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Partial answer so far... For the trajectories in your drawing the objects will miss the Moon. They pass much closer to Earth, so their orbital motion will be substantially faster than the Moon's so they will pass through the interception points days before the Moon does and definitely miss it. However if those are elliptical orbits with the same semi-...


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Community Wiki answer: I've incorporated "translations" from existing comments on the question; feel free to edit/improve: 1. RTCC SS1 and SS2 RTCC = real time computer complex; the Mission Control computer system 2. MSFN free flight trajectory MSFN = Manned Space Flight Network, the network of ground tracking stations. 3. Langley model Langley would ...


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What the heck are "space-fixed coordinates"? To what in space can a coordinate system be fixed? That's two questions. The answer to the first is that those are Earth-centered inertial or Moon-centered inertial coordinates as indicated by the "Ref. body" column. Look at the velocities. 25600 ft/sec is orbital velocity for a vehicle in low Earth orbit while ...


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Yes! The good folks at AGI (Makers of STK) simulated the entire Apollo 11 trajectory in STK: "Based on original NASA documentation, our engineers have built a high-fidelity, eight-day digital mission in Systems Tool Kit (STK). They modeled precise trajectories using STK Astrogator" The web site has tutorials on how to do some of the simulation using STK/...


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To answer the question literally: you'd be looking for NASA Apollo Trajectory (NAT) data files. The report Apollo Mission 11, Trajectory Reconstruction and Postflight Analysis Volume 1 (PDF) provides a summary for Apollo 11 and mentions that the raw NAT data is available in Volume 2 of the report. I have yet to find Volume 2 though, perhaps because The ...


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What the other answers fail to mention is that the mass of your orbiting object actually cancels out. It does not matter. See these two equations: (1) F1 = F2 = G*m1*m2 / r^2 (2) F1 = m1 * a1 Where F is force, G is the universal gravitational constant, m is mass, and r is distance between centers of mass of the orbiting and orbited bodies in question. The ...


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The acceleration due to gravity will be identical regardless of mass, assuming the mass of your spacecraft is negligible compared to mass of the object you're orbiting. For example the Earths moon is large enough to effect the motion of the earth so it doesn't orbit the centre of the earth, but instead it orbits the shared centre of mass of the Earth and ...


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Approximately, yes. The gross gravitational effects on the trajectories of the spacecraft and the other object will be the same. The force of gravity between two objects is proportional to the product of their masses; by $F = m a$, the acceleration of each object cancels out its own mass ( $a = \frac {F} {m}$ ) and so depends on the mass of the other object....


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This paper by Dan Adamo states that their fate is unknown. Other Apollo Program hardware certainly accompanied some of the components cited here into interplanetary space. Unfortunately, there are no empirical data relating to these objects' trajectories. Likely the largest such undocumented disposed components are four spacecraft/LM adapter (...


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The four pieces of the SLA fairing would follow a similar path as the third stage S-IVB of the Saturn V. The SLA pieces were separated by explosive devices only, so their additional acceleration was very, very small. See apollomaniacs. After Wikipedia, the S-IVB stages of the missions 8 to 12 are in heliocentric orbits now and of the missions 13 to 17 were ...


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The full answer is beyond the scope of StackExchange. A quick Astrogator tutorial for earth departure to Mars orbit can be found here. Ultimately you will need to use b-plane targeting if you want to get anywhere close to a gravity assist. Even for just simple targetting you will need to know the rough departure asymptote (normally from a Pork Chop plot) ...


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