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The diagram you show is the digital version of a drawing by someone with an Etch-a-Sketch: completely inaccurate. The diagram below is accurate, showing Pioneer 10 & 11 and Voyager 1 & 2 trajectories in a heliocentric, inertial reference frame, of course with the ecliptic N-S dimension collapsed. No retrograde, no dog-legs between planets. Every now ...


5

As $\Delta v$ is just change in velocity, we can just integrate the norm of the acceleration function over time: $$\Delta v = \int|\mathbf{a}(t)| dt$$ You're out of luck getting a closed form of that integral though. As far as analytical solutions goes, we can note that at $t = \frac{\pi}{2}$, all of $a_x$, $a_y$ and $a_z$ are maxed out, and hence $\Delta v &...


4

Converting between different frames of reference is pretty straight forward. You subtract the velocities. That's it. If you have the spacecraft's velocity relative to the Earth, you can just subtract the Moon's velocity relative to Earth. Then you have the spacecraft's velocity relative to the Moon. In the case of a lunar transfer orbit, the spacecraft is ...


3

Transfer Regardless of how escape from the Earth system is achieved, orbital mechanics pose some restrictions on travel time. The minimum velocity transfer possible is an elliptical transfer orbit touching Earth's orbit when closest to the Sun, and Jupiter's orbit when farthest from the Sun. The transfer time is then 940 to 1060 days, depending on where in ...


2

$$v_{in} = v_{out}$$ That's the rule when entering a single body system. You are going to leave with exactly as much velocity as you entered with. There are two notable exceptions: More than one body in the system. The flybys of both bodies can very well change how much velocity you leave with. Earth-Moon is such a system. Venus is not. Very small ...


2

The calculation uses the following model for "total propulsive delta-v": $$\Delta v_{total} = \Delta v_{spacecraft} + \Delta v_{launcher}$$ Here, $\Delta v_{spacecraft}$ is what propulsive capabilities the probe has by itself after leaving the Earth system entirely, and is presumed to be a known value that can be looked up. $\Delta v_{launcher}$ is ...


1

While the overall scope is too broad, let me address the "low energy transfer" part. Firstly, the idea of "gradually increasing apogee" doesn't save you any fuel. Chandrayaan-2 did that because of limited thrust. The only saving to be had here is picking an engine with a slightly lower mass. Secondly, a "ballistic capture", as ...


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