Surely, even Flat Earthers can understand it's not LITERALLY a pear right?
What he actually SAID was that it was "like pear-shaped" because "it's slightly wider below the equator than above the equator". Note how Flat Earth liars exclude and ignore the "like" and "slightly" parts.
So when you spin pizza dough it kind of flattens out, gets wider in the middle. So, Earth, throughout it's life, even when it formed it was spinning and it got a little wider at the equator than it does at the poles. So it's not actually a sphere, it's an.. it's oblate. Officially it's an oblate spheroid, that's what we call it. But not only that it's slightly wider below the equator than above the equator...
A little chubbier?
Yeah, cubbie is a good word, it's like pear-shaped.
Because the Earth is not a perfectly smooth ellipsoid the exact latitude of maximum actual circumference is slightly below the equator due to hills and mountains. Consider a latitude 1 mile south of the equator-the spheroid width at that point is just 1.3 meters smaller - so even a small hill, anywhere in between, could increase the actual circumference MORE THAN the reduction in circumference. That is what Neil deGrasse Tyson is referring to. The exact latitude of maximum circumference lies slightly below the exact equator due to simple terrain differences.
According to the WGS 84 (reference model for the shape of the Earth) the Earth's ellipsoidal approximation has an equatorial radius is 6378137 m (and the polar radius is 6356752.3 m, we will use this below).
Earth equatorial circumference is then given by:
C = 2π × 6378137
= 40075016.7 m [wolfram|alpha]
We can find our (spherical approximation) distance around at latitude from:
latitude circle = C × cos(λ)
= 40075016.7 × cos(0.014555°)
= 40075015.4 m [wolfram|alpha]
So that's a mere 1.3 meters smaller around (I also checked it by simply reducing the radius by 8 inches, the approximate "drop" over 1 mile). It is not at all improbable that the terrain could vary more than 1.3 meters over a 1 mile band below the equator. In fact, once you think about it, it seems extremely improbable that the largest geodetic circumference would be exactly at the equator.
Figure Of The Earth
For this exercise we're going to be looking at the 11000 x 11000 pixel image from the Japanese weather satellite Himawari 8 (and image found here) that orbits between 35782 km and 35789 km above the Earth. What we want to do is measure how many pixels wide the Earth is versus how many pixels high it is, and then compare that to our radial measurements as a ratio.
However, we're just not going to see see a 1.3 meter high bump 1 mile south of the equator in an 11000 pixel image -- (11000/(2 × 6378137)) = 0.00086 pixels per meter. There is no way this would be observable in a photograph, we're looking for the oblateness here.
The oblateness of the earth is very slight as we discussed above, 6378137 m versus 6356752.3 m - or a total difference of 21384.7 meters (or 1 part in 298.257223560), therefore we should expect something on the order of 37 pixels difference.
The ratio between the equatorial and polar radii is approximately 1.0034:
6378137 / 6356752.3 = 1.00336409207
As the image was taken in December the Northern-most section was very dark so had to play around with the image contrast and very carefully find the Northern most extent.
Doing this I found the dimensions to be 10896 x 10856, giving me a ratio of :
10896 x 10856 = 1.00368459838
So this means that each pixel is about (6378137 / 10896) = 585 meters across - so figure at least 3 pixels of error and you have about 1500 meters of accuracy here at best (zoom into the edge of the image and you'll likely find the probable error is even higher). Plus the atmosphere is blurring things out the edge and anti-aliasing smears that out even more. But we see roughly the amount of 'bulge' around the horizon we would expect.
I have zero interest in finding exactly which line of latitude has the greatest circumference but if you wanted to attack such a thing you would need to get the full resolution geoid data and let a supercomputer run the calculations for you, you'd to check about 161,000 paths per mile for 1 cm resolution and even that is only an estimate -- are trees supposed to count? Individual rocks? It's a pointless exercise. But at the same time, if anyone does it please post your results :)
You cannot just look at a photograph and expect to pick out a difference of 1 part in 298 - you have to carefully measure it. We've done that here and found extremely good agreement with the model. The figure of the Earth in the Himawari 8 images, although it sure looks like a circle when viewed in whole, is measurable not a perfect circle, it is indeed wider in the middle.
And before you ask "ok, but where are the satellites", please read my blog post on that for the details but, in short, they are vastly too small to detect. No, really. I calculated it. Hint: they are about the size of a bus. Remember that pixel size, each pixel is showing 585 x 585 meters of space - you are NOT going to see something the size of a bus at that distance.
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