Why does Earth's tilt produce summers and winters?

Ajanta wrote:
[snip]

The solar wind blows across the poles and cools them off.
Socks

In sci.physics Ajanta <[Only registered users see links. ]> 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.

This is also why fixed, i.e. non-tracking, solar panels are mounted
at an angle equal to the latitude.

--
Jim Pennino

Remove .spam.sux to reply.

[Only registered users see links. ] wrote:


To simplify this, take a fistful of maybe a dozen pencils
wrapped and held together by a rubber band. Hold them
perpendicular to a sheet of paper and press against
the paper. The marks are close together.

Now hold them at a 45 degree angle to the paper. Allow
the pencils to slide past one another till all pencils
have made a mark.

Compare the two. Each dot represents some unit of heat
that hits the paper (earth.) The more they're spread
out the less the heat density is.

Now take the pencils to a globe, and repeat for the
equator and the poles. holding the pencils on a line
where the erasers point towards the sun.

This is a typical Kibology question. Whether or not my
answer is satisfactory, I'm done playing.

unsettled <[Only registered users see links. ]> wrote:


This works for pencils but I am not sure about the Sun.

In case of pencils, there is only one lead running from the eraser end.
Then pencils have a thickness, so their writing ends (one per pencil)
do spread out.

By contrast, each point on the Sun's surface is sending a "ray" to each
point on the Earth's surface. Plus the "rays" have no thickness and
don't spread out like pencils.

But I am willing to consider a mere point on Earth's surface, say a
very small drop of water. Surely, so far away from the Sun, it is just
like a point and surface issues should not arise. Why is it receiving
less heat at the pole than at equator?

<[Only registered users see links. ].sux.com> wrote:


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?

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

Actually you can but the effect is small.


The problem is that your body doesn't just experience heat radaited directly
from the sun but it also gets heat conducted through the air from the
ground. The latter is bigger than the former so you tend not to noice the
variation.

John Popelish <[Only registered users see links. ]> 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!

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?!?!

In sci.physics Ajanta <[Only registered users see links. ]> wrote:



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

Remove .spam.sux to reply.

In sci.physics Ajanta <[Only registered users see links. ]> wrote:






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


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

--
Jim Pennino

Remove .spam.sux to reply.

Ajanta wrote:

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.


Yes.


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.