Calculating stellar temperature at a distance?

On 13/12/2013 23:43, Yousuf Khan wrote:
On 02/12/2013 4:53 AM, Martin Brown wrote:
I think you need to show your workings...

The suns photosphere is about 5800K.

Wolfram-Alpha showed that the temperature ranges between 5600K to 8900K
(which I rounded upto 9000K):

temperature of the Sun's surface - Wolfram|Alpha
http://www.wolframalpha.com/input/?i...un%27s+surface

Hmmm! What an incredibly badly written and misleading page!

No mention that that average effective temperature of the photosphere is
tightly constrained to ~5770K so that TSI varies systematically by only
around 0.1% with the solar cycle. You can get transient variations when
a large area of faculae or a naked eye sunspot is present.

Using the highest possible temperature in that range, I was hoping that
I could get somewhat higher temperatures at a large distance (e.g.
Earth's orbit) using the now-understood-to-be over-simplistic inverse
square law that I was assuming to be the rule for temperature radiation.

That is your fundamental error. I don't know where you got this idea.

The flux arriving per unit area falls off with the square of distance.

Back of the envelope I get:
Total flux escaping from a surface at radius R is determined by kr^2T^4

E = (7x10^5)^2 x 5800^4 = (1.5x10^8)^2 x t^4

Hence t^4 = 49x10^10 x 5800^4/(2.25x10^16) = 49/2.25 x 10^-6 x 5800^4

t = (22/10^6)^(1/4) x 5800 = 0.0685 x 5800 = 397K

Which for a back of the envelope sum compares favourably with the peak
daytime temperature reached on the lunar surface of about 110C.

Seeing as even this temperature is higher than the boiling point of
water, how do astrophysicists determine what the range of the Goldilocks
Zone of a solar system is? Obviously they have to make some adjustments
for a specific planetary body within that zone altering the final
temperature at its surface.

As soon as you have an atmosphere and liquid water there is a powerful
heat pump aka weather moving heat around the planet and smoothing out
the variations. Venus rotates very slowly but the temperature difference
between the sunlit side and the "dark" side is quite small.

Latent heat of evaporation is a very effective heat shunt.

I assume they all assume that there is an Earth-like planet there which
has an Earth-like atmosphere. But we're seeing many exoplanet
Super-Earths at these ranges, wouldn't that affect its Goldilocks zone
calculations?

I think they use some nominal atmosphere and require liquid water to be
present somewhere on the surface to facilitate chemistry. I suspect that
to support life you actually need all three phases of water - so
effectively clamping any excursions. Having a decent sized moon might
also be important so that there is a variation in tidal range.

We will only ever know when we find another planet that harbours life -
even if it is just green or orange slime.

Allowing for time averages and geometry factors on the Earth its
incident energy is about 1/4 of the full sun normal plane incidence

t(Earth) ~ t/sqrt(2) = 280K

A bit on the high side but then I have been very approximate here.

Is that simply due to the spherical geometry of the Earth, or is it due
to atmospheric insulation?

Spherical geometry viz:

Surface area of sphere = 4.pi.R^2
Surface area of circle = pi.R^2

The entire surface of the Earth is available to radiate the heat away
but the area of the disc determines the sunlit capture cross section.

It is also interesting to consider the equilibrium surface temperature
of the unfortunate comet ISON as it faded out at 3 solar radii.

E = 1^2 x 5800^4 = 3^2 x T^4

Hence T = 5800 / sqrt(3) = 3350K

Where basically only tantalum and tungsten have not melted.

I checked out the following table, and if there were diamonds or
graphite on this comet, then it won't have melted either.

The chemical elements of the periodic table sorted by melting point
http://www.lenntech.com/periodic-cha...ting-point.htm

It's somewhat weird and funny that the highest melting point elements
are typically much higher on the periodic table than iron, but then
right there amongst them there is Carbon (atomic number 6) and Boron
(atomic number 5). Something magical about those two light elements.


