[QUOTE=Leaffan]
In all due respect, either I’m missing something or you’re not explaining things in clear terms.
I have no idea what you mean by anything quoted above.

[/QUOTE]
This is going to be harder than I thought.
You said you didn’t understand anything you quoted so we’ll start at the beginning.
Force (denoted by the symobl F) is a push or pull that moves things. When I push my car I’m exerting a force on the car.
Acceleration (denoted by the symbol a) is the rate of change in velocity as time passes. Velocity (denoted by the symbol v) is the rate of change of distance traveled in a particular direction as time passes. That is, if you change either the rate at which you are covering ground or the direction you are going, or both, you have changed your velocity or accelerated.
Mass (denoted by the symbol M) is the resistance of a physical object to acceleration.
It turns out that these three things are connected by a simple equation F = Ma which is Newton’s third law.
By a bit of mathematical legerdermain we can rewrite Newton’s third law as a = F/M. (I think in the post you quoted I wrote M/F and I’m sorry about that.) As I said, the F is the net force that actually results in acceleration. On launch there is a drag force that is low at first because the velocity is low. This drag builds up rapidly with velocity in an atmosphere of constant density. As I said, the launch is vertical and the rocket’s path stays vertical to get out of the densest atmosphere as soon as possible. The drag is low at first because the velocity is low. It builds up to some maximum and then decreases as the atmosphere thins and eventually falls to essentially zero in low earth orbit.
Call the rocket thrust force F[sub]r[/sub] and the drag force f[sub]d[/sub]. The net force on launch is the rocket thrust, F[sub]r[/sub] - F[sub]d[/sub].
That makes the equation for the acceleration on launch a = (F[sub]r[/sub] - F[sub]d[/sub])/M
In a vacuum there is no drag force so the equation for acceleration is merely a = F[sub]r[/sub]/M.
The launch trajectory is configured so as to reduce F[sub]d[/sub] to a very low number as rapidly as possible and then the two equations are essentially equal to each other.
So it takes nearly the same amount of fuel to slow the shuttle down as it now takes to speed it up. And, as I said, if it now takes 4,474,000 lb. of fuel to put it in orbit from a standing start it might easily take 4,400.000 to slow it to 2000 mph. That means that we have to lift 4,400,000 lb. into orbit and we just can’t do that with present technology (I’m pretty sure.)