I've found that the area between ≈30° north and south contains exactly half of Earth's surface area. Is there an exact number out there?

Question

This doesn't just apply to Earth (I first discovered it when calculating the area of the Celestial Sphere), but to any perfect sphere.

Think of it this way: The area contained between 80°N and 90°N is far less than the area contained between 0°N and 10°N, despite both being bands 10° wide.

Anyway, I've found that the area inclosed between 29.85799272 degrees (those are the digits that I'm 1000% sure of, got them via desmos trial and error$\text{*}$) north and south of any sphere contains exactly half of the sphere's total surface area.

My question is: does this number have a nicer form, such as a fraction involving radicals? Or is it just a simple digit? And if so, has anyone else previously noticed this?

*I can't be arsed to list all the math I used to find this result, but I guarantee you it's 1000% accurate. Desmos only showed the first 3 digits, so I zoomed in as much as I can to see when the line denoting the area intersected y=50 (denoting 50%), and came to the aforementioned digit after trial-and-error trying other smaller digits. Just trust me lmfao I haven't entirely lost my sanity.

Also, I tried figuring this out in WolframAlpha but gave up after multiple attempts proving futile due to exceeding the character limit and computing time. So any help would be appreciated in trying to figure out if this value has a name or a nicer representation.

Answer 1

Some helpful formulae for curve surface areas on a perfect sphere with radius $R$ (not the Earth):

1. Each spherical / polar cap with $60^\circ$ polar angle: $2\pi R^2\left(1-\cos 60^\circ\right) = \pi R ^2$ ;

2. The spherical segment between $\pm 30^\circ$ latitudes, with distance $h = 2R\sin 30^\circ = R$ between them: $2\pi R\cdot h = 2\pi R^2$ ;

3. The whole sphere: $4\pi R^2$ .

So the required fraction of the surface area between $\pm 30^\circ$ latitudes is

$$1-\frac{2\cdot \overbrace{\pi R^2}^{\text{Cap}}}{4\pi R^2} = \frac{\overbrace{2\pi R^2}^{\text{Segment}}}{4\pi R^2} = \frac 12$$

Answer 2

The area of a spherical cap is linearly proportional to $H$ cap height.

$$ A= 2 \pi R H $$

The total H can be composed of or be the sum of any number of segments height $h_i$.

$$ A= 2 \pi R~ \Sigma h_i =2 \pi R H $$ This result can be found by integrating

$$ \int 2 \pi r ds $$ between required limits.

For a portion of the total area to be a half as in a hemisphere, the height should be i.e.,

$$ \frac12 4\pi R^2= (R) . {2\pi R}$$

The single spherical cap should have latitude $$ \sin^{-1} \frac{R/2}{R}= 30^{\circ}$$ exactly.The shaded area sum in each case is $ {2\pi R^2}.$