Does that picture have an elevation and a location that it was shot at?

I am not computer smart enough to access it if it had. maybe you are. I am old.

If we can find those then maybe we can figure out how long that horizon is.

At 8 in drop per mile then 10 miles of horizon should have a 6.66ft drop. (number of the beast)

I would think that the horizon in the photo is at least 100 miles long.

If it is then that is a 66.6ft difference.

We can estimate it pretty easily. The horizon is 950 pixels long, with 6 pixels of drop on either side. That picture appears to be from low Earth orbit, which means the distance to the horizon is about 1,400 miles.

Breaking out the simple geometry, you can find the radius of an arc based on height and width via:

r = H / 2 + W^2 / 8H

Plugging in my pixel-counting values:

r = 6p / 2 + (950p)^2 / 8(6p) = 18,805 p

This lets us turn pixels into distance since we have the radius in both values:

1,400 mi / 18,805 p = length / 950 p

Length = 70 miles

We should then see about 815 ft of curvature on either side of the picture away from the center given that both are 35 miles away from the middle. 815ft / (70mi*5280ft/mi) = 0.2% of the total distance.

That's why it's barely visible.