Some photos in this article at Ars technia:
Brian
Yeah, that issue with the head reminds us, because of the expense, budget precariousness and operational ponderousness of Artemis as a whole, they had to pack in all what in the 60s would have been unmanned system test flights in the one flight and all manned shakedown flights in the next. So no first test flight in LEO a-la Apollo 7 to test habitability, just go directly for lunar flyby as soon as we can.
Why we haven’t returned to the moon.
Yes, thanks, I finally figured it out. Essentially the indicated velocity supplied by NASA was the velocity relative to earth – how fast it was moving away or towards earth – and did not reflect its speed relative to the moon.
And NASA is also catering to public misconceptions when it refers to Orion’s speed as “velocity”. Velocity is a vector quantity and strictly speaking Orion’s velocity should currently be of the opposite sign of its departing velocity. But since the public doesn’t care except about how fast it’s going, they should properly just call it “speed”.
Yup. The reading declined to zero at apogee from Earth then began increasing again. It was just the relative earth ↔ capsule distance rate of change AKA radial speed and nothing more.
Dumb data for a dumb public.
It does not have to be too complicated, but there is, nevertheless, some trigonometry worth going through in order to understand the capsule’s velocity (or speed) with respect to the Earth at any given moment. As a first approximation, you can consider the spacecraft’s orbit as an ellipse, so there is a simple relation between the speed and the distance from the Earth at any given time. What about the Moon? What you can do is change coordinates when inside the Moon’s “sphere of influence”, and consider that you are in orbit around the Moon. The same relation will hold. If you want to know whether you are slowing down or speeding up with respect to the Earth, and by how much, you have to do som, I asumee vector addition.
Zero??? You mean the distance, not the orbital speed, I assume.
If I understand correctly and the reading was “speed relative to Earth” at the apogee it will always be zero, if it was more it would continue to get away from Earth, if it was less it will be falling back.
Yes, but, ignoring the Moon for a second 
if your velocity were zero, you would drop straight down towards the center of the Earth.
Not if “velocity” was measured in terms of moving away or towards the earth. Eccentric orbits aside, everything in a stable circular orbit around the earth has a velocity of zero (more or less) as defined by what NASA was telling us about Orion’s velocity.
The fact that Orion had a positive velocity (moving away from the earth) and now has a negative velocity as it returns means irrefutably that it had a zero velocity (as NASA was reporting it) at some point. That point was behind the far side of the moon so we never saw it.
You can see that this is false since if the velocity were zero, the object (assumed to be in a circular orbit) would hang in place and never complete an orbit!
The correct relation would be v=\sqrt{\mu/r} where \mu is about 4 × 1014 m3/s2
E.g. to stay at the Moon’s distance, let’s say 400000 km, you need to be going approximately 1 km/s along your orbit, at a right angle to the Earth.
We’re disagreeing on a point of semantics which is the basis of the whole argument about NASA’s misleading “velocity” number. If you define “velocity” as the rate at which an object is receding from – or approaching – the earth, or the gravitational center of the earth if you want to be more precise, then an object in a perfectly circular orbit has a velocity – by that definition – of precisely zero.
This was the whole basis of the misunderstanding of the NASA “velocity” stats for Orion. How fast it was moving away from the earth did not reflect how fast it was moving along other axes.
Jumping in here with this. Here is the moon and Artemis II on the Las Vegas Sphere. Quite well done!
➜ https://www.instagram.com/reel/DW0ZX9jgCl9 
In case you don’t have Instagram, here is a screen capture from it.
I get all that, but the “velocity” readout found, for instance, here: NASA: Artemis II appears to reflect the total speed along all axes, combined
I don’t know what that was supposed to prove, but all I got was a NASA logo, a black screen, and a crashed browser, and lost all my tabs which I’ve just recovered from the history log.
If that was supposed to refute what I said, it couldn’t have been accurate, as the stats showed Orion noticeably slowing as it came very close to the moon, which is like a baseball noticeably slowing as it falls to the ground.
Not velocity. Radial component of velocity. Distance change rate between center of Earth and center of capsule.
They have the same thing on Youtube, here: https://www.youtube.com/live/m3kR2KK8TEs
The “ground” is supposed to be the Earth, not the Moon, and it did not slow to zero.
Not what is on that feed. E.g, right now it says 1464 mph
Can you explain the magic by which an object moves away from the earth at high speed, slows down, and then begins falling back to earth, without ever going through a point of zero velocity with respect to the earth when it transitions from positive to negative velocity?
Because it didn’t lift off from the center of Earth. It lifted off from one side of the surface and then started going around the Earth.
Sure. First of all, note that there is a case when the velocity momentarily equals zero, namely, the degenerate case of on object radially moving away from the center of the Earth, reaching some zenith, and falling straight back down to the center (well, crashing into the surface). It is a good example, though.
More generally, consider an elliptical orbit with perigee radius r_p and apogee radius r_a. By conservation of energy, v^2/2-\mu/r is constant along the orbit. We want to know the velocity at apogee (or any other point). We have v_a^2/2-\mu/r_a = v_p^2/2-\mu/r_p and r_pv_p=r_av_a, which gets you \frac{v_a^2}{2}\bigl(1-\frac{r_a^2}{r_p^2}\bigr)=\mu\bigl(\frac{1}{r_a}-\frac{1}{r_p}\bigr). Doing a bit of algebra to solve for the velocity, \frac{v_a^2}{2}=\mu\cdot\frac{r_p}{r_a(r_p+r_a)}, or, in terms of the semi-major axis a=(r_p+r_a)/2, v_a^2/2 = \mu(1/r_a-1/2a).
In case r_a=2a, you would get a velocity of zero. However, let us plug in numbers for r_p = 6400 km and r_a = 400000 km. The velocity at apogee would then be 177 m/s. If you plot the trajectory, it looks like an ellipse with the Earth at one focus.
If you go the JPL Horizons web site, you can download the Artemis data, or just get a table online of the distance friom Earth, rate of change of the distance from Earth (which does go to zero at one point), and the velocity with respect to the Earth (which you can verify does not).
NB the Moon is going to have some effect on the speed compared to if it were not there while you are near it— imagine bouncing off a point mass— which was asked before.
Maybe this has been addressed, but y’all have GOT to see what happened when the POTUS called the astronauts. (TL : DW - they ignored him.)