20090316, 13:13  #1 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
17·251 Posts 
area by percent of sun in sky
I'm wondering how much of the sky is taken up by the sun, as seen from earth. From earth, the sun appears approximately 0.5 degrees across. This means the question is equivalent to asking what is the fraction of a disk that is 0.5 degrees on the surface of a sphere, to the surface area of the sphere.
I think it might be 1/720th, or .13888...%, since a half degree line on a circle would be 1/720th of the circumference of the circle, and you could equate a sphere to a series of circles such that each would have 1/720th taken up by this line. Is that right? It seems too low, but then so does that the sun is only half a degree, and that's right. Last fiddled with by MiniGeek on 20090316 at 13:14 
20090316, 13:43  #2 
"Ben"
Feb 2007
2·3^{2}·191 Posts 
Surface area goes as R^2, so I make it to be 1/518400 or about 1.9e4 %

20090316, 14:28  #3 
Aug 2006
3^{2}×5×7×19 Posts 
The surface area of a sphere is 4*pi*r^2, so the unit halfsphere has surface area 2pi. Approximating the sun as a circle, it has area 1/720^2 * pi/4. The ratio is thus 1/(720^2*8) = 1/4147200 ~= 0.0000241%.

20090316, 16:40  #4 
"Richard B. Woods"
Aug 2002
Wisconsin USA
1111000001100_{2} Posts 
The number of square degrees on a sphere's surface is 4 * pi * (180 / pi)^2 = 129600 / pi = ~ 41253. ([URL]http://en.wikipedia.org/wiki/Square_degree[/URL]) The Sun's disc is an area of about pi * (1/4 degree)^2, which is about (pi * 1/16 square degree) / (129600 / pi) = pi^2 / 2073600 = ~ 4.76e6 of the celestial sphere. Now, if [I]sky[/I] is taken to mean only half of the celestial sphere, the the Sun occupies ~9.52e6 of the sky.
Lessee ... are we all in agreement? Apparently not, since we have four different answers. Back to the calculator ... Let's try a different method. The Sun is ~865,000 miles wide. Its distance is ~93 million miles. A circle with diameter of 865,000 miles has surface area pi * (865000 / 2)^2 = pi * 187056250000 square miles. A sphere with radius of 93 million miles has surface area 4 * pi * (93,000,000)^2 = pi * 34596000000000000 square miles. The ratio is 187056250000 / 34596000000000000 = 5.4e6. Hmmm... My two answers are 4.76e6 and 5.4e6. Aha!! The first one assumed a solar diameter of 1/2 degree; it's actually more like 32 arcminutes = 8/15 of a degree. Correcting the first figure by multiplying by (8/7.5)^2 gives ~5.4e6, which agrees with the second figure (= ~1.08e5 of the halfsphere sky). Last fiddled with by cheesehead on 20090316 at 17:27 
20090316, 17:15  #5  
Cranksta Rap Ayatollah
Jul 2003
641 Posts 
Quote:
Last fiddled with by Orgasmic Troll on 20090316 at 17:15 

20090316, 17:16  #6 
Cranksta Rap Ayatollah
Jul 2003
641 Posts 
Cheesehead: Your initial analysis and mine agree

20090316, 17:25  #7  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
10AB_{16} Posts 
Quote:
I'm thinking that, possibly, equating a sphere to an infinite number of 0 thickness lines with the same radius as the sphere does not result in consistent arithmetic. I think if wanting to equate a sphere to circles with the same radius as the sphere, we would have to instead give them each a finite, however small, thickness based on degrees, which gives us many arcs that are probably harder to calculate for than a sphere, and so are useless. Last fiddled with by MiniGeek on 20090316 at 17:31 

20090316, 17:37  #8 
"Phil"
Sep 2002
Tracktown, U.S.A.
2137_{8} Posts 
I was going to post, but Orgasmic Troll beat me to it  I agree with him.

20090316, 17:40  #9  
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}×3×641 Posts 
Quote:
And, yes, you have to use only parallel circles (which will have smaller and smaller radii as you go north or south, until reaching zero at 90 degrees above or below the Sun), or you'd be doublecounting the (individually infinitesimal, but significant in total) areas where they cross and overlap. Last fiddled with by cheesehead on 20090316 at 17:51 

20090316, 18:25  #10  
Jun 2003
7·167 Posts 
Quote:
The problem is that you can't legitimately reason from length to area in this way. Instead, consider an infinitesimally thin wedge bounded by two radii an infinitesimal angle apart. Just as before, it intersects with the smaller circle over about half its length, but because it's the thin end, that's only a quarter of its area. 

20090316, 22:33  #11  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
10253_{8} Posts 
Quote:
I know that my original statement was wrong, and I think I get why (thanks for the example Mr. P1), but it's a bit of a brain twister. 

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