What is the correct way to calculate the area of the sky in square degrees?

In summary, the person is trying to explain how the area of the sky is calculated. They say that the area is 360*180 square degrees, but this is not the correct answer because the images might be square, but that doesn't mean the sky is. Furthermore, the area of Earth is smaller than 40 000 km x 20 000 km.
  • #1
Phys12
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TL;DR Summary
How do you measure the area of the sky? Why is the answer not 360*180 square degrees?
Most of the calculations that I have seen that measure the area of the Sky involve doing this:

2*pi*r = 360. => r = 57.295 degrees. And then 4*pi*(57.295)^2 = 41251.83 square degrees. Now the units check out fine, but here are the places where I am having trouble understanding this derivation:
1) When you talk about the formula 2*pi*r = circumference, r and circumference are usually in length. And this makes sense because if you write pi=circumference/2*r, then it's apparent that they should be since the definition of pi is the length of circumference by the diameter. But what exactly is your r if you have written your circumference in units of angle (like degrees/arcminutes)? Moreover, how can you then plug that into an equation that takes length 4*pi*r^2 to get the surface area?

2) I would naively expect the area of the sky to be 360*180 square degrees. Because if you look at the HUDP, say, the angular area would be 1/20deg * 1/20deg since each size of the square enclosing the image is 1/20 deg. Now if I were to extend this idea to the entire sky, I imagine drawing a circle around the sky of 360 and then rotating it by 180 degrees to make a sphere of it and get the angular area of the sky in square degrees. But this comes out to be 360*180 = 64800. Why is that not the correct answer?
 
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  • #2
Do you know what a steradian is?
 
  • #3
Phys12 said:
2) I would naively expect the area of the sky to be 360*180 square degrees. Because if you look at the HUDP, say, the angular area would be 1/20deg * 1/20deg since each size of the square enclosing the image is 1/20 deg. Now if I were to extend this idea to the entire sky, I imagine drawing a circle around the sky of 360 and then rotating it by 180 degrees to make a sphere of it and get the angular area of the sky in square degrees. But this comes out to be 360*180 = 64800. Why is that not the correct answer?
The images might be square, but that doesn't mean the sky is. It should be pretty clear that if you go outside and look around, the sky is a projected sphere. I'm not sure how that could be confusing, but if it is, look straight up and spin yourself 360 degrees. How many blocks of 1/20th degree do you see at the zenith? (answer: 1).
 
  • #4
On a globe, you have lines of latitude and longitude. If you do the observation at or near the equator,(i.e. from the center of the globe, looking out to the equator), you have ## dA=R^2 \, d \theta \, d \phi ##, (angles measured in radians). Otherwise ## dA=R^2 \sin{\theta} \, d \theta \, d \phi ##, where ## \theta ## is the polar angle, with ## \theta=0 ## at the north pole.
I think this ## \sin{\theta} ## term might be the source of the puzzle. It is a result of the spacing of the longitude lines, so that ## ds=R \sin{\theta} \, d \phi ## for the distance traveled when moving along a latitude line.
 
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  • #5
Take, for example, area of Earth.
The same naive argument about 360x180 degrees would suggest Earth´ s area as 40 000 km x 20 000 km.
However, while Equator is 40 000 km long, the other parallels, even though they also cross 360 degrees of longitude, are smaller, until they become quite small circles around poles.
So the area of Earth is smaller than 40 000 km x 20 000 km, but what exactly?

If you put a sphere into a barrel and compare the surface of sphere to the side surface of the barrel, they turn out to be equal. Every latitude away from equator is tilted relative to the barrel sides, so the true surface of a latitudinal band exposed to sky is bigger than the surface turned towards the barrel would be; but the latitude parallels are also shorter. Those two effects cancel exactly, so the small circles along the poles are precisely equal in area to long but narrow bands along the sides of the barrel.
 
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  • #6
Do the top and bottom objects look to be the same area?
1599683962117.png

If you took a sheet of paper the size of the top rectangle and tried to fold it to make the bottom shape, what do you think would happen to the paper?
 

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  • #7
For an easy 3D explanation, look at a Globe (a model Earth) and notice how the shapes formed by the lines of latitude and longitude are 'almost' trapeziums (slightly curved sides) . They correspond to (say) ten degrees steps in angle but the ones near the Poles are a lot 'thinner, although they are the same height at all latitudes. Maps on flat paper are 'projections' and they either get the directions right and the areas wrong or the areas right and the directions wrong.

A noddy explanation with no diagrams and no maths.
 
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  • #8
Or, even simpler, this is what one "square" degree looks like near the poles.
1599690054495.png


A degree of latitude is always about 67 miles "tall" (24,900/360).
A degree of longitude is 67 miles "wide" only at the equator; it's 0 miles "wide" at the poles.
 
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  • #9
The 180 units are measured from north pole to south pole=over the whole sphere=referring to post 6, also in reference to post 5.
 
  • #10
Thanks for all your responses! I have the answer to my 2nd question, my naive assumption was grossly incorrect. But what about my first point?
Vanadium 50 said:
Do you know what a steradian is?
Vaguely, let me look up more on that and it might solve my problem. Thanks!
 
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  • #11
Yes, the solid angles you need are measured in steradians. Solid angle ## \Omega=A/r^2 ##. The solid angle of the whole sphere is ## 4 \pi ## steradians.
In the direction of the equator, you do have ## \Delta \Omega=(\Delta \theta )(\Delta \phi ) ##. See post 4.
Essentially, you can set up coordinates so that viewing overhead has ##\Delta \theta ## and ##\Delta \phi ##, but it doesn't work for a whole sphere, how you tried to do. One of the angles needs to be referenced to longitudinal type lines, and somewhere you need to have poles.
 
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  • #12
To understand the calculation of the area of the sky in square degrees, don't think of the circumference of the sky as 360 degrees. Think of it as 360 feet instead. Then, as you point out, r = 57.296 feet and the area of the sphere is 41252.96 square feet. In other words, don't think of "360 degrees" as an angular measurement, but rather as a unit of length around the circumference.
 

1. How do scientists measure the area of the sky?

Scientists use a variety of tools and techniques to measure the area of the sky, including telescopes, cameras, and computer software. They also take into account factors such as atmospheric distortion and light pollution when making their measurements.

2. Why is it important to measure the area of the sky?

Measuring the area of the sky allows scientists to accurately map and understand the distribution of stars, galaxies, and other celestial objects. It also helps in studying the evolution of the universe and identifying patterns and structures within it.

3. How do scientists account for the curvature of the Earth when measuring the sky?

To account for the curvature of the Earth, scientists use specialized techniques such as spherical trigonometry and spherical geometry. These methods allow them to accurately measure the positions and distances of celestial objects in the sky.

4. Can the area of the sky change over time?

Yes, the area of the sky can change over time due to the movement of celestial objects and the expansion of the universe. Scientists continuously update their measurements to account for these changes and gain a more accurate understanding of the sky.

5. How do scientists measure the area of the sky in different wavelengths of light?

Scientists use different types of telescopes and instruments to observe the sky in different wavelengths of light, such as infrared, ultraviolet, and radio waves. They then combine these observations to create a comprehensive map of the sky in all wavelengths.

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