LorgSkyegon

19th Feb 2021

Apollo 13 (1995)

Corrected entry: As the Apollo 13 Command, Service and Lunar Modules near the Earth they make a comment about "shallowing" and this is due to a "couple hundred pounds of moon rocks" that the crew failed to collect. In reality, this configuration (CM, SM and LM) does not exist on return to the earth as the LM descent stage is left on the moon and the ascent stage is jettisoned after the crew returns from the moons surface. They may be shallowing but it's not due to moon rocks, it's due to the LM (5000-10000 lbs).

Correction: They've already accounted for the weight of the LEM. But they took the prescribed weight of the other modules for the return trip and forgot that they have previously included the weight of the rocks in that weight.

LorgSkyegon

4th Jan 2020

Apollo 13 (1995)

Corrected entry: When the crew are on their way back to Earth, Fido states "Flight. We're still shallowing up a bit in the re-entry corridor. It's almost like they're under weight." After some discussion in Mission Control, the crew is advised "We gotta get the weight right. We were expecting you to be toting a couple of hundred pounds of moonrocks." The amount of propellant in the LEM (since they didn't land, or return to lunar orbit) plus the fact that the LEM still had the descent stage attached, which, if they had landed would have been left on the moon would have more than made up for the weight of the rocks they were supposed to be carrying. But far, far worse than that is the fact that how much they weigh makes no difference at all, as Commander David Scott proved on Apollo 15 when he showed that a hammer and a feather dropped at the same time on the moon land at the same exact instant proving Galileo's law of gravity, that all objects fall at the same speed regardless of mass. (01:53:30)

Correction: That refers solely to things falling by means of gravity without air resistance. An object with a larger mass will require more force to move. Under the principle of F=M*A, it is also true that A=F/M.

LorgSkyegon

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