Or putting it another way the melting point of the toughest rocks is
about 2000K so once the comet is closer than about 70 solar radii its
surface can potentially melt to a glass if the subsurface is unable to
keep up a supply of steam and other volatiles to cool it.

NB equilibrium radiation temperature at distance r scales as 1/sqrt(r)

Is it different because it's much closer to the Sun at that point?

Yousuf Khan

At some point atmospheric extinction in the suns atmosphere and the
local intensity of the solar wind and worse still for the comet the
Roche limit trying to tear it apart will see the thing off.

I am actually quite curious how ISON expired. It seems to me that it
must have quite literally ran out of steam before it got ripped apart by
tidal forces from the sun. Otherwise you would have expected the
disintegration to have been accompanied by rapid brighting as fresh
interior material form the core would be exposed by a Roche limit
disintegration due to gravitational effects.

It is still a puzzle to me why it did fizzle out like that. Did it
disintegrate or melt on the surface and end up glassified?

--
Regards,
Martin Brown

On Tuesday, December 10, 2013 9:14:04 PM UTC-8, Yousuf Khan wrote:
On 09/12/2013 5:39 PM, Steve Willner wrote:

Martin's post is extremely helpful. In general, one can calculate an

answer for any specified thermal properties, but rotation also has to

be taken into account. (A fast rotator is essentially the same as a

good conductor.) The Moon is an example of a slowly rotating

insulator; the Earth, because of its rotation, is closer to a

conductor though not a perfect one. (It's typically colder at night

than in the daytime, but the difference on Earth is nowhere near as

large as on the Moon.)



Yet, Venus is an extremely slow rotator, and it's very warm on both its

day and night sides, although I don't know what the temperature drop off

is between the day and night on Venus.


Yousuf Khan

With the geothermal upwelling and heating its atmosphere from each and every m2 of its surface, plus thousands of geothermal vents adding energy and substance, there's only a 25~50 K difference unless you go for either polar zone where the near surface atmospheric temperature can drop off by as much as another 100 k.

On 16/12/2013 2:46 AM, Martin Brown wrote:
On 13/12/2013 23:43, Yousuf Khan wrote:
On 02/12/2013 4:53 AM, Martin Brown wrote:
I think you need to show your workings...

The suns photosphere is about 5800K.

Wolfram-Alpha showed that the temperature ranges between 5600K to 8900K
(which I rounded upto 9000K):

temperature of the Sun's surface - Wolfram|Alpha
http://www.wolframalpha.com/input/?i...un%27s+surface

Hmmm! What an incredibly badly written and misleading page!

It's not really a written page, it's a generated page from data within
Wolfram's database.

No mention that that average effective temperature of the photosphere is
tightly constrained to ~5770K so that TSI varies systematically by only
around 0.1% with the solar cycle. You can get transient variations when
a large area of faculae or a naked eye sunspot is present.

So the 8900K upper temperature might just be the temperature around or
inside sunspots? I thought sunspots are supposed to be cooler areas of
the Sun's surface?

Using the highest possible temperature in that range, I was hoping that
I could get somewhat higher temperatures at a large distance (e.g.
Earth's orbit) using the now-understood-to-be over-simplistic inverse
square law that I was assuming to be the rule for temperature radiation.

That is your fundamental error. I don't know where you got this idea.

Just a hypothesis since I didn't know any better. Didn't know what the
real formula would be.

Seeing as even this temperature is higher than the boiling point of
water, how do astrophysicists determine what the range of the Goldilocks
Zone of a solar system is? Obviously they have to make some adjustments
for a specific planetary body within that zone altering the final
temperature at its surface.

As soon as you have an atmosphere and liquid water there is a powerful
heat pump aka weather moving heat around the planet and smoothing out
the variations. Venus rotates very slowly but the temperature difference
between the sunlit side and the "dark" side is quite small.

Latent heat of evaporation is a very effective heat shunt.

The triple point of water occurs at different temperatures depending on
atmospheric pressure. They must've made an assumption about pressure and
temperature at the surface of an Earth-like or Super-Earth planet to get
their calculations done.


