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Yes, the dark side of the moon is mostly hidden from earthbound observers. This is due to tidal locking: A tidally locked body takes just as long to rotate around its own axis as it does to revolve around its partner. This causes one hemisphere constantly to face the partner body. In other words, one moon-year is just as long as one moon-day. The ...

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tl;dr: Perfectly full moon lasts, never more than, approximately 2 minutes. Estimating an answer from the point of view of Plain Geometry: If we approximate, that the sun is at an infinite distance, then exactly half of the moon is illuminated. Next, how much of the Moon do I see when I look at it from Earth? (This has nothing to do with illumination of ...

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Running the math for a 5 meter long pendulum and 1 kg mass, I get an amplitude of 0,017 mm. You are off by quite a bit. There is essentially no horizontal deflection when the Moon is at the horizon. The maximum horizontal deflection occurs when the Moon is about 45 degrees above or below the horizon. The tidal acceleration at some point on the surface ...

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The Isaac Newton Group of Telescopes has a web resource for plotting the altitude of celestial bodies, given a date and your latitude, longitude, altitude, etc. It is used for general RA, Dec. angles, but it automatically plots the altitude of the moon as well, for any plot that you generate. It's called Staralt.

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The US Naval Observatory has a web page which lets you print a table of moonrise/moonset times for any location, together with twilight times. If you use this approach, you would print the data and then examine it for a suitable date. I don't know of a computing resource where you would put in your requirements it would report the appropriate date and time.

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Small point, but Ganymede is slightly larger than Titan as moons are measured by their solid surface, though with atmosphere, Titan appears slightly bigger. The drawing is probably accurate, but slightly misleading. It's worth adding that Titan is out-gassing so it may have, quite some time ago, been larger than Ganymede. Source According to wikipedia, ...

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This website has a mathematical in depth analysis of the Apollo 11 translunar trajectory. It looks to be a reasonably reliable resource but I've only had a quick skim through so you should check on the sources cited. There should be enough information there to make a computer program to calculate the orbits.

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You seem to be referring to the computation of $m$ in equation (15) for the extinction as the light from the source passes through the atmosphere. The author intends to say that $m\approx \frac{1}{\sin(\alpha_s)}$, where $\alpha_s$ is the altitude of the source (in radian), so restricted to the range $0$ to $\pi/2$. (Note this is not the inverse of the sine ...

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