Calculating stellar temperature at a distance?

"Dr J R Stockton" wrote in message
nvalid...
In sci.astro message , Sun, 1 Dec 2013
09:03:06, Yousuf Khan posted:

How does one calculate the temperature of the space surrounding a
stellar body at a certain distance from it? I was just using an inverse
square relationship between temperature and distance, but that comes up
with non-sense results.

Not entirely like that. Temperature is not the sort of thing which
radiates outward in a manner resembling lines of force.

Let's take the Earth and Sun as an example. If the Sun's surface
temperature is 9000K, and its radius is 700,000 km, and the Earth is 1
AU (1.5E+8 km) away from the Sun. At that distance using an inverse
square relationship, I get 0.2K as the answer. Obviously the Earth is
much warmer than that. What's the real way to obtain temperature here?

You need the T^4 (http://en.wikipedia.org/wiki/Stefan-Boltzmann_law)
law.

The energy density from unit area of the Sun is proportional to a
9000^4. That is diminished by (7e5/1.5e8)^2, and take the fourth root,
getting about 615K. Now correct the solar temperature to 5800, getting
396K. Now recall that the Earth's radiating surface is four times
bigger in area that the area of sunshine intercepted, so halve the
result to 198K. Now get the details right, include the "greenhouse
effect" and other terrestrial details, and the answer should work out to
agree with observation.

Or ask any real physicist.

--
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hanson wrote:
Here's a discussion we had a decade ago, in s.p.
on a similar issue.
It might help you to clarify the aspects you are addressing:
"Jean Radius" and the "Phillips Ignition Temp" equation.

https://groups.google.com/forum/#!msg/sci.physics/usXytJCZN0Q/qqDvIsAch_wJ
or same in
http://tinyurl.com/Star-Ignition-Temp
Read Prof. Martin Hardcastle's next 2 posts in that link.

Yousuf Khan wrote:

How does one calculate the temperature of the space surrounding a
stellar body at a certain distance from it?

not at all. There is nothing to calculate. A temperature of the
electromagnetic radiation field is only defined for an equilibrium
state. That's why Planck was talking about hollow body radiation - the
radiation field within a hollow body is in equlibrium with the body walls.

For the radiation emitted by a stellar body, there is no equilibrium,
because there's is a hot stellar body surrounded by a cold environment.
Only directly a the stellar body's surface, the radiation field can be
approximated as being in equilibrium, enabling one to describe it using
Planck's radiation law that indicates some spectral composition and
energy density for a given temperature.

However, far away from the surface, equlibrium approximation is no
longer appropriate. The spectral composition is still the same as near
the surface, but the energy density is much lower. Therefore, one cannot
define a temperature for the radiation field any longer. If one defined
temperature by spectral composition, energy density wouldn't match, and
if one defined temperature by energy density, spectral composition
wouldn't match vice versa.

There is a quantity they call color temperature, which is defined by the
spectral composition, independent from energy density. But that is not a
temperature. E.g. the sun light has always a color temperature of 6000
°K (degree Kelvin), no matter how far away from the surface of the sun.
Even beyond the trajectory of Pluto, the sun light has this color
temperature, although its energy density is that low there that is has
no appreciable heating effect.


Let's take the Earth and Sun as an example. If the Sun's surface
temperature is 9000K, and its radius is 700,000 km, and the Earth is 1
AU (1.5E+8 km) away from the Sun. At that distance using an inverse
square relationship, I get 0.2K as the answer. Obviously the Earth is
much warmer than that. What's the real way to obtain temperature here?

You have to distinguish between the temperature of the radiation field,
that is only defined in equilibrium state, and the temperature of a body
that is heated by absorbing radiation from the radiation field.

To calculate the latter, you need to consider how much energy the body
absorbs per time interval from the radiation field, and how much energy
the body emits per time interval as function of its own temperature. The
body's temperature approaches a value where both energy flows are the same.

