Why does Earth's tilt produce summers and winters?

Dear Randy Poe:

Randy Poe wrote:
....

The largely transparent, poorly-absorbing, poorly-radiating
atmosphere... *isn't* for all wavelengths. Ionization provided by
direct solar irradiance (equator) and solar wind (via the Van Allen
belts, poles) keep the ionosphere very "active" at wavelengths
corresponding to IR near (guessing a number here) -30°C as a
threshold. This allows the atmosphere to be a very good absorber /
radiator / conductor of heat... at some upper bound that is still
danged cold. Now you have the CMBR sink at 3K and the ionospheric
"waveguide" at 243K that establish the balance.

It isn't just transport that atmosphere provides, but storage and
insulation as well.

Should anyone care to delve more into radiative heat transfer, a Google
search phrase might be:
"shape factor" radiation OR radiative

With a "point source" like the Sun, however, it is still as simple as
the angle between the surface normal and the line to the Sun that
describes local irradiance. The light reflected / emitted by the Earth
to the Sun is inconsequential, compared to the bounty we receive. Our
global sink is the ~3K CMBR background, which is everywhere the Sun
isn't.

David A. Smith

"AlexZ" <[Only registered users see links. ]> wrote in message
news:[Only registered users see links. ]...

Two things cause all to happen.

[a] Earth's tilt
[b] Earth going around the Sun

And of course the fact that earth rotates itself.

As the tilt is retained as what it is throughout Earth's traversal around
the Sun at one point the tilt makes one pole [the end of Earth's rotational
axis] exactly tilted towards the Sun and the other pole away from it. This
is a mathematically precise instant when the line joining the center of Sun
to the center of Earth is coplanar with the Earth's rotating axis. Let this
be our starting point.

At this point one pole gets sunlight 24 hours and the other pole none at all
as the rotation of earth does nothing to the poles , ie it cannot ever move
the polar point into or out of the sunny side or dark side caused by sun's
rays on earth. Regions away at latitudes closer to the equator are fortunate
to get some sunlight [12 hours of sunlight and 12 hours of night at the
equator, more than 12 hours of day and less than 12 hours of night in the
particular polar hemisphere towards the Sun and the reverse in the case of
the other polar hemisphere]
and this differential illumination of the Sunlight on the earth causes
summer in one hemisphere and winter in the opposite. The Sun appears to be
centered not on top of the equator but on a latitude towards that pole
tilted closer to Sun. The Sun's positional point can be marked at the
intersection of the line drawn between the center of earth and center of sun
and earth's surface. Because of the tilt it happens to be on the Cancer or
Capricorn. Of course the shadow can be viewed in a dark room by lighting up
an earth-like ball by a torchlight and viewing the effect of rotation.

Having said this it is easy to describe the others. At 90 degrees clockwise
or anticlockwise the sun to earth line is perpendicular to the earth's axis
which can only position the sun on the equator. Both poles get their typical
quota of slant sunrays 12 hours each and are treated equal by the Sun.

90 degrees around the sun again gets us the reverse of the last but one
paragraph.

This could be as simple as it can get

Researcher




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In sci.physics, AlexZ
<[Only registered users see links. ]>
wrote
on Thu, 02 Nov 2006 07:06:24 GMT
<[Only registered users see links. ]>:

I'm probably late to the party, but never mind.
For various reasons I'll take a point near Boston
with coordinates 42 degrees North, 71 degrees West.
(Apparently this is in the middle of a road in a housing
development -- convenient for my purposes.)

Mark out a level flat area of 1 m^2 or so. In the summer
(actually right near or at the summer solstice), the sun
will be at an angle of maybe 42 + 22.5 = 64.5 degrees. At
noon the heat falling on that area will be roughly
sin(64.5) * 1000 W = 903 W.

Now wait about six months, till winter solstice. The sun
will now impinge that area at an angle of 42 - 22.5 =
19.5 degrees, and the energy on that area will only be
sin(19.5) * 1000W = 334 W.

The effect is even more dramatic if one goes about 19.5
degrees north, at the Arctic circle. (This puts one near
a glacier, lake or sea, apparently. Google indicates
Kennedy.) The summer sun in that case will never set,
although it will get very close to the horizon. The winter
sun will never rise, although one might see a rosy hue
for a significant portion of the day.

