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Here on Earth, we have polar nights and midnight suns at our planet's poles. In terms of Sols, what would the day/night cycle on Mars be for a colony situated at the north pole?

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I'll expand on the other answers to address the duration of the sunrise and sunset at Mars' poles.

Wikipedia sez for Mars: Axial tilt: 25.19° to its orbital plane and for Earth: Axial tilt 23.4392811° which are pretty similar.

This answer puts a year at 668.6 days, which means that the Sun's elevation at the exact pole moves as

$$\text{El}(t) = 25.19° \cos \left( 2 \ \pi \ \frac{t-t_0}{ \ \text{668.6 days}}\right)$$

where $t_0$ is the time where a given pole points towards the Sun; i.e. the Summer solstice for that hemisphere. This answer might help towards figuring out what day that would be on various Martian calendars.

The rate of change is a maximum at the horizon at about 0.237 degrees/day.

The distance from Mars to the Sun ranges from 249.2 to 206.7 million kilometers and the Sun's diameter is 1.392 million kilometers, which means that the Suns' angular diameter ranges from 0.320 to 0.386 degrees.

Ignoring atmospheric refraction effects from Mars' atmosphere (less than 1% of Earth's) that means that twilight could between 1.35 and 1.63 days, depending on the orientation of Mars' axis relative to its argument of perihelion.

That turns out to be 286.502 °, which is close enough to 270 ° and the orbit close enough to circular that I'll punt and split the difference and say that the duration of sunrise and sunset are roughly equal and about half way between 1.35 and 1.63 days or about a day and a half.

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Since a Martian year works out to 668.5991 Sols, the Martian north pole would get a half a year of daylight, about 334 Sols, and a little less than half of a Martian a year of night. I don't know how many Sols of twilight the pole would get.

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Same as on Earth's poles, long nights and long days, depending on the season.

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