On 03/12/2013 23:18, Gregor Scholten wrote:
Yousuf Khan wrote:

How does one calculate the temperature of the space surrounding a
stellar body at a certain distance from it?

not at all. There is nothing to calculate. A temperature of the
electromagnetic radiation field is only defined for an equilibrium
state. That's why Planck was talking about hollow body radiation - the
radiation field within a hollow body is in equlibrium with the body walls.

For the radiation emitted by a stellar body, there is no equilibrium,
because there's is a hot stellar body surrounded by a cold environment.
Only directly a the stellar body's surface, the radiation field can be
approximated as being in equilibrium, enabling one to describe it using
Planck's radiation law that indicates some spectral composition and
energy density for a given temperature.

Although what you say is strictly true it is not a helpful way of
interpreting the OPs question. Temperature in a vacuum with radiation
and solar wind passing through it is not well defined.

He should have asked:

What is the equilibrium temperature of a test particle placed in orbit
at some distance from a star?

And that question is answerable. The answer assuming a black body also
depends on whether is is a perfect insulator or a conductor.

Basically the test object is in thermal radiation equilibrium with the
sun T~5800K occupying the solid angle S it subtends at its surface and
the rest of the universe 4K occupying 2pi-S (or 4pi-S for a conductor).

However, far away from the surface, equlibrium approximation is no
longer appropriate. The spectral composition is still the same as near
the surface, but the energy density is much lower. Therefore, one cannot
define a temperature for the radiation field any longer. If one defined
temperature by spectral composition, energy density wouldn't match, and
if one defined temperature by energy density, spectral composition
wouldn't match vice versa.

There is a quantity they call color temperature, which is defined by the
spectral composition, independent from energy density. But that is not a
temperature. E.g. the sun light has always a color temperature of 6000
°K (degree Kelvin), no matter how far away from the surface of the sun.
Even beyond the trajectory of Pluto, the sun light has this color
temperature, although its energy density is that low there that is has
no appreciable heating effect.

Pluto is about 40AU away. Insolation at the Earth is 1362 W/m^2 so at
Pluto the insolation is a shade under 1W/m^2. That gets thermalised.


Let's take the Earth and Sun as an example. If the Sun's surface
temperature is 9000K, and its radius is 700,000 km, and the Earth is 1
AU (1.5E+8 km) away from the Sun. At that distance using an inverse
square relationship, I get 0.2K as the answer. Obviously the Earth is
much warmer than that. What's the real way to obtain temperature here?

You have to distinguish between the temperature of the radiation field,
that is only defined in equilibrium state, and the temperature of a body
that is heated by absorbing radiation from the radiation field.

To calculate the latter, you need to consider how much energy the body
absorbs per time interval from the radiation field, and how much energy
the body emits per time interval as function of its own temperature. The
body's temperature approaches a value where both energy flows are the same.

Agreed. And that radiation balance gives a solution that is of the form

Teq = k/sqrt(R)

Taking average temperatures

Tearth ~ 288K @ 1AU
so
Tpluto ~ 288/sqrt(40) ~ 45K

Which is in good agreement with measurements 33K-55K see for example:

http://en.wikipedia.org/wiki/Pluto
Bottom RHS table.

--
Regards,
Martin Brown

In article ,
Martin Brown writes:
Although what you say is strictly true it is not a helpful way of
interpreting the OPs question. Temperature in a vacuum with radiation
and solar wind passing through it is not well defined.

He should have asked:

What is the equilibrium temperature of a test particle placed in orbit
at some distance from a star?

And that question is answerable. The answer assuming a black body also
depends on whether is is a perfect insulator or a conductor.

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.)

Basically the test object is in thermal radiation equilibrium with the
sun T~5800K occupying the solid angle S it subtends at its surface and
the rest of the universe 4K occupying 2pi-S (or 4pi-S for a conductor).

To clarify this, in equilibrium, exactly as much power leaves the body
as it absorbs.

A complication is that emission and absorption in general occur at
different wavelengths, so the body's emissivity as a function of
wavelength is important. The Earth's greenhouse effect is an
example, but even bodies with no atmosphere will typically have
different temperatures than the simple calculation would give. A
familiar example is that in sunlight on Earth, black objects will
usually become hotter than white ones. For very small objects ("dust
particles"), the wavelength dependence of emissivity means the
equilibrium temperature goes as the -0.4 or -0.33 power of distance
rather than the -0.5 for large objects.

