Why the sky is dark in the night

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.
 
Daecon said:
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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The distance between the Earth and sun and diameter of the sun are two sides of a triangle...
 
So the hypotenuse of that triangle would be barely any higher than 150 million as well? The square root of 22,501.96 million?
 
Daecon said:
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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Right, so you needn't bother with that -- just apply a trig function to get the angle.
 
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?
 
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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?
 
Daecon said:
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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Heh, you corrected yourself just as I was typing it; Yes, about half a degree. 0.57 by my math.
 
Yeah, I did the division but totally forgot to actually use the tan-1 function. A silly error.
 
Daecon said:
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. ...I'm totally lost.
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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)
 
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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.
 
Billy T said:
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)
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So for Alpha Centauri it would be 360 * 1.4*10^6 / (4.1*10^13 * 2 pi)?
 
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.
 
Daecon said:
The problem is that until today I didn't even know what a radian was.
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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).
 
Russ_Watters said:
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).
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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.
 
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Russ_Watters said:
Heh, you corrected yourself just as I was typing it; Yes, about half a degree. 0.57 by my math.
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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...
 
Daecon said:
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.
Click to expand...

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