I know the orbital period of my planet and the rotational period of my planet. Now how do I calculate the solar day? It would only be a small difference to be sure, but do remember that if we didn't use solar days instead of sidereal days, noon would be 12:00 AM in July and 12:00 PM in January.

EDIT: It was staring at me in the face. If the phase is half a rotation period for half a year, it means the solar day is exactly one day slower than a sidereal day and it's a simpler math problem.

EDIT2: As far as I can tell the formula is the following:

(P/(Y-P)) + P = D

Where P = Sidereal Rotational Period, Y = Sidereal Orbital Period, and D = Solar Rotational Period.

  • $\begingroup$ I'm not sure that this is entirely on-topic here... but a first cut at a solution to your question is to determine how far along the orbit the planet goes in one siderial rotation, as a radial fraction of a full circle, and then multiply that fraction by the length of the siderial rotation and add it to the length of the siderial rotation. For example, if the orbital period ("anno") is 360 siderial rotations ("stels"), and a stel is 360 whatevers, then each stel, the world moves 1/360 of the orbital circumference, and must rotate for an additional 1/360 stel to get the solar rotation ("sol"). $\endgroup$ Jun 26, 2017 at 19:59
  • $\begingroup$ So, the sol is 361 whatevers. Remember, that's only the first cut at the solution. To be perfectly correct, you have to take into account the additional distance along the orbit that the world moves in that 1 additional whatever, and add that fraction to the sol, and so on. $\endgroup$ Jun 26, 2017 at 19:59
  • $\begingroup$ (For what it's worth, I think the example anno of 360 stels ends up being 359 sols.) $\endgroup$ Jun 26, 2017 at 20:03

1 Answer 1


Each orbital revolution effectively adds (for retrograde rotation) or subtracts (for prograde rotation) one from the number of sidereal rotations to get the number of apparent solar rotations.

This is easy to demonstrate if you play around with spirographs (or poi, or spinning staffs, or any similar toy involving compound circles). For prograde rotations, you get one less "flower petal" in the spirograph pattern than the number of rotations, and for retrograde you get one more "flower petal" than the number of rotations.

So, the number of solar days is given by the length of the sidereal year divided by the length of a sidereal day, plus or minus one; and the length of a solar day is then the length of the solar year divided by that number. I.e.,

$L_{D-Solar} = \cfrac{L_{Year}}{\cfrac{L_{Year}}{L_{D-Sidereal}}\pm1}$

The intuitive logic behind this formula kinda breaks down if the sidereal day is longer than the sidereal year, though, so you may prefer a slightly different form:

$L_{D-Solar} = \cfrac{L_{Year}L_{D-Sidereal}}{L_{Year} \pm L_{D-Sidereal}}$

This can be derived from the first equation simply by multiplying both numerator and denominator by $L_{D-Sidereal}$. The $\pm$ again indicates retrograde vs. prograde rotation, and negative values for solar day length indicate that the sun appears to move backward, rising in the west and setting in the east.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .