Why does Earth's tilt produce summers and winters?

In sci.physics Ajanta wrote:
wrote:

The amount of energy received per unit area on a surface from the Sun
is a function of the sine of the angle of the surface to the Sun.

At the equator, the angle is near 90 degrees (sin 90 = 1) and at the
poles the angle is near 0 (sin 0 = 0). Add a correction angle for the
tilt of the Earth's axis for the actual 0 and 90 degree points.

Still not sure about this explanation. Let us consider a mere point on
Earth's surface: say a very small droplet of water. Surely, so far away
from the Sun, the droplet is just like a point and surface issues
should not arise. Why is it receiving less heat at the pole than at
equator?

You are confusing yourself with points, it is energy per unit area.

The energy per unit area is essentially fixed.

Take a 1 foot square piece of paper and look at it 90 degrees to
the surface.

How much area do you see? 1 square foot.

The paper would intercept all the energy in 1 square foot if the
energy were coming from your face.

Turn the paper so it is at 0 degrees, that is you are looking at the
edge.

How much area do you see? Essentially 0.

The paper would intercept none of the enery in 1 square foot if the
energy were coming from your face.

Now turn the paper to 45 degrees.

How much area do you see? .707 square feet, which is the sine of 45 deg.

The paper would intercept .7 of the enery in 1 square foot if the
energy were coming from your face.

If you don't believe the last one, take a 1 foot square piece of paper
in one hand, a ruler held at 90 degrees in the other hand, and measure
the apparent length and width you see with the paper held at 45 degrees.

Multiply the two together and it should be close to .7 square feet
depending on how accurately you do this.

--
Jim Pennino

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In sci.physics Ajanta wrote:
John Popelish wrote:

Picture yourself standing on the western goal line of a foot
ball field at the moment when the setting Sun is just above
the western horizon. You cast a shadow all the way the
eastern goal line. The sunshine feels somewhat warm on your
body.

But your body is intercepting all the solar heat that would
have been deposited in that long (and very much larger)
shadow area your body is casting, if you hadn't been here.
When the heat you are feeling is spread out over that long
(and large) shadow area, it is very much weaker at any spot,
than when it strikes you, almost perpendicular to your surface.

I have is no problem with the shadow, but with the fact that my body
itself receives less heat at the pole than at the equator!

To clarify my confusion:

Take two identical objects, say two ice-cubes. Place one at the pole
and the other one at the equator.

Now remove the Earth!

We have two same size ice-cubes at an almost same distance from the
Sun. They should receive the same heat from the Sun.

Assuming they have the same relative angles to the Sun, yes.

Now, why does it become different when we slip the Earth back in behind
those ice-cubes?!?!

The Earth is radiating heat, nothing more, nothing less, and has no
effect on the heat received directly from the Sun.

--
Jim Pennino

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Ajanta wrote:
John Popelish wrote:

Picture yourself standing on the western goal line of a foot
ball field at the moment when the setting Sun is just above
the western horizon. You cast a shadow all the way the
eastern goal line. The sunshine feels somewhat warm on your
body.

But your body is intercepting all the solar heat that would
have been deposited in that long (and very much larger)
shadow area your body is casting, if you hadn't been here.
When the heat you are feeling is spread out over that long
(and large) shadow area, it is very much weaker at any spot,
than when it strikes you, almost perpendicular to your surface.

I have is no problem with the shadow, but with the fact that my body
itself receives less heat at the pole than at the equator!

To clarify my confusion:

Take two identical objects, say two ice-cubes. Place one at the pole
and the other one at the equator.

Now remove the Earth!

And also remove the 10 miles or so of atmosphere between the Sun and
the equatorial ice cube, and 100 miles or so of atmosphere between the
Sun and the polar cube.

We have two same size ice-cubes at an almost same distance from the
Sun. They should receive the same heat from the Sun.

Yes.

Now, why does it become different when we slip the Earth back in behind
those ice-cubes?!?!

If the Earth had no atmosphere, the intensity of the direct radiation
on each would be the same, on the side facing the Sun. But the energy
falling on each square mile of horizontal area would be very different,
because the equatorial square mile would intercept a lot more total
energy arriving perpendicular to it than the polar square mile, that is
almost edge on to the Sun, would intercept.

(Except for the larger atmospheric absorption at the pole) the
intensity of sunlight on a surface perpendicular to the Sun's direction
receive the same energy. There are just a lot more perpendicular
surfaces at the equator thant there are at the poles.

Ice cues laying on a horizontal equatorial surface each receive their
own sunbeams, while ice cubes sitting on a horizontal polar surface are
shaded, to a large extent, by their neighbors. Remove the Earth,
but retain the positions of many nearby ice cubes in both locations,
and this fact remains. The polar region is not a single ice cube, but
a multitude of ice cubes, mostly in the shade of other ice cubes. The
problem is not low intensity (though the atmospheric absorption does
lower the intensity) so much as excessive shade.

Dear Ajanta:

Ajanta wrote:
....
I have is no problem with the shadow, but with the fact that
my body itself receives less heat at the pole than at the
equator!

Basic heat transfer.
Insolation from the Sun is one factor. (Heat in.)

Convection, evaporation, reflection, and radiation to the environment
are the ways a body can lose heat. These all require a temperature
difference between the body and the "environment".

The poles are cooler than the equator... at least on Earth.

To clarify my confusion:

Take two identical objects, say two ice-cubes. Place one
at the pole and the other one at the equator.

Now remove the Earth!

We have two same size ice-cubes at an almost same
distance from the Sun. They should receive the same heat
from the Sun.

Without an atmosphere, yes.

Now, why does it become different when we slip the Earth
back in behind those ice-cubes?!?!

You have already been told that incident radiation on the "environment"
is a function of the angle of incidence. Whether or not you "get
it"...

You have a longer path length through air to go along with the angle of
incidence, which tends to decrease solar radiation intensity still
further. Then you have a 24 hour rotation, that commonly removes all
direct solar heating once a day. Then you have varying amounts of air
currents, relative humidity, and local terrain to provide enhanced or
retarded cooling capacity.

David A. Smith

"Ajanta" wrote in message
...
| John Popelish wrote:
|
| Picture yourself standing on the western goal line of a foot
| ball field at the moment when the setting Sun is just above
| the western horizon. You cast a shadow all the way the
| eastern goal line. The sunshine feels somewhat warm on your
| body.
|
| But your body is intercepting all the solar heat that would
| have been deposited in that long (and very much larger)
| shadow area your body is casting, if you hadn't been here.
| When the heat you are feeling is spread out over that long
| (and large) shadow area, it is very much weaker at any spot,
| than when it strikes you, almost perpendicular to your surface.
|
| I have is no problem with the shadow, but with the fact that my body
| itself receives less heat at the pole than at the equator!
|
| To clarify my confusion:
|
| Take two identical objects, say two ice-cubes. Place one at the pole
| and the other one at the equator.
|
| Now remove the Earth!
|
| We have two same size ice-cubes at an almost same distance from the
| Sun. They should receive the same heat from the Sun.
|
| Now, why does it become different when we slip the Earth back in behind
| those ice-cubes?!?!

The air blanket between one ice-cube and sun is 100 km thick,
between the other ice-cube and sun is 2000 km thick.
To see the effect that has on sunlight, note the colour of sunset.

Ajanta wrote:
AlexZ wrote:

One explanation I have never understood is that we have Summers and
Winters because Earth is tilted.
...

I don't have an answer but a similar question: Why are the poles so
much colder than the equator?

I imagine that the Earth's size is much smaller than Earth-Sun
distance, so the poles and the equator are more or less equidistant
from the Sun...

Moreover, one of the poles is always facing the Sun and therefore does
receives sunlight. Then, why is this Sun-facing pole colder than the
Equator?


It is as much due to the distance as it is the sunlight reflecting
off the atmosphere. Think of the atmosphere like a peice of glass. When
you shine a light through it at a perpendicular angle the little of the
light is reflected. The more you tilt the glass the more reflective it
becomes. Therefore the more light (so more heat) is reflected off the
atmosphere.

Sam Wormley wrote:


The scientific evidence strongly suggests that the moon was formed
as the result of a major collision with the young earth. The impact
certainly also contributed to the degree of tilt in the earth's axis
of rotation.