Yousuf Khan

On 17/12/2013 04:45, Yousuf Khan wrote:
On 16/12/2013 2:46 AM, Martin Brown wrote:
On 13/12/2013 23:43, Yousuf Khan wrote:
On 02/12/2013 4:53 AM, Martin Brown wrote:
I think you need to show your workings...

The suns photosphere is about 5800K.

Wolfram-Alpha showed that the temperature ranges between 5600K to 8900K
(which I rounded upto 9000K):

temperature of the Sun's surface - Wolfram|Alpha
http://www.wolframalpha.com/input/?i...un%27s+surface

Hmmm! What an incredibly badly written and misleading page!

It's not really a written page, it's a generated page from data within
Wolfram's database.

It is worse than useless since although what it says is not actually
wrong it is extremely misleading unless you already know the answers.

I am not sure where they got 5600K or 8900K from - the true range is
more like 3000K (sunspot) to 7000K (faculae) with the odd outlier.

I couldn't find an authoritative number for faculae online.

No mention that that average effective temperature of the photosphere is
tightly constrained to ~5770K so that TSI varies systematically by only
around 0.1% with the solar cycle. You can get transient variations when
a large area of faculae or a naked eye sunspot is present.

So the 8900K upper temperature might just be the temperature around or
inside sunspots? I thought sunspots are supposed to be cooler areas of
the Sun's surface?

Sunspots are cooler than the main photosphere and appear dark (although
they are still a lot hotter than red hot on Earth). Faculae are brighter
areas often associated with sunspots on an active sun and usually
occupying a larger area (hence on average an active sun is very slightly
brighter since the mix of sunspots (cooler) and faculae (hotter) results
in a net increase in output most of the time. See:

http://solarscience.msfc.nasa.gov/feature1.shtml

(not a very good picture of faculae)

Here is a slightly better one during the transit of Venus from ESO
http://vt-2004.astro.cz/fotogalerie/...to-02-rost.jpg

Seeing as even this temperature is higher than the boiling point of
water, how do astrophysicists determine what the range of the Goldilocks
Zone of a solar system is? Obviously they have to make some adjustments
for a specific planetary body within that zone altering the final
temperature at its surface.

As soon as you have an atmosphere and liquid water there is a powerful
heat pump aka weather moving heat around the planet and smoothing out
the variations. Venus rotates very slowly but the temperature difference
between the sunlit side and the "dark" side is quite small.

Latent heat of evaporation is a very effective heat shunt.

The triple point of water occurs at different temperatures depending on
atmospheric pressure. They must've made an assumption about pressure and

No it doesn't. The triple point of water is a well defined unique point
on the phase diagram of water with coordinates.

http://en.wikipedia.org/wiki/Phase_diagram

The freezing point decreases slightly with increasing pressure and the
boiling point increases considerably. This does put bounds on what
atmospheric pressure can be as does the requirement to avoid spontaneous
combustion of organic materials on its composition.

temperature at the surface of an Earth-like or Super-Earth planet to get
their calculations done.

Anything between about 10mBar and 100Bar will do in terms of having all
three phases present with ice at below about 0C and the boiling point
varying from 10C to 300C. My guess is that for life at least 500mBar and
probably less than 10Bar (and partial pressure 0.1 O2 1Bar).

I don't know what assumptions if any they use in the exoplanet searches.
I suspect they would settle for finding any planetary atmosphere that
was clearly not in chemical equilibrium.

--
Regards,
Martin Brown

In sci.astro message , Mon, 16 Dec 2013
07:46:23, Martin Brown posted:


At some point atmospheric extinction in the suns atmosphere and the
local intensity of the solar wind and worse still for the comet the
Roche limit trying to tear it apart will see the thing off.

I am actually quite curious how ISON expired. It seems to me that it
must have quite literally ran out of steam before it got ripped apart
by tidal forces from the sun. Otherwise you would have expected the
disintegration to have been accompanied by rapid brighting as fresh
interior material form the core would be exposed by a Roche limit
disintegration due to gravitational effects.

