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With a hypothetical system, where the moon would be always on the opposite side of the planet than the sun, in a way that the moon would only be visible at night on the planet.

I don't know if this is possible, but if it was possible how they would behave. For example, I think that if this being possible, maybe at dusk/dawn both moon and sun would appear on the horizon at opposite sides.

And the last thing, could this system have a periodic event where both star and satellite would appear entirety above the horizon line.

Another thing I am not sure, is if the place at the planet would affect this, but if it doesn't matter I can work with both scenarios.

Old question:

Is there any way to calculate how a planetary system would behave, where the satellite and the star appear simultaneously in the sky only once a year?

I'm writing a novel where this day would be of great importance, but would like to have some facts as realistic as possible, for example the duration of one day and one year for this scenario to be possible.

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I could only think of the case, where a planet is orbiting a star, and meanwhile there is another object, about the size of Pluto (or a comet), with a very eccentric orbit, such that once every year that object may be observed from the planet. This is like Halley's comet, which appears once every 75 years, though in your case it would be once every year. However, this isn't a satellite of that planet, since it isn't orbiting around that planet; it's orbiting around the sun. But the effect would be just as you describe: a small celestial body and a star that appear simultaneously only once a year (the star, however would appear every day, as does our Sun.)

Nothing has to be different re: duration of day and year in this case, as the small eccentric body has a small effect on the orbit of the planet. Hell, you may even DEFINE a year to be the length between the days that the comet/star appear!

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Is there any way to calculate how a planetary system would behave, where the satellite and the star appear simultaneously in the sky only once a year?

To be precise, you're saying that a moon orbiting a planet should be more or less at a Conjunction and that too you put a constraint saying that this should happen only once a year? There are a few things to be addressed...

1) "From where do you see it?" - Because, a conjunction visible in my place isn't visible to you (sometimes the objects won't be there at all...)

2) In reality, configuration of 1 year orbital period for both planet and satellite doesn't (won't) exist, because such a configuration can be perturbed easily by other celestial objects (of course, a comet can perturb the moon because of the long 1-year orbital radius it's circling)


Assuming identical conditions like you're at a position far from where the objects are aligned perfectly (i.e) you're in your pod somewhere along the axis of alignment so that you can watch the conjunction live, this can be calculated from classical Newtonian version where Kepler's third law helps a lot.

Let's take our Sun-Earth-Moon system and remove all other things (including Ceres, Pluto and their siblings to avoid perturbations) from the solar system. Since the orbital period makes use of the effective mass $M\approx m_1+m_2$, it's simply $T^2\approx r^3$.

The Earth is already at 1 AU and has orbital period of an year. Now, let's shift moon from its current orbit (at a distance of 0.00255 AU having orbital period of 27 days) to the 1-year orbit (at around 0.0145 AU) or you can reduce the mass of something on the system and bring the orbit closer.

I've used center-to-center distances above (robbed from this calculator), and though Kepler's law is used, I didn't make use of his elliptical orbit proposal - all orbits are approximated to circles...

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  • $\begingroup$ While this don't completely answer my question, it helped me resolve the problem, so I accepted it. Thank you. $\endgroup$ May 28, 2013 at 23:27
  • $\begingroup$ @PhilipiWillemann: Pardon me that I didn't notice your revision. You're welcome BTW ;-) $\endgroup$ May 29, 2013 at 2:43

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