area of a sphere

tony873004

Can I use degrees as a unit like inches or meters for the purpose of computing the total area of the sky?

I want to figure out what percentage of the sky the full Moon takes up. Can someone verify my logic?:

There's 360 degrees in a circle, so the circumference of the sky (both the half we see and the half that's set) is 360 degrees. Therefore, the radius of the sky is 360 / 2pi = 57.2957795130824 degrees. (This is the part that seems strange to me, using degrees as a unit for radius, which is a straight line). So the entire sky is 4 pi r^2 = 4 * pi * 57.2957795130824^2 = 41252.9612494193 square degrees.

The Moon subtends 1/2 degree in the sky. So the radius of the Moon is 1/4 degree. So the Moon occupies pi r^2 = pi * 0.25^2 = 0.196349540849362 square degrees.

Therefore the Moon takes up 0.196349540849362 / 41252.9612494193 = 4.75964718416732E-06, or (multiply by 100 for percentage) ~0.0005 %. Since the sky we see is only 1/2 the total sky, the Moon occupies ~0.001% of the visible sky. Seems a little small.

mathmike

circumfrance is 2 * pi * r

how did u get 360 deg

moose

It's then relative area, since if you get closer it will change.....

tony873004

Circumferance is a full great circle. 360 degrees is a full circle. So I get radius starting out with circumference and use the formula you list to solve for r.



The sky is at infinity. But unlike expressing the area of a sphere in inches or meters, degrees also get smaller as you get closer. So even if I stood in the middle of a sphere with radius 100 meters, the sphere should still have the same area in square degrees as a sphere with radius 200 meters, or radius infinity.

HallsofIvy

The surface area of a sphere of radius one is [itex]4\pi[/itex]. The appropriate measure for 3-dimensional angles is the "steradian" (combination of "stereo" and "radian") defined so that the entire spherical angle has measure [itex]4\pi[/itex] steradians. One octant, kind of like a "right angle" would have measure
[tex]\frac{4\pi}{8}= \frac{\pi}{2}[/tex]
steradians.

SpaceTiger

It's common to underguesstimate the angular size of the moon because it's usually the only sizable thing in the sky.

As best I can tell, your calculations are correct, units and all.

tony873004

I never knew about the steradian. Thanks, HallsofIvy. Actually, maybe I've seen it before in a topo atlas that divided California up into things almost square-shaped. I think the lines of latitude were straight, but lines of longitude were not quite parallel, closer at top. I forget what they called it. Is this a steridan? How big would the Moon be in steradians? (1/2 degree).

ST, even though you say my calcs and units are correct, do you mean that I arrived at the correct answer, despite some creative math, or is it correct to say "square degree". In astronomy, you often hear stuff like "this picture is 3 degrees by 2 degrees." But is it correct to say this picture covers "6 square degrees of sky?" I'm guessing not. Although this term may be used in everyday talk, I'm guessing that it would be more correct to refer to a visual area sky in steridans rather than square degrees.

SpaceTiger

To my knowledge, lines of longitude are never parallel. Really these units are just a way of defining an angular sphere. It's the same as with the physical area of a sphere, but not scaled to a particular length. For example, the upper half of a balloon would, as measured from its center, subtend the same number of steradians as the upper half of the earth (as measured from its center), but would have a dramatically different surface area.


Just take the fraction of the sky taken up by the moon (you already calculated it) and multiply by the total number of steradians in the sky ([itex]4\pi[/itex]).


Square degrees are a perfectly acceptable and standard unit. You should check it out on google or wikipedia sometime.


No, but not because there's something wrong with the units. You have to keep in mind that the picture is "projected" onto a flat surface, but the sky itself is not flat. I could just as easily take a giant image of the entire night sky and project it onto a piece of paper. We know that the night sky is half of a sphere, [itex]2\pi[/itex] steradians, but if I were to simply calculate the apparent angular area of the projected image (it would look like a circle), then I would find:

[tex]\pi r^2 = \pi * (\frac{\pi}{2})^2 = \frac{\pi^3}{4}[/tex]

That's not the same as [itex]2\pi[/itex]. If the angles involved are small, however, it turns out this calculation will give you an answer that's nearly right (and good enough for much of astronomy). Why? Well, it's basically the same reason that I can calculate the area of my room by multiplying the lengths of the sides. Despite the fact that it sits on a curved surface (the earth), the radius of curvature is much larger than the dimensions of my calculation and the region is, for all intents and purposes, flat.


Nope, both are equally correct and commonly used by astronomers, but I think it could be safely argued that radians and steradians carry more mathematical elegance. This is probably why HallsofIvy was advocating their use.

George Jones

I'm guessing that you're refering to people who are looking directly at the moon. When the moon isn't in the sky, I find the opposite usually happens.

On moonless nights, I often ask people to look at the pointer stars of the Big Dipper and estimate how many full moons (or suns) would fit between them. Because they overguestimate the size of the moon, most people underestimate this by about a factor of 2.

Regards,
George

SpaceTiger

Haha, I was really out of it when I wrote that first post (long night). I meant to say they overguesstimate it...which also makes more sense in this context.