I do not understand what you are discussing.
This is what I know:
Newton:1 - If a body is not subject to any net external force it either remains at rest or continues in uniform motion.
Newton:2 - F = ma (Law of acceleration)
Newton:3 - When two particles interact, the force on one particle is equal to and opposite to the force on the other.
So, when dealing with real life stuff you must consider air resistance as a force that wants to slow things down per N:3. The force D working on the object is calculated by the drag equation:
D = 1/2(rho)CA(v^2)
C is the drag coefficient for that particular object. This is not constant but varies with its shape and size - and of course what kind of air flows around it (wind conditions etc).
The force T acting on a falling object in air would then be:
T = F+D
which is dependent on mass to some extent (greater size usually means greater mass).
As for the debate about varying gravitational forces, I do not understand what you are fuzzing about. The gravitational coefficient g is calculated like:
g = GM/R^2
The interesting part here is R^2 which means the radius squared. As the earth is elliptical and not a sphere, this value will vary, depending both on altitude and geographical position relative to the earth's core. Thus, the 'g' will vary across the earth. But it is decided for all practical engineering reasons that g=9,81 uniformly. (Actually sometimes engineers use g=10 just to build in extra margins of safety and make calculations easier.)
Tidal water again, is explained by the gravitational forces distributed between earth and moon. Newton summed it up in his Law of gravitation:
F = G(Mm)/r^2
The moon will therefore exert some gravitational force on that part of earth that is closest to it. Since land does not move that easily, water is the main reason we notice this mutual force between out space home and its moon. The water is "attracted" to the force and moves in the direction of the moon which causes tidal flows.