Calculating stellar temperature at a distance?

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

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?

Use a tomographic algorithm

On Sunday, December 1, 2013 6:03:06 AM UTC-8, 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

Only a millionth of the mass of Earth gets directed heated by the sun at any given time. If the sun were to turn off, we'd survive quite nicely, though as mostly underground.

Brad Guth wrote:
On Sunday, December 1, 2013 6:03:06 AM UTC-8, 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

Only a millionth of the mass of Earth gets directed heated by the sun
at any given time. If the sun were to turn off, we'd survive quite
nicely, though as mostly underground.

The amount of energy coming to the surface from the interior is only about
1-2 W/m^2, vs the solar heating at the equator of around 1.4 kW/m^2. If the
Sun were to turn off, it would get cold pretty fast--within a year the land
would be frozen solid, though it would take the oceans a while to lose their
heat. Eventually the atmospheric gases would liquefy and eventually freeze.

The mantle and crust are very good insulators.

--
Mike Dworetsky

(Remove pants sp*mbl*ck to reply)

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?

--
Mike Dworetsky

(Remove pants sp*mbl*ck to reply)

On 01/12/2013 14:03, 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?

I think you need to show your workings...

The suns photosphere is about 5800K. 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.

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.

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.

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)

--
Regards,
Martin Brown

Dear Martin Brown:

On Monday, December 2, 2013 2:53:03 AM UTC-7, Martin Brown wrote:
....
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.

I would suspect that "silicon dioxide" also has a triple point, and such would be exceeded upon approach. So the various states could help keep the central mass cooler.

Excellent analysis. Thank you very much for showing how it should be done. I like the quality checks too.

David A. Smith

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.

--
(c) John Stockton, near London. Mail
Web http://www.merlyn.demon.co.uk/ - FAQish topics, acronyms, and links.
Correct = 4-line sig. separator as above, a line precisely "-- " (RFC5536/7)
Do not Mail News to me. Before a reply, quote with "" or " " (RFC5536/7)

"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?

Is the Moon much warmer than that? It's the same distance.



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.

--
(c) John Stockton, near London. Mail
Web http://www.merlyn.demon.co.uk/ - FAQish topics, acronyms, and
links.
Correct = 4-line sig. separator as above, a line precisely "-- "
(RFC5536/7)
Do not Mail News to me. Before a reply, quote with "" or " "
(RFC5536/7)

On Sunday, December 1, 2013 2:32:11 PM UTC-8, Mike Dworetsky wrote:
Brad Guth wrote:

On Sunday, December 1, 2013 6:03:06 AM UTC-8, 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



Only a millionth of the mass of Earth gets directed heated by the sun

at any given time. If the sun were to turn off, we'd survive quite

nicely, though as mostly underground.



The amount of energy coming to the surface from the interior is only about
1-2 W/m^2, vs the solar heating at the equator of around 1.4 kW/m^2. If the
Sun were to turn off, it would get cold pretty fast--within a year the land
would be frozen solid, though it would take the oceans a while to lose their
heat. Eventually the atmospheric gases would liquefy and eventually freeze.


The mantle and crust are very good insulators.

--

Mike Dworetsky

Only the outer most millionth mass of Earth would get near cryogenic cold. However, in certain locations situated over geothermal vents, open water could still exist.

A km below the icy surface, the surrounding temperature might only drop by a few degrees.