Under this section of FW Wiki it states the following:
The apparent view of rising and setting are caused by perspective, just as a flock of birds overhead will descend into the horizon as they fly into the distance.
Same principle can be applied to an aircraft. Lets say it is flying level at 35,000ft. If seen passing through the observers zenith or overhead point, the distance between the observer and the aircraft will be as stated, 35,000ft will it not. To the pilot, situated 35,000 ft directly above the observer, the horizon is 229.3 miles away according to ringbell.co.uk. To the observer on the ground however, lets say they are 6ft tall, the visible horizon to them is only 3 miles away.
If RE is correct and the aircraft is flying level w.r.t the surface (constant height) then the aircraft will follow a curved path which remains parallel with the circumference of the Earth. That would mean I would have thought from the observers point of view as the angle between the aircraft and the observer changes, so it would appear to sink lower and lower in the sky until it meets the horizon. The higher the aircraft the longer it remains in the observers line of sight.
If the Earth is flat then I would suggest for a given height the aircraft would remain in view for a much longer time than it does in the real world because the path of the aircraft and the surface of the Earth would be straight, parallel lines and so in effect the aircraft would
never actually reach the horizon let alone disappear below it. Only the limits of the transparency of the air would limit the visibility of the aircraft.
How long does the ISS remain visible from the same observation spot assuming it passes through the zenith?