 1/T(final)^{3 –} 1/T(hot)³

Hence, 1/T(hot)³ is neglected with regards to 1/T(final)³

Therefore it is stated:

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime2.html#c1

"If the hot temperature is more than 3.16 times the final temperature, then the contribution of the high temperature term is less than 1%. For example, for the 300K ambient on the earth, an original temperature higher than 1000K would make the second temperature in the expression above negligible".

Now the T(hot) had been neglected and this is a real paradox!

There must be an error in this process. We have to find it.

One possibility is that it was forbidden to neglect the T(ambient) from Stefan-Boltzmann law.

I would like to verify what might be the outcome of the Integral without neglecting the T ambient (although its contribution is less than 1%).

I can prove that there is a paradox in the formula of T(cooling) as follow:

Let's assume that there are two identical stars.

The T(hot) of the first one is 20,000K and the T(hot) of the other one is 1000K

Based on the T(cooling) formula, both should have almost the same cooling time.

Is it real???

Edited April 2, 2015 by David Levy

[Strange]

Based on the T(cooling) formula, both should have the same cooling time.

Is it real???

 

Very nearly the same, yes.

 

For the reasons you have clearly explained in your post. Or did you really type all that and not understand it?

[John Cuthber]

there is no paradox.

It's perfectly simple.

 

1/T(final)^{3 –} 1/T(hot)³

is the difference between two parts

1/T(final)³

and

1/T(hot)³

So, if T(hot) is big then T(hot)^3 is very big

so 1/T(hot)³ is very small

So, subtracting it from

1/T(final)³

doesn't really change 1/T(final)³ much.

And, compared to the other approximations made in the calculation, it's just not worth doing the extra maths.

[swansont]

Now the T(hot) had been neglected and this is a real paradox!

 

No, not a paradox. What has happened is that you have taken two separate examples and tried to combine them. What is being demonstrated is that if T>> T_ambient, you can ignore T_ambient and that cutoff is quantified. Separately, it is shown that if T_hot >> T_final, that you can ignore T_hot. Again, quantified (but I would argue it's a factor of 4.64 to get 1%; he just copied 3.16 that applies to the S-B law. No matter in the big picture.)

 

You are correct that if the ambient temperature is not small, then the analysis will be inaccurate. But the ambient temperature in this problem (i.e. the cooling of a big rock in space) is about 3K. That's modified slightly by the presence of the sun, but the sun only subtends 0.00055% of the sky. That won't become important in the cooling time until fairly late in the game.

 

So, to review: as long as T_final >> T_ambient and T_hot >> T_final you can ignore T_hot

 

I can prove that there is a paradox in the formula of T(cooling) as follow:

Let's assume that there are two identical stars.

The T(hot) of the first one is 20,000K and the T(hot) of the other one is 1000K

Based on the T(cooling) formula, both should have almost the same cooling time.

Is it real???

 

This is easy to check. Calculate the time it takes to cool from 20,000K to 1000K using the formula, compared to cooling from 1000K to 300K. While you're at it, (for some insight) compare the ratio of radiated power at those two higher temperatures.

[David Levy]

 

This is easy to check. Calculate the time it takes to cool from 20,000K to 1000K using the formula, compared to cooling from 1000K to 300K. While you're at it, (for some insight) compare the ratio of radiated power at those two higher temperatures.

 

 

Thanks

Based on the Cooling time formula, there is virtually no meaning for the starting temperature. (As long as it's high enough with relative to the final temp).

Hence, the requested cooling time for T(final) at 1000K is:

30 My x (300/1000)³ = 30 x 0.027 = 81,000 Year

I still estimate that there is a paradox in this calculation.

Therefore, I would like to verify what might be the outcome of the Integral without neglecting the T ambient (although its contribution is less than 1%).

Edited April 2, 2015 by David Levy

[Robittybob1]

...

And can we have a shout across to the thread that claims that maths is not essential - until you get to grips with the T^4 power in Boltzman and the integration of the 1/t^4 that gives us inverse cubes then the answers are completely insane.

 

Over a period of 12,000 years something which is 100,000K will cool to about 2.5K higher than something that started at 1000K.

 

....

Doesn't that suggest to you there is a problem with the maths? Well maybe not. He did the research not me.

Edited April 2, 2015 by Robittybob1

[Strange]

Therefore, I would like to verify what might be the outcome of the Integral without neglecting the T ambient (although its contribution is less than 1%).

 

If you don't ignore the second term then, because its contribution is about 1%, the cooling time will be about 1% shorter.

[David Levy]

There is one more explanation for this paradox:

Let's assume that we don't know the temperature on Earth.

So, we already know that for T(final) = 300K the T(cooling) time is 30 My.

