My question is during the initial deceleration, is that simply applied
opposite the current orbital velocity vector? And when the satellite
arrives at the new orbit, what direction is the correctional
acceleration/deceleration made? Is it the difference of the desired
tangential direction that it wants to go and the current direction it
is going? or simply opposite its current vector?
The two directions are the same. The impulses are done at the apoapsis and the periapsis of the transfer orbit, where the velocity is also tangential to the circular orbits (the larger one at apoapsis and the smaller at periapsis).
I understand that the point of the Hohmann transfer orbits is to use
as little velocity change (and thus fuel) as possible, but can an
orbital transfer be faster or slower if you're willing to burn more
fuel to affect the velocity change?
[snip]
Finally, if you had unlimited fuel, and time was more a consideration,
would you even bother with this process versus something more "point
and shoot, turn and burn"?
If you have enough delta-v and thrust, you can just "point and burn". You will be doing a hyperbolic transfer orbit that, in the limit of very large velocities, will tend to a straight line ($e \to \infty$).
It appears you made a few mistakes.
The formula $E_P = mgh$ is only an approximation for objects near the ground. The more complete formula is
$ E_P = -\frac{\mu m}{r} $
where $\mu = 398600.44$ is Earth's standard gravitational parameter, and $r$ is the distance between the object and the Earth's center of gravity.
Especially note the negative sign; this has to do with the definition of potential energy in the context of orbits. This is where I think you went wrong.
Also, where did you find $V = 11.068$ km/s for a geostationary orbit? That looks more like an escape speed than a normal orbital speed...Indeed, if you look up the altitude for a geostationary orbit you see that it is $35768$ km above the equator. That means the total pathlength traversed by the satellite in one stellar day is
$2\pi \cdot ( 35768+R_E) \approx 264,811 $km
making the speed
$264,811 \mathrm{\ km} / 86164 \mathrm{\ seconds} \approx 3.07 \mathrm{\ km/s}$
so much much slower than the ~11 km/s you stated. Lumping all this together:
$E_P^{GEO} \approx -\frac{398600.44}{42164} = -9.45 $ kJ/kg
$E_K^{GEO} \approx \frac{3.07^2}{2} = 4.71 $ kJ/kg
$E_{tot}^{GEO} = 4.71-9.45 = -4.74 $ kJ/kg
while for the other orbit
$E_P^{alt} \approx -\frac{398600.44}{45000} = -8.86 $ kJ/kg
$E_K^{alt} \approx \frac{2.98^2}{2} = 4.44 $ kJ/kg
$E_{tot}^{alt} = 4.44-8.86 = -4.42$ kJ/kg
which is indeed higher than the GEO orbit.
This makes sense -- you need to input a lot more energy to let anything escape from Earth's gravity than, say, an apple falling to the ground (which is also in an "orbit", albeit one far closer to the Earth, and not exactly on an escape trajectory).
If what you say would be true, everything would simply fall up and escape the Earth. There are a few experiments that will show that that is not actually what happens :)
With regard to your statement about the moon: the moon is indeed slowly escaping from the Earth. The mechanism here is that the Moon is gaining orbital speed at the expense of Earth's rotational momentum, through tidal interaction.
Roughly translated: as Earth's rotation slows down, the Moon speeds up, making the Moon progress farther away from the Earth, towards a lower speed.
The total energy in that higher orbit is higher, because the drop in speed is disproportionally small in relation to the gain in potential energy. Eventually, after a few million years of repeating the above, the moon will have gained enough energy to escape the Earth and orbit the Sun on its own.
Best Answer
Consider this an orbit of a satellite.
Angular momentum $\vec{L}=\vec{r} \times \vec{p}$ where $\vec{r}$ and $\vec{p}$ are vectors.
Angular momentum is a conserved quantity. In a satellite with a decaying orbit due to atmospheric drag , as $r$ becomes smaller by the loss of energy, conservation of angular momentum means that $p$ has to increase and therefor the velocity, since $\vec{p}=m \vec{v}$.
It cannot fall straight down because conservation of angular momentum would be violated.