Artemis 2 launch - humans return to the Moon after 54 years

[Charles Link]

and a follow-on to post 90: It appears the Nasa engineers really had to thread the needle on this one to get the rocket to do what it did. It is basically a textbook scattering problem, sending the rocket in the vicinity of the moon, but of all the possible trajectories, they needed to have one that looped around so the rocket could make its way back to earth. They managed to get what is an almost symmetric hyperbola, but there are many, many hyperbolas with very different trajectories.

 

[PeterDonis]

It appears the Nasa engineers really had to thread the needle on this one to get the rocket to do what it did.

There's a fairly narrow band of trajectories that will give a free return to Earth, yes. They plan for mid-course correction burns to make sure the spacecraft stays within that band. Note, though, that doing so was possible even with late 1960s computers and technology; the first few Apollo missions that went to the moon were on free return trajectories (in case an issue arose that would prevent them from inserting into lunar orbit), and Apollo 13 was put on one after the mishap.

 

[Charles Link]

One thing that comes out of the scattering theory is the scattering (in the rest frame of the moon) makes two asymptotes, (incoming and outgoing), where the total scattering angle is $2 \arctan(GM/(v_o^2 b ))$ where $v_o$ is the asymptotic speed and $b$ is the impact parameter which comes about with the conservation of angular momentum. I had no luck in showing though why this scattering angle is such a simple expression, but I was able to show that the scattering angle would simply depend upon some form of $GM/(v_o^2 b )$, from the differential expression involving the energy and angular momentum.

Note: M=mass of moon.

and to expound upon this, the complete scattering angle comes out of the solution in polar coordinates $r=r_o/(1+\epsilon \sin{\theta})$ of the differential equation, where the asymptotes are easily determined. I thought there should be a good reason for this that might show up in x-y coordinates, but so far it has eluded me. Note: $\epsilon=\sqrt{1+(v_o^2 b/GM)^2}$.