Let's try to verify the requested time for T(final) = 200K

As : (300/200)³ = 3.375

Then, the T(cooling) time for T(final) at 200K is:

30 x 3.375 = 101.25 My.

However, we know that after at least 4.4 By the temperature on Earth is still 300K.

Therefore, we have to look again on

Stefan-Boltzmann law .

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime.html#c1

There is no error in this formula!

We just need to understand it and use it correctly.

The answer is already integrated in this formula.

It is called - Emissivity.

The biggest error of the modern science is that they have used an ideal case for the Emissivity and set it to 1.

This is a severe error!

The value shouldn't be one ever and never.

This sign represents the effects of the Atmosphere and Crust heat conductance and other factors which I have started to highlight at thread no. 54

For 4.4 By, this emissivity is in full balance with the temperature of the Earth.

However, it is expected that the emissivity of the earth is changing over time. Therefore, the science must understand what might be the emissivity of the Earth from its first day.

By finding the correct value of the emissivity we can get more realistic cooling time verification.

Edited April 2, 2015 by David Levy

[imatfaal]

Thanks

Now it is clear, however it seems that there is a real paradox in the final formula for Cooling time - T(cooling).

Let's start with Stefan-Boltzmann law .

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime.html#c1

"Here P is the power emitted from the area, and E is the energy contained by the object. For very hot objects, the role of the ambient temperature can be neglected. If the hot temperature is more than 3.16 times the ambient, then the contribution of ambient terms is less than 1%. For example, for 300K ambient on the earth, an object of temperature higher than 1000K can be treated like a pure radiator into space."

Hence, based on Stefan-Boltzmann law :

P = dE/dt = function of: T(hot)⁴ – T(ambient)⁴

Hence, based on Stefan-Boltzmann law the : T(ambient)⁴ is neglected with regards to T(hot)⁴

If T(hot) = 1000K and T(ambient) = 300K we will neglect the T(ambient).

Therefore, it is assumed (by error of less than 1%) that:

P = dE/dt = function of: T(hot) ⁴

Hence, the power is a direct function of the high temperature T(hot), without any influence of T(ambient).

However, in order to calculate the Cooling Time it is needed to set an integral from T(hot) to T(final).

The result of this integral is that:

Cooling time is function of: 1/T(final)^{3 –} 1/T(hot)³

Hence, 1/T(hot)³ is neglected with regards to 1/T(final)³

Therefore it is stated:

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime2.html#c1

"If the hot temperature is more than 3.16 times the final temperature, then the contribution of the high temperature term is less than 1%. For example, for the 300K ambient on the earth, an original temperature higher than 1000K would make the second temperature in the expression above negligible".

Now the T(hot) had been neglected and this is a real paradox!

There must be an error in this process. We have to find it.

One possibility is that it was forbidden to neglect the T(ambient) from Stefan-Boltzmann law.

I would like to verify what might be the outcome of the Integral without neglecting the T ambient (although its contribution is less than 1%).

I can prove that there is a paradox in the formula of T(cooling) as follow:

Let's assume that there are two identical stars.

The T(hot) of the first one is 20,000K and the T(hot) of the other one is 1000K

Based on the T(cooling) formula, both should have almost the same cooling time.

Is it real???

 

David - stop thinking you can rewrite hundred year old physics and approach the maths with a bit of humility.

 

 

eg

 

T_{final}=300

T_{hot}=10000

 

[latex]\frac{1}{T_{Final}^3}-\frac{1}{T_{Hot}^3}=3.70360370*10^{-8}[/latex]

 

[latex]\frac{1}{T_{Final}^3}=3.70370370*10^{-8}[/latex]

 

Difference = 0.0001*10^-8 ie 0.01 percent differnce

 

 

Or for Stefen Boltzman

[latex]T_{hot}^4 - T_{final}^4=9.9999919*10^{15}[/latex]

 

[latex]T_{hot}^4=1*10^{16}[/latex]

 

Difference = 8.1*10^9 - ie a minuscule difference tens of thousandths of one percent; it is raised to the fourth now so the difference is more marked

[John Cuthber]

 

 

Therefore, I would like to verify what might be the outcome of the Integral without neglecting the T ambient (although its contribution is less than 1%).

Why?

Do you realise that redoing the calculation with both terms will not necessarily give you a more accurate answer?

Do you understand that there are other approximations in that calculation that will make more difference?

For example, the assumption is made that the emissivity is exactly 1 (i.e. that it's a perfectly black body).

Well I don't know how well that applies to the early earth, but at the moment the figure is probably more like 0.61 (based on data from the name of a Vangelis album).

If the cooling earth resembled the moon then the emissivity might be nearer 1.

