I've recently detailed a world orbiting a M2 V star and have run into a conundrum I hope someone here can help me puzzle out.
The planet orbits in the star's habitable zone (orbit 0, or 0.1 AU). Using CTs Grand Survey by DPG, I generated an orbital period of just over 6 standard days and a rotational period of over 40 standard days.
Does this mean the planet's effective (ie. according to the guy standing on the surface) local day is around the same length of a local year? Or is the 960+ hour day the experience from the surface?
Rotational Period and "local day" (from the surface) are two different things, because when the rotational period is large compared to the orbital period, the effect on the local day as perceived from the surface can't be ignored.
For example, Earth's Rotational period is 23hrs 56min 4sec, but its local day is
exactly 24hrs, by definition. Why? Because it takes the Earth 23h 56m 4s to make one complete rotation on its axis, but in that time it has also moved 1/365th of its orbit around the Sun. So if I measure a day from high-noon to high-noon (sun directly overhead): Starting at high-noon, after 23h 56m 4s, the Earth will have made 1 complete rotation, but becuase the Earth has moved in its orbit, the sun will not be directly overhead, but the Earth will have to rotate for an additional 3m and 56s before the sun is directly overhead again (i.e. total time elapsed: 24hrs). This discrepancy is minor because the length of Earth's rotation period is small compared to its orbital period.
When the difference between the rotational period and orbital period are not small, however, it has a significant effect, and potentially makes for an interesting "local day".
Venus has an orbital period of 225 earth-days, but a (retrograde) rotational period of 243 earth-days. From the surface (minus cloud cover), the Sun would rise in the west and the local day would last 116 earth-days because of the large rotational period relative to the orbital period.
Mercury has an orbital period of 88 earth-days and a rotational period of 59.5 earth-days (3:2 resonance). It actually experiences a "double sunrise" from the surface (its local day) due to the orbital-rotational period difference (and due to its orbit's high eccentricity). The sun rises, then sets again not long after (relatively speaking), and then rises a second time (Locals might refer to "first" and "second" dawn). Its total local day length is 176 earth-days.
I know that doesn't specifically address your question about the mechanics in
DGP:Grand Survey, but it is tangentially relevant to your question.