What scientific evidence?

NPS

dlzc wrote:
Dear Ajanta:

Ajanta wrote:
...
I have is no problem with the shadow, but with the fact that
my body itself receives less heat at the pole than at the
equator!

Basic heat transfer.
Insolation from the Sun is one factor. (Heat in.)

Convection, evaporation, reflection, and radiation to the environment
are the ways a body can lose heat. These all require a temperature
difference between the body and the "environment".

The poles are cooler than the equator... at least on Earth.

To clarify my confusion:

Take two identical objects, say two ice-cubes. Place one
at the pole and the other one at the equator.

Now remove the Earth!

We have two same size ice-cubes at an almost same
distance from the Sun. They should receive the same heat
from the Sun.

Without an atmosphere, yes.

Now, why does it become different when we slip the Earth
back in behind those ice-cubes?!?!

You have already been told that incident radiation on the "environment"
is a function of the angle of incidence. Whether or not you "get
it"...

You have a longer path length through air to go along with the angle of
incidence, which tends to decrease solar radiation intensity still
further. Then you have a 24 hour rotation, that commonly removes all
direct solar heating once a day. Then you have varying amounts of air
currents, relative humidity, and local terrain to provide enhanced or
retarded cooling capacity.

Uh... What happens to the solat energy that disappears in the longer
atmospheric path? Doesn't it just contribute to atmospheric heating
within at most 200 km or so?

The ice-cube point is a valid one. Ignoring atmospheric absorbtion, a
spherical absorber (a black ball) on the earth will receive the same
solar energy each second from the sun , no matter how high the sun is
above the horizon. What must be included is the blocking effect of
that black ball on all the other surrounding black balls. (And the
length of daylight)

As a previous poster neatly observed, the sunlight on your back stays
the same, what changes is the length and thus the size of your shadow,
the area you are depriving of its solar energy...

wrote:
..

Uh... What happens to the solat energy that disappears in the longer
atmospheric path? Doesn't it just contribute to atmospheric heating
within at most 200 km or so?

Oops... Make that at most 1200 km or so...

wrote in message
oups.com...
|
| dlzc wrote:
| Dear Ajanta:
|
| Ajanta wrote:
| ...
| I have is no problem with the shadow, but with the fact that
| my body itself receives less heat at the pole than at the
| equator!
|
| Basic heat transfer.
| Insolation from the Sun is one factor. (Heat in.)
|
| Convection, evaporation, reflection, and radiation to the environment
| are the ways a body can lose heat. These all require a temperature
| difference between the body and the "environment".
|
| The poles are cooler than the equator... at least on Earth.
|
| To clarify my confusion:
|
| Take two identical objects, say two ice-cubes. Place one
| at the pole and the other one at the equator.
|
| Now remove the Earth!
|
| We have two same size ice-cubes at an almost same
| distance from the Sun. They should receive the same heat
| from the Sun.
|
| Without an atmosphere, yes.
|
| Now, why does it become different when we slip the Earth
| back in behind those ice-cubes?!?!
|
| You have already been told that incident radiation on the "environment"
| is a function of the angle of incidence. Whether or not you "get
| it"...
|
| You have a longer path length through air to go along with the angle of
| incidence, which tends to decrease solar radiation intensity still
| further. Then you have a 24 hour rotation, that commonly removes all
| direct solar heating once a day. Then you have varying amounts of air
| currents, relative humidity, and local terrain to provide enhanced or
| retarded cooling capacity.
|
| Uh... What happens to the solat energy that disappears in the longer
| atmospheric path? Doesn't it just contribute to atmospheric heating
| within at most 200 km or so?
|
| The ice-cube point is a valid one.


No it isn't.
If he thought about area instead of a point and accepted reality
he'd see why.
Androcles





Ignoring atmospheric absorbtion, a
| spherical absorber (a black ball) on the earth will receive the same
| solar energy each second from the sun , no matter how high the sun is
| above the horizon. What must be included is the blocking effect of
| that black ball on all the other surrounding black balls. (And the
| length of daylight)
|
| As a previous poster neatly observed, the sunlight on your back stays
| the same, what changes is the length and thus the size of your shadow,
| the area you are depriving of its solar energy...
|