In first-year astronomy courses, the reverse problem is often
presented as "Explain how to measure the Sun's temperature using a
thermometer and a stopwatch." The stopwatch gives the angular
diameter of the Sun (as it rises or sets) and the thermometer the
temperature of the Earth. From there, it's just geometry and the
Stefan-Boltzmann law. Obviously the result is very rough indeed
(ignoring the greenhouse effect entirely), but it illustrates the
principle.

--
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 , Mon, 9 Dec 2013
22:39:34, Steve Willner posted:

In first-year astronomy courses, the reverse problem is often
presented as "Explain how to measure the Sun's temperature using a
thermometer and a stopwatch." The stopwatch gives the angular
diameter of the Sun (as it rises or sets) and the thermometer the
temperature of the Earth. From there, it's just geometry and the
Stefan-Boltzmann law. Obviously the result is very rough indeed
(ignoring the greenhouse effect entirely), but it illustrates the
principle.

If the thermometer is sufficiently like the traditional mercury-in-glass
thermometer, the stopwatch is not needed.

Just use the thermometer, held at arm's length, as a ruler. Measure the
angular diameter of the Moon in apparent degrees C (apparent degrees F
in the USA), and also measure in apparent degrees C the known angular
distance of 2 pi radians (in at least 20 segments) between somewhere on
the horizon and itself. From that one gets the angular size of the Moon
in radians. Anyone reaching that stage of an astronomy course should
know that the Moon and the Sun are, near enough, of the same angular
size as seen from any location currently accessible to first-year
astronomers.

The stopwatch method, if used to measure the time during which the Sun
is partly obscured by the apparent surface of the Earth, requires the
Sun's "motion" to be near enough perpendicular to the relevant part of
the apparent surface - which cannot always be easy for the Inuit.

--
(c) John Stockton, near London. Mail
Web http://www.merlyn.demon.co.uk/ - FAQish topics, acronyms, and links.
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Do not Mail News to me. Before a reply, quote with "" or " " (RFC5536/7)

On 01/12/2013 5:33 PM, Mike Dworetsky wrote:
Yousuf Khan wrote:
How does one calculate the temperature of the space surrounding a
stellar body at a certain distance from it? I was just using an
inverse square relationship between temperature and distance, but
that comes up with non-sense results.

Let's take the Earth and Sun as an example. If the Sun's surface
temperature is 9000K, and its radius is 700,000 km, and the Earth is 1
AU (1.5E+8 km) away from the Sun. At that distance using an inverse
square relationship, I get 0.2K as the answer. Obviously the Earth is
much warmer than that. What's the real way to obtain temperature here?

Yousuf Khan

How are you doing the calculation? Do you have a correct figure for the
total luminosity of the Sun?

Just using the surface temperature of the Sun at its surface radius
(700,000km), and then taking the inverse of the distance squared at the
average orbit of the Earth (i.e. 1AU). Don't have any figures on the
luminosity of the Sun.

Yousuf Khan

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

On 11/12/2013 05:14, 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

Venus atmosphere is so dense that there is a runaway greenhouse effect -
the small amount of natural sunlight that does get down to the surface
finds it very difficult to escape again after being thermalised.

Without its atmosphere the planet would have a similar range of
temperatures to the unlucky Mercury (corrected for solar distance).

--
Regards,
Martin Brown

In article ,
Yousuf Khan writes:
Yet, Venus is an extremely slow rotator, and it's very warm on both its
day and night sides,

As Martin implied, Venus' atmosphere is a very good conductor. (It's
really winds, technically "advection," and not conduction, but the
effect is the same.) That makes the temperature very nearly the same
everywhere on the planet.

Venus' atmosphere has lots of absorption features at different
wavelengths, making the simple calculation inaccurate. In
particular, the atmosphere is highly opaque in the infrared, where
heat would "like" to escape, giving rise to the runaway greenhouse
effect that is seen.

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

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

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.

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.

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?

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?

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