DC426913, thank you for those renderings. wish I could reply in kind.
1) Could we agree, that the first one, # 674 is correct in showing the strength of the gravitational field (or energy required to overcome it) in the empty inside of the shell? because it has to be at the same level as at infinity, O. and not at Y.
No, we cannot. Do you understand
what the Y axis on this graph
is?
It is not net gravity. It is gravitational
potential.
At infinity, its value is zero and at the bottom of the well, its negative Y.
Here - I've made it a little more intuitive for you.
I've also given the shell a non-zero thickness (your "gravity wall" is not vertical).
So:
To rise out of the well, rocket R must burn fuel, increasing its potential energy (climbing
up the graph).
If it turns its engine off, it will fall back into the well - which trades potential energy for kinetic energy as it falls (gaining speed).
According to your (flawed) diagram, the rocket would need to burn fuel to go
into the shell (rise out of the well), increasing its potential energy.
If the rocket were inside the shell, coasting toward the wall, it would suddenly fall down a
very steep slope, converting its potential energy to kinetic energy. If your diagram were correct,
the rocket would come out of the hollow and be instantly accelerated to escape velocity (by converting all that potential energy to kinetic energy).
Essentially, your graph turns any hollow shell into a
rocket launcher that uses no energy.
So let's correct a few of those misunderstandings before moving to anything more complex.