 

So there's an easy way to find errors that are something like 10% , or even 40%.

What do you think you gain from worrying bout that error that we know is less than 1%?

 

Even if you carefully account for all the factors, it' irrelevant. The radiative age of the earth isn't its age by a factor of many thousands.

Edited April 2, 2015 by John Cuthber

[swansont]

 

 

Thanks

Based on the Cooling time formula, there is virtually no meaning for the starting temperature. (As long as it's high enough with relative to the final temp).

Hence, the requested cooling time for T(final) at 1000K is:

30 My x (300/1000)³ = 30 x 0.027 = 81,000 Year

I still estimate that there is a paradox in this calculation.

Therefore, I would like to verify what might be the outcome of the Integral without neglecting the T ambient (although its contribution is less than 1%).

 

So go ahead and verify it. What's stopping you? You just have to include the ambient temperature term in the integral. Doing that for the energy loss calculation is trivial.

Do you understand that there are other approximations in that calculation that will make more difference?

 

And there's another couple of approximations specifically mentioned in the calculation at hyperphysics. One approximation that will make the time estimate shorter, and one that would make it longer. Plus two others that we've mentioned in the thread.

There is one more explanation for this paradox:

Let's assume that we don't know the temperature on Earth.

So, we already know that for T(final) = 300K the T(cooling) time is 30 My.

Let's try to verify the requested time for T(final) = 200K

As : (300/200)³ = 3.375

Then, the T(cooling) time for T(final) at 200K is:

30 x 3.375 = 101.25 My.

However, we know that after at least 4.4 By the temperature on Earth is still 300K.

Therefore, we have to look again on

Stefan-Boltzmann law .

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime.html#c1

There is no error in this formula!

We just need to understand it and use it correctly.

Yes, "we" do.

 

The answer is already integrated in this formula.

It is called - Emissivity.

The biggest error of the modern science is that they have used an ideal case for the Emissivity and set it to 1.

This is a severe error!

Nope. You missed it.

 

Maybe you should tackle a less ambitious set of problems while you learn this. Though to be honest, that an engineer is bumbling around and making such simple errors in a thermodynamics problem is sort of frightening.

[David Levy]

I think that I start to understand the solution for this enigma.

So far we only discuss about "The rate of radiative energy emission from a hot surface is given by the Stefan-Boltzmann law ."

Hence, if we take the Earth outside the solar system in a dark place without any protection (Atmosphere as an example), then yes – after 30 My the earth should decrease its temperature from any starting temp to 300K. This is fully correct.

Never the less, this isn't applicable for real Earth.

We must take in account the effect of the Sun radiation, Atmoshere and several other factors.

So, let's try to understand the impact of a real surface and the effect of sun radiation by the following article:

RADIATION HEAT TRANSFER

http://nptel.ac.in/courses/112108149/pdf/M9/Student_Slides_M9.pdf

"Real Surfaces

Thus far we have spoken of ideal surfaces, i.e. those that emit energy according to the Stefan-Boltzman law:

Eb = σ·Tabs4

Real surfaces have emissive powers, E, which are somewhat less than that obtained theoretically by Boltzman. To account for this reduction, we introduce the emissivity, ε.

so that the emissive power from any real surface is given by:

E = ε·σ·Tabs4

However:

Radiosity

We have developed the concept of intensity, I, which let to the concept of the view factor. We have discussed various methods of finding view factors. There remains one additional concept to introduce before we can consider the solution of radiation problems.

Radiosity, J, is defined as the total energy leaving a surface per unit area and per unit time. This may initially sound much like the definition of emissive power, but the sketch below will help to clarify the concept.

J ≡ ε·Eb + ρ·G

 

Receiving Properties:

Targets receive radiation in one of three ways; they absorption, reflection or transmission. To account for these characteristics, we introduce three additional properties:

Absorptivity, α, the fraction of incident radiation absorbed.

Reflectivity, ρ, the fraction of incident radiation reflected.

Transmissivity, τ, the fraction of incident radiation transmitted.

 

We see, from Conservation of Energy, that:

α + ρ + τ = 1

 

So, let's go back to Radiative Cooling Time:

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime.html#c1

P = dE/dt = ε·σ·A·(T(hot)⁴ –T(ambient)⁴)

Here P is the power emitted from the area, and E is the energy contained by the object.

However, we must add to this formula the contribution of the Sun radiation as well as the effect of real surface.

Therefore:

E(updated total energy) = E(energy contained by the object) + E(energy from the sun)

Hence:

P(updated) = dE(updated total energy)/dt =dE(energy contained by the object)/dt + dE(energy from the sun)/dt

If we do it correctly, we will be able to extract the real cooling time!

Edited April 3, 2015 by David Levy