The calculation with the moon's gravitational potential does indeed work in the rest frame of the moon. To compute it, we can to a good approximation simply ignore the effect of the earth when the rocket is near the moon. For initial conditions, we can put the rocket 6000 miles from the path of the moon, moving forward with ## v_o=1000 ## m.p.h. and the moon to the right a distance of 20,000 miles and moving right to left at a speed of ## v_m=2200 ## m.p.h. That makes it so the moon is a distance (measured from its center) of ## r_{mo}=21,000 ## miles. In the reference frame of the moon, the rocket is approaching the moon at a speed of ## v_{mo}=2400 ## m.p.h.
We have with conservation of energy ## \frac{mv_{m1}^2}{2}- \frac{GM_m m}{r_{m1}}=\frac{mv_{mo}^2}{2}-\frac{GM_m m}{r_{mo}} ##. Nasa gives that at closest approach ##v_{m1}= 3139 ## m.p.h. at 4067 miles from the surface, so that with a radius of the moon of about 1100 miles, ## r_{m1}=5,170 ## miles (at closest approach). Putting in the known constants, I get ## v_{m1} \approx 3200 ## m.p.h. ,(at closest approach), in good agreement with Nasa.
(It may be worth repeating what was mentioned in post 73 above that this conservation of energy expression is not valid if the speeds used are those measured in the reference frame of the earth. The results of doing the calculation in that manner are completely incorrect).
Since at closest approach the rocket is now moving approximately left to right, (at 3200 m.p.h. w.r.t. the moon, with the moon moving right to left at 2200 m.p.h. w.r.t. the earth), the speed of the rocket in the rest frame of the earth will still be ## v=1000 ## m.p.h. approximately. Thereby the speed of the rocket measured in the reference frame of the earth didn't change appreciably during the last part of the approach to the moon. The above calculation is in complete agreement with this. (Note: The direction changed, but the speed didn't).
It may be worth noting that to do the arithmetic above, I used ## v_{esc}=25,000 ## m.p.h. and ## v_{esc}^2=2GM_e/R_e ## with ## R_e=4000 ## miles, and ## M_m=M_e/81 ## approximately.
