I came across a relatively simple formula for estimating
the emitting surface temperatures of planets based on the
Stefan–Boltzmann law which states that total thermal
radiation fromis proportional to the fourth power

I accidently posted this before it was finished.

I came across a relatively simple formula for estimating
the emitting surface temperatures of planets based on the
Stefan–Boltzmann law, which states that total thermal
radiation is proportional to the fourth power of the absolute
temperature.

R1 is the radius of a star.

R2 is the radius of a planet's orbit.

A is the planet's albedo or reflectance.

T1 is the absolute temperature of the star's surface.

T2 is the average absolute temperature of the planet's
emitting surface, where thermal radiation from the planet
can escape into space. For a planet with an atmosphere
this is often referred to as the "optical surface".

T2= (the fourth root of ((((R1/R2) squared) (1-A)) x.25)) x (T1)

In the case of Earth:

R1, the Sun's radius is 696,000 km.

R2, the Earth's orbital radius is 149,600,000 km.

A, the Earth's albedo is .296

T1, the absolute temperature of the Sun's surface is 5,778 K

Putting these values into the formula, R1/R2 = .0047

Squaring this gives .0000216, the ratio of the radiation intensity
at the Sun's surface to the radiation intensity at the Earth's orbit.

Dividing this by 4 corrects for the fact that the Earth absorbs
solar radiation over its circular cross section but emits infrared
radiation over its entire surface area,

..0000216/4 = .00000541

Multiplying this by (1 - A) corrects for the fraction of solar
radiation that is reflected back into space as light and does
not contribute to warming the planet:

..00000541 x (1 - .296) = .00000381

This is the ratio of the intensity of the infrared radiation
emitted from the Earth's surface to the radiation intensity
at the Sun's surface.

Taking the fourth root of this ratio gives the ratio of the
absolute temperature of the Earth's emitting surface,
or optical surface, to the temperature of the Sun's surface:

The fourth root of .00000381 = .0442

..0442 x 5,778 K = 255 K, which is about 18 degrees below zero
Celsius, or zero degrees Fahrenheit. 255 K is in fact a commonly
accepted value for the emission temperature of the Earth. As
they say, that's close enough for government work. Of course,
temperatures further down in the atmosphere average higher
than this due to the atmospheric greenhouse effect.

This presumes a near-circular orbit; a planet with a notably
eccentric orbit would be subject to thermal lag. Hal Clement
had to take this into account in creating the planet Mesklin
for his novel _Mission of Gravity_.

On 10/12/22 18:49, peterwezeman@hotmail.com wrote:
> I accidently posted this before it was finished.
>

I preferred the previous version.

Zero Fahrenheit. Happy Holidays.

On Saturday, December 10, 2022 at 12:49:22 AM UTC-5, peterwezeman@hotmail.com wrote:
> I accidently posted this before it was finished.
> I came across a relatively simple formula for estimating
> the emitting surface temperatures of planets based on the
> Stefan–Boltzmann law, which states that total thermal
> radiation is proportional to the fourth power of the absolute
> temperature.
>
> R1 is the radius of a star.
>
> R2 is the radius of a planet's orbit.
>
> A is the planet's albedo or reflectance.
>
> T1 is the absolute temperature of the star's surface.
>
> T2 is the average absolute temperature of the planet's
> emitting surface, where thermal radiation from the planet
> can escape into space. For a planet with an atmosphere
> this is often referred to as the "optical surface".
>
> T2= (the fourth root of ((((R1/R2) squared) (1-A)) x.25)) x (T1)
>
> In the case of Earth:
>
> R1, the Sun's radius is 696,000 km.
>
> R2, the Earth's orbital radius is 149,600,000 km.
>
> A, the Earth's albedo is .296
>
> T1, the absolute temperature of the Sun's surface is 5,778 K
>
> Putting these values into the formula, R1/R2 = .0047
>
> Squaring this gives .0000216, the ratio of the radiation intensity
> at the Sun's surface to the radiation intensity at the Earth's orbit.
>
> Dividing this by 4 corrects for the fact that the Earth absorbs
> solar radiation over its circular cross section but emits infrared
> radiation over its entire surface area,
>
> .0000216/4 = .00000541
>
> Multiplying this by (1 - A) corrects for the fraction of solar
> radiation that is reflected back into space as light and does
> not contribute to warming the planet:
>
> .00000541 x (1 - .296) = .00000381
>
> This is the ratio of the intensity of the infrared radiation
> emitted from the Earth's surface to the radiation intensity
> at the Sun's surface.
>
> Taking the fourth root of this ratio gives the ratio of the
> absolute temperature of the Earth's emitting surface,
> or optical surface, to the temperature of the Sun's surface:
>
> The fourth root of .00000381 = .0442
>
> .0442 x 5,778 K = 255 K, which is about 18 degrees below zero
> Celsius, or zero degrees Fahrenheit.

This is known as a zero-dimensional energy balance model. The next step in the EBM hierarchy is a one-d model, which assumes zonal and temporal
symmetry, and has a north-south dimension. You need an approximation for the variation of solar input and albedo with latitude, and a transport
mechanism (usually diffusive) but now you have a model with more than one equilibrium solution, and you can begin to think about stability issues.

255 K is in fact a commonly
> accepted value for the emission temperature of the Earth.

As you have shown it must be by conservation of energy - unless the earth is rapidly warming or cooling.

The earth's longwave output, seen from orbit, is in fact near 255K for a large part of the spectrum. In areas where
the atmosphere is more transparent to radiation, e.g. the "atmospheric window", the emission follows a
warmer Planck curve, and colder where there is more absorption.

As
> they say, that's close enough for government work. Of course,
> temperatures further down in the atmosphere average higher
> than this due to the atmospheric greenhouse effect.

Many of the massive "Earthlike" worlds seem likely to have surface temperatures far higher
than their emissive temperatures. More Venuslike than Earthlike.
>
> This presumes a near-circular orbit; a planet with a notably
> eccentric orbit would be subject to thermal lag. Hal Clement
> had to take this into account in creating the planet Mesklin
> for his novel _Mission of Gravity_.

Also Tony Rothman for "The World is Round". For very long period orbits, as in Helliconia, lag is
not as important.

You can produce a time-dependent version of any of these simple models. EBMs with
diffusive transport do generally overestimate lags on large continents, to get that
right you have to move to a different class of model.

If you can travel back in time to 1995 you can take my course in climate models. Also, then-me
would appreciate any stock advice you have.

William Hyde