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 ($e \to \infty$).
4.Without air resistance, how much energy would be required to get a given mass to this altitude?
Energy Needed ($E$) = Potential Energy at L1 ($V_{L1}$) - Potential Energy at Earth's Surface ($V_e$)
$$V_e = -Gm(\frac{M_e}{r_e} +\frac{M_l}{LD - r_e}) $$
$$V_{L1} = -Gm(\frac{M_e}{d_{L1}} +\frac{M_l}{LD - d_{L1}}) $$
where $m$ is the transported mass, $M_e$ is Earth's mass, $M_l$ is the Moon's mass, $LD$ is the center to center Earth-Moon distance, $r_e$ is Earth's radius, and $d_{L1}$ is the distance from the center of the Earth to L1.
5.What does the formula for this look like (getting to a specified altitude of a given astronomical body)?
The formula above it for a two body system, along the center line of such a system. You could replace $d_{L1}$ with a smaller distance if you want to go less than all the way to L1 along this line.
6.If you were to launch from, say, 100km up with a few weather balloons, how would the formula need to be adjusted?
In the formula for $V_e$ substitute $r_e + 100km$ for $r_e$
7.What type and how many rocket engines would be required to get a tiny model rocket to this altitude, regardless of feasibility?
This subquestion doesn't have a specific answer. Even if a zero mass payload is assumed, the propellant and structure holding the propellant have mass. The chemical nature and mass of the propellant, mass of the inert sturcture, and arrangement of stages would be major factors. Multiple stages are advantageous to reduce the mass during flight by eliminating no longer needed inert sturcture.
8.Other than the issues of landing safely, would this be a more efficient way to transport supplies to the moon if they could survive the crash landing?
If fuel is not used to make a soft landing on the Moon, this would definitely increase efficiency.
Addtional considerations:
None of the above considers the gravitational potential of the Sun. If you don't mind crashing into the Moon, launching when the Moon is between the Earth and the Sun would minimize the amount of energy needed. This would add a third term involving the Sun's mass and distances to the Sun to each of the potential energy equations.
As previously suggested by user "I like Serena", since Earth is rotating about its own axis, a rocket launched from Earth will have an initial velocity component due to this rotation. It is optimal to launch from near the equator to take advantage of the maximum velocity due to rotation, as well as greater Earth diameter/less gravity. Launch should be timed such that the rotational velocity component is directed to the Moon as much as possible, at which time the direction of the Moon will be generally eastward.
The L1 point is calculated considering gravitational potential and centrifugal force of a body in the rotating frame. The point of maximum gravitational potential along a line joining the Earth and Moon would be a somewhat different point. It would be more correct to find the maximum gravitation potential along this line and use that potential energy value, although it should be similar to the potential energy to get to L1.
For more information on low energy transport to the moon, without using Hohmann transfer, and without crash landing, see Low Energy Transfer to the Moon. An alternative low-energy approach was used by the 1991 Japanese Hiten mission.
Best Answer
For an object in low earth orbit (at 100+ miles above the earth's surface) the speed needed is about 17,000 miles per hour. Even if a trebuchet could achieve that speed on the earth's surface, you would have at least three problems:
The object would IMMEDIATELY burn up in our dense atmosphere. Think about the space shuttle which is going at orbital speed when it encounters the very tenuous atmosphere at very high altitudes. It needs special heat resistant ceramic tiles due to the heating caused by a very tenuous atmosphere. If the angle at which the first encounter the atmosphere were too steep it would completely incinerate. So there is no material that you could use to build the satellite that would prevent it from immediately burning up.
If you could magically make all the atmosphere disappear, you still could not launch a satellite with a trebuchet from the surface of the earth. Well you could, but it would only complete less than one orbit. If you got the right speed, it would start out on a nice elliptical orbit, but the ellipse would bring you back to the launch point coming up through the crust of the earth. In other words the ellipse will pass through the earth such that in less than one orbit you will impact the earth's surface again. To successfully launch, the satellite would need to have some kind of rocket motor onboard so that once it got to an appropriate altitude, it could change the velocity direction to be in an orbit that doesn't intersect the surface of the earth.
The last problem that will make this Trebuchet impossible is the mass and required strength of the arm that will connect the heavy weight to the pivot point to the satellite. I suspect that making this arm strong enough will make it too heavy to work. So, for now let's assume the arm has zero mass and infinite strength. Then if we assume the heavy weight falls in say, about 1 second at about 1G, then to get the satellite to 17,000 miles per hour, the acceleration of the satellite would have to be 25,000 ft/sec^2 which means it would accelerate at 780Gs (so humans would be killed for sure). That would mean that the length of the arm to the satellite would have to be 780 times longer than the short arm to the heavy weight. So if the short arm were 10 feet, the long arm would have to be 7,800 feet which is 1.5 miles. I think you can see that the arm requirements would make this totally impractical if not impossible. For this to even work, the heavy weight would have to be greater than mass of the satellite times the long arm length divided by the short arm length by a very large factor (to insure the heavy weight falls at about 1G). If we assume a 100kg satellite, then in this case that means the heavy weight would have to be something like 10 or 100 times (7800/10)*100 kg - thus something like 780,000kg to 7,800,000kg. Imagine the strength of the arm that is required. Then think about how heavy the arm would be and how that would make all of these requirements even more impossible since a heavy arm would greatly decrease the acceleration of the satellite.
So, no it CANNOT be done...