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Attiyah Zahdeh
2006-Sep-05, 05:19 AM
What is the difference in the length of the path of the solar light in the troposphere (say it is 15 km thickness) beween viewing the Sun at the horizon and viewig it overhead?

In other words, how many kilometers the path of the solar light in the troposphere when the Sun is at the horizon is longer than the same path when the Sun is overhead?

tusenfem
2006-Sep-05, 06:14 AM
Dear Attiyah!

Or drawing a pic and doing some math to get to the answer of this question?

Attiyah Zahdeh
2006-Sep-05, 06:55 AM
Dear Attiyah!

Or drawing a pic and doing some math to get to the answer of this question?
Hello Tusenfem,
Answers via the specialists give one reliable, and referable to statements. Of course, asking via BAUT gives the opprtunity to who do not know to know and allows the possibility for fruitful discussions and arguments.
(1) That you do not know the answer or are not sure of your knowledge about it.
(2) That you know the answer well and ready to reply. If this case was yours, why did you aviod answering my question?
(3) That you only want to prevent others from asking.

astromark
2006-Sep-05, 08:20 AM
In aviation altitude is measured in feet. Every where. This is to avoid any confusion of terminology. For the sake of your question I will use feet. At just 12,000 feet our air is so thin that we need extra oxygen to breath comfortably. With out added oxygen or a pressurized vessel we can not venture beyond 20,000 feet. So how deep is our atmosphere? Straight up you are in space at just 50,000 feet but, there is still a little air pressure and still air is present right up to 100,000 feet. and even then some particles of air are to be found. None of this has much to do with your question except that I want you to see the thinning but still present atmosphere reaches out into space a little more than most of us think. At 300,000 feet where the shuttle and the ISS can be found there is still sufficient air to require the space craft to be nudged up back to its higher orbit. Its those random air particles that slow the ISS. We know the diameter of Earth, so working out the angle of light must travel to reach you at sun set is clear. How much air is that, about 600,000 feet plus the 100,000 feet so as my best guess,700,000 feet of significant air to travel through. Thats why its not so bright. In fact the more you think about this the greater the slice of air at sunset seems to be. The same can be said of the twinkling of stars. As they reach those lower latitudes nearing the horizon the twinkling seems to become positively Chaotic. So to your question,? Troposphere of 15 km becomes hundreds at sunset.

antoniseb
2006-Sep-05, 12:04 PM
I agree with Tusenfem that it is surprising that you don't seem to have the geometry knowledge to do this yourself.

You are looking at two concentric circles, one with a radius 6378 km and the outer one with radius 6393 km. You want to know the half-length of the chord of the outer circle that is tangent to the inner circle. Draw this and you will see a right angle triangle with the right angle at the tangent point, and so you can use the Pythagorean Theorum to compute the length:

63932 = 63782 + x2

grant hutchison
2006-Sep-05, 12:11 PM
The question is actually badly designed, since it draws a random line as the upper boundary of the atmosphere for the purpose of calculation.
Better to think in terms of air masses. If you look straight up at the zenith, you look through "one air mass"; if you look at the horizon (and beyond, to outer space), you look through 38 air masses.

Grant Hutchison

tusenfem
2006-Sep-05, 12:17 PM
Hello Tusenfem,
Answers via the specialists give one reliable, and referable to statements. Of course, asking via BAUT gives the opprtunity to who do not know to know and allows the possibility for fruitful discussions and arguments.
(1) That you do not know the answer or are not sure of your knowledge about it.
(2) That you know the answer well and ready to reply. If this case was yours, why did you aviod answering my question?
(3) That you only want to prevent others from asking.

Well I did not know the answer immediately, but I saw how you can calculate it simply.
1. Draw a circle of 1 Re (okay do not do this 1:1)
2. Draw a circle of 1 Re + 15 km (the 15 km do have to be on scale with 1 Re)
3. Draw a radius line from the center of the circle to the surface of the Earth 1 Re
4. Draw a tangent to the circle where the radius line intersects the 1 Re circle and make that it intersects the larger circle 1 Re + 15
5. Draw a radius line to the point where the tangent intersects the circle 1 Re + 15
6. Now you have a triangle with a right corner and one side 1 Re, one side you want to know and the hypothenuse is 1 Re+15km.