In textbooks it is said that it would be better to launch a spacecraft towards the east to take benefit from the Earth's self-rotation. However, in TV we see that all rockets are launched vertically. It seems that it is because of technical difficulty in launching a rocket non-vertically. But is it true that after getting to some height, the rockets will turn their directions to the east?
[Physics] Should a spacecraft be launched towards the East
rocket-science
Related Solutions
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.
When launching into a low Earth orbit only your velocity relative to the Earth matters, as seen from the not-rotating reference frame of the Earth. Your velocity relative to the sun does not matter, because once you are in the orbit your velocity vector relative to the Earth will oscillate between pointing towards and away from the velocity vector of the Earth relative to the sun.
When performing an interplanetary transfer the Earth's velocity does matter. Usually such transfer is performed when in low Earth orbit. So if you want to travel to space outside Earth's orbit, then you want to leave Earth's "gravity" in the same direction as its velocity relative to the sun, also called prograde. But because the Earth will also slightly curve your escape trajectory you will have to burn while near trailing side of the Earth (where the sun is setting) such that you pass behind Earth's night side. The opposite is true when you want to go to space inside Earth's orbit.
Best Answer
I think you are mis-understanding. It dosn't matter which way you launch; it matters which way you orbit. No matter what direction you launch you still have Earth's eastward rotational energy. The reason you launch vertically is you want to get out of the thickest part of the atmosphere as quickly as possible to reduce your atmospheric drag; the fastest way to do that is vertically. It is fairly complex calculus to decide exactly how quickly to pitch over.