Solar day formula for moons

[garwel]

So, I need to calculate duration of a solar day (i.e. time between two sunrises or two sunsets) of a moon. Let's ignore inclination and assume near-circular orbits. Is there any handy formula for that?

[Green Baron]

1 hour ago, garwel said:

 Is there any handy formula for that?

I fear not.

Only the trivial case for a single moon (like ours), being tidally locked to its planet, it is one orbit (~27 days) + the time earth moves around the sun during that period. In total that is one synodic period, time between oppositions of planet and moon towards the sun, if i am not erring. The lunar day so is considerably longer than its orbital period around its planet (wikipdeia says 29.5 days for our moon, but forgets to take it into account for Titan i see).

For a non tidally locked moon, or if there are more than one moons in resonances or not it clearly goes over my head, i must pass.

Edited September 6, 2018 by Green Baron

[K^2]

The only necessary assumptions are that parent object (planet) is in near-circular orbit, and the moon's axis of rotation is always roughly perpendicular to direction towards the star, id est, inclination of moon's axis is small with respect to plain of orbit of the planet; and moon's orbit is relatively small compared to distance from the star. In that case, you have a very simple formula.

T_day = 1 / (1 / T_rotation - 1 / T_year)

 

Here:

T_day - mean solar day on the moon

T_year - sidereal orbital period of the planet

T_rotation - sidereal rotational period of the moon

Negative values can be used to denote retrograde motion of either body.

This gives you mean solar day. The exact time of sunsets and sunrises will drift a bit either seasonally or monthly, depending on configuration, so even sunrise-sunrise time isn't going to be exactly constant, but it will not vary wildly, and will average to the above.

If you need exact sunrise/sunset times, that's where things get a little complicated. There's nothing like a clean formula, but if you really need it, I can write you a script in C#, JS, or whatever. It will have a LOT of input parameters.

[garwel]

Cool, so it wasn't that hard after all. Thanks!

[Green Baron]

It is a good approximation for a tidally locked moon without other influences under the given constraints, the earth's moon is only a few hours off.

For the moons of the gas giants like Ganymede or Titan you can even omit the term (1/orbital period of planet), their daytime is roughly the same as their orbital time.

Edited September 7, 2018 by Green Baron

[K^2]

8 hours ago, Green Baron said:

For the moons of the gas giants like Ganymede or Titan you can even omit the term (1/orbital period of planet), their daytime is roughly the same as their orbital time.

Orbital time of moon around planet is irrelevant. It creates a small adjustment to relative position of the Sun, but it's periodic, so it has no impact on mean solar day. If you wanted a precise sunrise/sunset times, you'd have to take it into account. If you just want mean solar day, all you need to know is how long it takes you to go all the way around the Sun, which is planet's orbital period, and how long it takes the moon to rotate relative to the distant stars, which is sidereal rotational period.

And there are no assumptions about tidal lock here.

Edited September 7, 2018 by K^2

[Green Baron]

[own nonsense deleted]

Your formula does not take into account for an own rotation, an axis tilt, a non circular orbit or one that is inclined towards the planets ecliptic. It only works for a given point on the moons surface if the moon rotates together with its orbital period. That's tidal locking, or not ?

Edited September 7, 2018 by Green Baron

[K^2]

6 hours ago, Green Baron said:

Your formula does not take into account for an own rotation, an axis tilt, a non circular orbit or one that is inclined towards the planets ecliptic. It only works for a given point on the moons surface if the moon rotates together with its orbital period. That's tidal locking, or not ?

Sidereal rotation is own rotation. That's taken into account. Axis tilt is not accounted for, but so long as it's small with respect to plain of planet's orbit, mean solar day is not impacted on most of the surface. Only polar regions are going to be off. The size of these regions depends on the degree of the tilt. So again, small tilt means small regions where this doesn't apply. I've mentioned small tilt ('roughly perpendicular' bit) in my first post.

Orbital period and inclination of the moon's orbit is entirely irrelevant. It has zero impact on duration of mean solar day. So no, tidal locking is not necessary and is explicitly not an assumption.

[Green Baron]

Saying a day on earth lasts 24h is not totally incorrect, but on the surface the experience may be very different and time and location dependent.

 

Here's, some amateur level astronomy for an algebraic approach to estimate sunset and -rise times on a body and thus the time between two such events. I think that answers the question better and it gives some more insight. This was what went through my silly head when reading the op. Numeric implementations can be found anywhere on the internet, i should think. I may be off topic, though and i accept a verdict in that direction :-)

- a local coordinate system that describes the location of the observer on the surface and the corresponding horizon. A lunar centric azimuthal one servers here.

- a local coordinate system that describes the location of the other bodies relative to the observer. A lunar centric equatorial one is the choice.

- transformation between these two.

- To simplify the following steps a fixed coordinate system in which to describe the positions of the bodies would be a good idea. A sun-centric one with the parent planet's ecliptic as plane is what i would choose here.

- transformations to one of the two above

- a location independent representation of time. Like julian date.

- a set of orbital elements for the planet and the moon in question. These can be taken from a nautical or astronomical almanac.

- a set of precalculated perturbations for at least (in earh's case(*)) venus, mars and jupiter to apply to

- the following number cruncher that "simply" applies the perturbations to the mean orbital elements for the given time and coordinates on the moon and solve it so that the sun appears on the local horizon.

... et voilá, that'll leave you with an accuracy in a few minutes range, enough for a game. Not enough for nautical (1NM = 1 minute on a great circle on earth) application or for use in a decent telescope to find a star.

 

Equatorial coordinate systems are linked to a limited time frame, as everything changes over time, and more than one thinks. So either a formula to describe the changes (precession, nutation, plane change through orbital influence of other bodies, ....) or a range of validity in the margins of an accepted error should be taken into account. This is done by naming an epoc for which the above numbers are assumed to be valid.

Phew :-)

(*) in moons case we shouldn't forget earth's influence ...

btw.: Sidereal period of the moon is colloquially and frequently in literature used as the period it takes for the moon to return to the same spot in the sky as seen from earth. Even my "beloved" wikipedia has it that way. Which would not include a proper rotation of the moon, that's why i objected.

We are still too earth-centric, we humans .... :-)

Edited September 8, 2018 by Green Baron

[K^2]

22 hours ago, Green Baron said:

btw.: Sidereal period of the moon is ...

Yeah, which is why I've been consistently referring to sidereal rotational period. Which is proper period of rotation of the body with respect to distant stars, regardless of its orbital motion. That is the quantity of interest if you are talking about solar day, because we then only need to find relative motion of the Sun, and we're in business.

And of course, if you want very precise duration of the day for any given location, you need to do precise math on relative sun position, and that involves dramatically more math. Fortunately, that level of precision is rarely warranted if you're looking at duration only, and mean day is quite sufficient.

[Green Baron]

Well, *cough*, maybe use wording everybody understands ?

*duckandcover*

For a mean day duration over a year that's correct, we agree that for a calculation between sunset and rise more math is necessary. That is good.

Edited September 9, 2018 by Green Baron