tl;dr: The chart is outdated and unsupported and we are operating deep into three-body territory where Kepler orbits don't address the reality and talking about geocentric $C_3=0$ doesn't even make sense.
fyi I've just added a bounty to A spacecraft leaves Earth exactly with escape velocity V2 - what trajectory it will have in Solar System?
I find that ...
For the purpose of $\Delta v$ planning, an orbit at Earth-Moon L4 or L5 is about identical to just any orbit with a radius of 380,000 km. At L4/5 distance to Earth and Moon are identical and as the gravitational force changes with $r^2$, the pull of the Moon is just 0.01% of that of Earth at this point and can be neglected. It's sufficient to keep the orbit ...
The simple theoretical delta V to achieve a particular orbit is constant, but in practice (or on more detailed analysis) Delta V is not constant for a number of reasons.
For launches from the surface of a moon or planet delta V will be
greater than the theoretical value because:
A rocket will not be able to achieve orbit instantly, it will need to
Another way of saying delta V to LEO = 10 km/s is this:
To be in orbit, a thing needs to move horizontally at a speed of at least 7.8 km/s
To get to orbit, the rocket delivering the thing will have to get up to that speed, and get out of the atmosphere
While it does that, the drag from gravity and air resistance make it have to exert as much force as if it ...
This is only a partial answer, but there are two major factors that make the performance of the three vehicles not comparable at all.
There are two reasons why Starship is so much worse for missions with a large C3, and both are design features:
First, it's built to land (repeatedly) on planets. This makes it a lot more sturdy than other second stages. The ...
To enjoy the greatest advantage from the Earth's rotation, you want two things:
Launch from as close to the equator as possible.
Launch as close to directly east as possible.
The first part is fairly simple. The rotational velocity is proportional to $\cos(latitude)$, so the Guiana Space Centre gets over 99% of the effect, Kennedy Space Centre around 88%, ...