It is still a puzzle to me why it did fizzle out like that. Did it
disintegrate or melt on the surface and end up glassified?


The "standard" Roche limit applies to a fluid body with zero internal
strength; and approximately to a rubble-pile of non-adhering particles.

ISTM that even if a comet aggregated gently from icy & rocky bits at
very low temperatures, a degree of mutual welding would occur during the
many many years before it got persuaded to sun-dive.

Roche will not disintegrate a sufficiently small body with non-zero
cohesion.

I have looked at Édouard Roche's original paper, "La figure d'une masse
fluide soumise á l'attraction d'un point éloigné", /Acad. des Sci. de
Montpellier/, vol.1 (1847-50) p.243, as quoted with a later full
calculation in /Phil. Trans. Roy. Soc./, 206A (1906) 161-248. That was
reprinted in GH Darwin, Papers III (CUP, 1910) p.436 /ff/, which I saw
long ago.

Does anyone have a URL for that paper? ... come to think of it, I do (in
the manner of Caesar's Gaul), in
http://www.merlyn.demon.co.uk/gravity8.htm#Roche.

--
(c) John Stockton, nr London, UK. Mail via homepage. Turnpike v6.05 MIME.
Web http://www.merlyn.demon.co.uk/ - FAQqish topics, acronyms and links;
Astro stuff via astron-1.htm, gravity0.htm ; quotings.htm, pascal.htm, etc.
No Encoding. Quotes before replies. Snip well. Write clearly. Mail no News.

On 17/12/2013 19:16, Dr J R Stockton wrote:
In sci.astro message , Mon, 16 Dec 2013
07:46:23, Martin Brown posted:


At some point atmospheric extinction in the suns atmosphere and the
local intensity of the solar wind and worse still for the comet the
Roche limit trying to tear it apart will see the thing off.

I am actually quite curious how ISON expired. It seems to me that it
must have quite literally ran out of steam before it got ripped apart
by tidal forces from the sun. Otherwise you would have expected the
disintegration to have been accompanied by rapid brighting as fresh
interior material form the core would be exposed by a Roche limit
disintegration due to gravitational effects.

It is still a puzzle to me why it did fizzle out like that. Did it
disintegrate or melt on the surface and end up glassified?


The "standard" Roche limit applies to a fluid body with zero internal
strength; and approximately to a rubble-pile of non-adhering particles.

ISTM that even if a comet aggregated gently from icy & rocky bits at
very low temperatures, a degree of mutual welding would occur during the
many many years before it got persuaded to sun-dive.

Although the dirty snowball model holds that a comet isn't really very
strongly bound. The argument (elsewhere) is that only dust exited
perihelion whereas I am inclined to think that the surface melted during
closest approach and it exited with a glassified surface.

It did perk up a bit as it got a few solar diameters away but nothing
like the brightness on the way in. My point was that had it
disintegrated as some have claimed then the interior would have been
suddenly exposed and significant brightening occurred (and it didn't).

Roche will not disintegrate a sufficiently small body with non-zero
cohesion.

I agree. That was the point I was making albeit from a slightly
different starting point. However, others think it did disintegrate.

I have looked at Édouard Roche's original paper, "La figure d'une masse
fluide soumise á l'attraction d'un point éloigné", /Acad. des Sci. de
Montpellier/, vol.1 (1847-50) p.243, as quoted with a later full
calculation in /Phil. Trans. Roy. Soc./, 206A (1906) 161-248. That was
reprinted in GH Darwin, Papers III (CUP, 1910) p.436 /ff/, which I saw
long ago.

Does anyone have a URL for that paper? ... come to think of it, I do (in
the manner of Caesar's Gaul), in
http://www.merlyn.demon.co.uk/gravity8.htm#Roche.


Agreed although you can put in the strength of the appropriate material
(if known) and the mathematics is the same. And it was awfully close to
the sun even if it was a bit more strongly bound.