If one is standing right at the North Pole, one will get
sin(45) * 1000 W = 707 W during high summer -- for the
entire 24 hour period -- but *nothing* during the dead
of winter. I'll leave the integration over the whole year
to other parties for the latitudes I've given here.
Suffice it to say that the area around Boston will get a
little more energy than the North Pole.

One can also look at the Equator. At W 71 this puts the
individual somewhere deep in the rain forest.

HTH.

--
#191, [Only registered users see links. ]
Linux. Because life's too short for a buggy OS.

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"Ajanta" <[Only registered users see links. ]> wrote in message
news:[Only registered users see links. ]...

OK - don't think of just one icecube, but think about 100 of them, arranged
in a 10x10 arrangement on the ground.

How many icecubes get the sun's radiation at the equator? All 100 of them of
course,
How many icecubes get the sun's radiation at the pole - only 10 of them,
because the other 90 are not in the direct light of the sun.

bonkers wrote:

It has little to do with distance or with the atmosphere. It has to do
with the angle of incidence of light. At noon on an equinox, a 1m wide
beam of light from the sun heats a 1m wide strip of land at the
equator. On the same date and time, a 1m wide beam of sunrays incident
on the earth at a latitude of 60 degrees (either north or south of the
equator) heats a 2m wide strip of land. So, at noon on an equinox the
Sun delivers only half the radiant energy per unit area of land at a
latitude of 60 degrees and even less energy per unit area at even
higher latitudes. At other dates and times, the calculations are more
complicated but let it suffice to say that the average angle of
incidence of sunlight is highest at the equator and lowest at the poles
and that the poles are therefore heated less (than the equator) by
radiant energy from the Sun.

"The Chief Instigator" <[Only registered users see links. ].com> wrote in message
news:[Only registered users see links. ].com...

Nothing to do with it. If the distance that light had to travel through the
atmosphere was the determinant, it would be warmer at high altitudes, not
colder.

"John Popelish" <[Only registered users see links. ]> wrote in message
news:[Only registered users see links. ] ...

Correct (and a good explanation).


Sort of true, but irrelevant. The reason the sun is dimmer is because it has
transferred more of its energy to the column of air a few tens of kilometers
long between you and the Sun. This actually increases local air temperature,
as more radiation is absorbed by the air.


As is true for every location on the earth.


I live in Sydney. 'nuff said.

Peter Webb wrote:
(snip)
(snip)

A "column of air a few tens of kilometers long" is not
"local". Any particular cubic meter of that column is not
warmed more because the total path through the air is longer.

"John Popelish" <[Only registered users see links. ]> wrote in message
news:[Only registered users see links. ]...

If you are considering the climate at a particular spot, a few tens of
kilometers is local. The light that lands of kilometers behind you went
directly over your head on its travels, warming that air.


Indeed, but there are more cubic metres of air transitted, so more transfer
of energy.

Imagine that the earth's atmosphere absorbed 50% of the light at the tropics
(ie with the Sun over head). If it was at a low incident angle (eg near the
Poles or at sunset) it would be far dimmer. More of the energy in the light
would have been absorbed by the atmosphere, which means more heating of the
air. (The energy has to go somewhere.). Direct warming of the air by the Sun
is not a significant contributor to the temperature of the atmosphere; if it
was, having the Sun shining through more atmosphere would increase the
percentage of the energy absorbed by the atmosphere, having the opposite
effect.

"Sorcerer" <Headmaster@hogwarts.physics_e> wrote in message
news:Bly3h.97162$[Only registered users see links. ].blueyonder.co. uk...

Because of angles of incidence.


Partially because more of its short wavelength light is absorbed by the
atmosphere, which has the effect of warming the air. (But mostly because of
refraction, which has no net effect on temperatures).



Of course. The Sun is weaker because more of its energy has been transferred
already to the atmosphere as heat.

The further distance through the atmosphere, the more energy is absorbed by
the atmosphere, and hence the warmer the resulting air temperature. By your
logic, planets with very thin atmospheres should be warmer (they are not),
and temperature should rise with altitude (which it doesn't).

Tell me, if the Sun appears dimmer at low angles, isn't this because the
atmosphere absorbs more of its energy? And if the atmosphere absorbs more
energy from the Sun, wouldn't that make the atmosphere warmer?