The REAL flight path is (for the most part) parallel to the flat earth. You would probably not want to be on a plane whose flight path was perpendicular to the flat earth. In any case, in terms of the FICTITIOUS heliocentric flight path, the issue under discussion was the latitude-dependent tangential velocity resulting from the earth's (alleged) rotation, not the (alleged) curvature of the earth per se. It's a pity that you find this so confusing.
I suggest you go back and read my post #909 which clearly explains how your example also inadvertently provides the case against a flat earth.
Since the story now involves flying parallel to a flat earth let’s look at the science and maths here and consider a commercial aircraft cruising at an altitude of 10,000 m.
In the case of the aircraft flying around a spherical earth, the gravitational field has a spherical symmetry and the acceleration ‘a' is directed radially downwards and varies with height according to the formula;
where:
G is the gravitational constant (6.67430×10⁻¹¹ N⋅m²/kg²)
M is the mass of the Earth (5.972×10²⁴ kg)
R is the mean radius of the Earth (6.371×10⁶ m)
h is the altitude above the Earth's surface.
At 10,000 m altitude;
a = 9.75 m/s².
This result is confirmed by accelerometer measurements.
For a flat earth gravity does strange things, it is only at the centre of a flat earth the gravitational field accelerates objects vertically downwards.
Since flat earthers are not terribly helpful on the physical shape of the flat earth such as its thickness, we mathematicians have had to make some assumptions.
(1) The flat earth is in the shape of a disk of radius R and has an infinitesimally small thickness.
(2) The mass of the earth is uniformly distributed throughout the disk.
(3) The disk sits in the x-y plane and the aircraft on the z axis which passes through the centre of the disk, in other words it is directly above the flat earth’s centre at a vertical distance or height z.
The acceleration equation is:
The acceleration for the aircraft at 10,000 metres is found to be;
a = 19.6 m/s² nowhere near the value of 9.75 m/s².
Things get worse if that was possible when the aircraft is off-axis where the equation becomes horrendously more complicated and requires the use of
elliptic functions.
As one moves away from the centre, there is a horizontal component of acceleration which becomes stronger as the vertical component weakens.
At the edge acceleration is a purely horizontal and an accelerometer on the aircraft would read zero at the edge as it measures acceleration purely in the radial (vertical) direction.
Gravity is yet another example which destroys this flat earth nonsense.