--
Regards,
Martin Brown

In article ,
Yousuf Khan writes:
Wolfram-Alpha showed that the temperature ranges between 5600K to 8900K
http://www.wolframalpha.com/input/?i...un%27s+surface

It says that but hardly "shows" it. As Martin wrote, the solar
effective temperature is very close to 5800 K (5777 K according to
AAQ 4th edition). There are many solar temperatures one could refer
to, but "effective temperature" is the one we want for this
calculation.

... using the now-understood-to-be over-simplistic inverse
square law that I was assuming to be the rule for temperature radiation.

Inverse square is correct for "solar flux," i.e., the total outbound
power per unit area. What temperature follows from that requires
equating energy absorbed by a body with energy emitted. If we do
that for a body just above the solar surface (imagining there is such
a thing), assuming a small, perfectly absorbing, thermally conducting
body with no input or output from conduction, the temperature would
be (5800 K)/(4^1/4) or about 4100 K. That's because energy is
absorbed by an area of pi*r^2 but radiated by an area of 4*pi*r^2.

One AU is about 214 solar radii, so the temperature at 1 AU is about
280 K or 7 deg_C as Martin derived. The Earth's mean temperature is
a bit higher than this (15 deg_C??) because of the greenhouse effect.

how do astrophysicists determine what the range of the Goldilocks
Zone of a solar system is? Obviously they have to make some adjustments
for a specific planetary body within that zone altering the final
temperature at its surface.

I don't know of any set of standard assumptions, but there have to be
some assumptions about how the planet emits and absorbs radiation and
how temperatures are distributed around the planet. Real planets may
depart from assumptions, but we have to start somewhere while
recognizing the approximate nature of the calculations.

--
Help keep our newsgroup healthy; please don't feed the trolls.
Steve Willner Phone 617-495-7123
Cambridge, MA 02138 USA

In sci.astro message , Wed, 18 Dec 2013
08:16:11, Martin Brown posted:

I have looked at Édouard Roche's original paper, "La figure d'une masse
fluide soumise á l'attraction d'un point éloigné", /Acad. des Sci. de
Montpellier/, vol.1 (1847-50) p.243, as quoted with a later full
calculation in /Phil. Trans. Roy. Soc./, 206A (1906) 161-248. That was
reprinted in GH Darwin, Papers III (CUP, 1910) p.436 /ff/, which I saw
long ago.

Does anyone have a URL for that paper? ... come to think of it, I do (in
the manner of Caesar's Gaul), in
http://www.merlyn.demon.co.uk/gravity8.htm#Roche.


Agreed although you can put in the strength of the appropriate material
(if known) and the mathematics is the same. And it was awfully close to
the sun even if it was a bit more strongly bound.


From what I recall of the paper, already sufficiently mind-boggling,
putting in the material strength would be a ***major*** effort,
certainly beyond me. On the other hand, it should not be too difficult
to consider a solid homogeneous non-rotating cubical satellite with a
gravitational field gradient everywhere constant and parallel to an edge
of the cube; and that should give a reasonable approximation for a small
cube.

--
(c) John Stockton, nr London, UK. Mail via homepage. Turnpike v6.05 MIME.
Web http://www.merlyn.demon.co.uk/ - FAQqish topics, acronyms and links;
Astro stuff via astron-1.htm, gravity0.htm ; quotings.htm, pascal.htm, etc.
No Encoding. Quotes before replies. Snip well. Write clearly. Mail no News.

On 17/12/2013 4:05 AM, Martin Brown wrote:
On 17/12/2013 04:45, Yousuf Khan wrote:
It's not really a written page, it's a generated page from data within
Wolfram's database.

It is worse than useless since although what it says is not actually
wrong it is extremely misleading unless you already know the answers.

I am not sure where they got 5600K or 8900K from - the true range is
more like 3000K (sunspot) to 7000K (faculae) with the odd outlier.

Well, I'm pretty sure there is probably a bit of debate right now about
what the exact temperatures of various features on the Sun are. Perhaps
they think faculae can get upto 8900K rather than 7000K like you say?

