# Why the sky is dark in the night

Discussion in 'Astronomy, Exobiology, & Cosmology' started by The God, Mar 23, 2016.

1. ### DaeconKiwi fruitValued Senior Member

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I'd be interested in learning how to figure out the maths required to determine how many degrees of an arc are we getting of light from the sun, compared to how many degrees of an arc are we getting of light from the other stars.

But I wouldn't even know where to begin the equations.

If the diameter of the sun is about 1.4 million km, and the distance to Earth is 150 million km, and the circumference of Earth's orbit is about 940 million km, do I divide 940 million by 360 to make about 2.6 million, and then do something else...? I'm totally lost.

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3. ### Russ_WattersNot a Trump supporter...Valued Senior Member

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The distance between the Earth and sun and diameter of the sun are two sides of a triangle...

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5. ### DaeconKiwi fruitValued Senior Member

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So the hypotenuse of that triangle would be barely any higher than 150 million as well? The square root of 22,501.96 million?

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7. ### Russ_WattersNot a Trump supporter...Valued Senior Member

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Right, so you needn't bother with that -- just apply a trig function to get the angle.

8. ### DaeconKiwi fruitValued Senior Member

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So... would that be the tangent? 1.4 divided by 150 to get 0.009 degrees (to three decimal places)?

So we get just over 0.009 degrees of sunlight from our local star, compared to the nearest non-sun star, Alpha Centauri, where the calculation would be (1.4*10^6) divided by (4.1*10^13)?

Where 1.4*10^6 would be the diameter of Alpha Centauri in kilometers, assuming the same diameter as Sol for convenience, and 4.1*10^13 would be the distance between Alpha Centauri and Sol in kilometers?

Which would be about 2.9*10^-7?

Last edited: Apr 5, 2016
9. ### DaeconKiwi fruitValued Senior Member

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Hold on, I think I messed up the tangent. It should be a bit over half a degree, shouldn't it, not just under a hundredth of a degree?

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10. ### Russ_WattersNot a Trump supporter...Valued Senior Member

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No, it is the inverse tangent.

11. ### Russ_WattersNot a Trump supporter...Valued Senior Member

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Heh, you corrected yourself just as I was typing it; Yes, about half a degree. 0.57 by my math.

12. ### DaeconKiwi fruitValued Senior Member

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Yeah, I did the division but totally forgot to actually use the tan-1 function. A silly error.

13. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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Work in radians, then it is simple (no trig functions needed). I. e. the sun's angle is then (using your data) 1.4 / 150 radians. Valid for small angles.
The full 360degree circle has 2pi radians, so if you want answer in degrees it is:
360 x 1.4 / (150 x 2 pi)

Last edited: Apr 5, 2016
14. ### DaeconKiwi fruitValued Senior Member

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The problem is that until today I didn't even know what a radian was.

But using the inverse tangent of 2.9*10^-7 would mean the nearest star shines about 17 millionths of a degree of starlight onto Earth?

Hmm, but that's not even taking into account the relative brightness and distances of all the other individual stars, and other variables.

I think it's too far above my skill level.

15. ### Russ_WattersNot a Trump supporter...Valued Senior Member

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Well, yeah - you made a mistake, fixed it, and learned something. And you didn't die or burst into flames or anything! The God should take note of how painless learning can be!

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16. ### DaeconKiwi fruitValued Senior Member

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So for Alpha Centauri it would be 360 * 1.4*10^6 / (4.1*10^13 * 2 pi)?

17. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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Yes if those are the diameter and distance numbers, but I don't think they are.

18. ### DaeconKiwi fruitValued Senior Member

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Yeah, I think I confused myself somewhere. I was only using rough figures from a quick browse of Wikipedia for an estimate of the differences in light from the other stars, relative to our sun.

19. ### Russ_WattersNot a Trump supporter...Valued Senior Member

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I prefer working in degrees anyway. They are better/more common for astronomy. Angular diameters are typically in degrees, arcminutes (1/60th of a degree) and arcseconds (1/60th of an arcminute).

20. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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That is very foolish. Stars all subtend a very small angle so the radian's small angle approximation is excelent - less than one part in a billion error I guess. (at least a 10,000 times more accurate than any trig table.)
Why use trig functions when you don't need to?
If you can only think in degrees, calculate the angle in radians (star diameter / star distance) and then multiply by 57.2958 to convert to degrees. Or if you want to go to arcminutes directly multiply the radian answer by 3,437.75

I. e. what could be easier than 3,437.75 x (star diameter / star distance) with no need of trig tables and accuracy limited by the distance ratio, not some short trig table. Multiply by 60 again to get arcseconds, or directly by:
20,626.45 x (star diameter / star distance) is the star's angle (from earth) in arcseconds.

PS just a 20626 factor is more than enough - no ratio of the distances is correct to even four significant figures.
That is a "nice number" you can even keep in your head, if you do this often.

Last edited: Apr 5, 2016
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21. ### The GodValued Senior Member

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Lets see whether you show the same mettle as shown by Daecon ?

How did you get 0.57 deg ? A simple math that is (1.4/150)*(360/2pi) gives you 0.53...

22. ### The GodValued Senior Member

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Thats lovable daecon.....I know that in your earlier posts you behaved rather strangely giving an impression that you understand all these stuff....but nonetheless its a great thing to learn and show such honest intent for the same..If I was unduly harsh to you, I withdraw the same.

23. ### The GodValued Senior Member

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making lot of noise, Russ Watters ? Empty cans do that very often...