For example, on Wolfram again, the only thing I could find for searching
on "sunspot" yields the number of sunspots visible on its surface right
now (apparently 97 today), but no temperature info:

sunspot - Wolfram|Alpha
http://www.wolframalpha.com/input/?i=sunspot

And as for searching for "facula", it yields just a definition of what
it means:

facula - Wolfram|Alpha
http://www.wolframalpha.com/input/?i=facula

So Wolfram has a bit of explaining to do.

I couldn't find an authoritative number for faculae online.

Which probably explains why Wolfram has such a different value for it
than you do.

The triple point of water occurs at different temperatures depending on
atmospheric pressure. They must've made an assumption about pressure and

No it doesn't. The triple point of water is a well defined unique point
on the phase diagram of water with coordinates.

http://en.wikipedia.org/wiki/Phase_diagram

Yeah, what I meant to say was that the range of any particular planet's
temperatures & pressures affects how close it is to the vicinity of the
triple point. Assumption being that life needs to be somewhere near the
triple point to develop.

The freezing point decreases slightly with increasing pressure and the
boiling point increases considerably. This does put bounds on what
atmospheric pressure can be as does the requirement to avoid spontaneous
combustion of organic materials on its composition.

temperature at the surface of an Earth-like or Super-Earth planet to get
their calculations done.

Anything between about 10mBar and 100Bar will do in terms of having all
three phases present with ice at below about 0C and the boiling point
varying from 10C to 300C. My guess is that for life at least 500mBar and
probably less than 10Bar (and partial pressure 0.1 O2 1Bar).

If a planet has a boiling point of water around 300C, then would that
imply that life on such a planet would be able to withstand a much
higher temperature than the ones here? And consequently would a planet
with a boiling point near 10C have unbelievably fragile life?

Let's say for the sake of argument that most standard lifeforms can
withstand temperatures upto 50% of the boiling point before they die. So
on a 300C boiling-point planet, the life can withstand 150C before
dying; on a 100C planet, that'll be 50C; and on a 10C planet, that'll be
just 5C! Or is life going to have the same temperature limits
everywhere, regardless of where the boiling point of water is?

I don't know what assumptions if any they use in the exoplanet searches.
I suspect they would settle for finding any planetary atmosphere that
was clearly not in chemical equilibrium.

What do you mean bt "not in chemical equilibrium"? How is the Earth not
in chemical equilibrium?

Yousuf Khan

In article ,
Yousuf Khan writes:
Well, I'm pretty sure there is probably a bit of debate right now about
what the exact temperatures of various features on the Sun are.

It's probably not meaningful to assign single temperatures to
"features." The "effective temperature" of the Sun, which is
relevant to this topic, is known to about three significant digits
from satellite "active cavity radiometer" observations.

If a planet has a boiling point of water around 300C,

thus very high atmospheric pressure

then would that
imply that life on such a planet would be able to withstand a much
higher temperature than the ones here?

I doubt anyone knows. If the actual temperature is high, then
obviously any life has to be adapted for that. However, life on a
cold planet with high atmospheric pressure wouldn't necessarily be
resistant to high temperature as far as I can see.

And consequently would a planet
with a boiling point near 10C have unbelievably fragile life?

Again probably unknown. Intracellular pressure might be higher than
atmospheric pressure, for example, so there could be stable liquid
inside cells even though the same liquid would boil at the low
surface pressure. Also liquids with dissolved solids boil at higher
temperatures than pure liquids.

What do you mean bt "not in chemical equilibrium"?

No spontaneous chemical changes occurring. Another way of saying it
is that forward and reverse reactions have the same rates.

How is the Earth not in chemical equilibrium?

Free oxygen and carbon-bearing material (plants, animals, exposed
coal, atmospheric methane) both present.

--
Help keep our newsgroup healthy; please don't feed the trolls.
Steve Willner Phone 617-495-7123
Cambridge, MA 02138 USA