The length of a day on different planets in the solar system varies a lot. For instance, Mars' day is about the same length as Earth, while a day on Venus is equivalent to 243 Earth days (source). And Jupiter rotates about 143 times faster than Mercury.

What determines the length of a day on a particular planet? Put another way, what factors determine a planet's period of rotation?

To make the connection to worldbuilding clearer, say I wanted a day on a planet to last two Earth weeks. What would the characteristics of this planet (such as mass, radius etc.) need to be to achieve this?

  • $\begingroup$ If a planet aquired a large moon the moons size, orbital period and direction could easily speed up or slow down a planets rotation by a large margin over a long enough period. $\endgroup$
    – Slarty
    Commented Aug 27, 2018 at 9:32

3 Answers 3


There is, as far as I know, only one hard factor, and that is for planets that are very close to the star (Mercury and Venus). It is believed that the long rotational period of these planets is caused by different but related mechanisms - tidal locking for Venus (gravity interacting with it's thick atmosphere) and gravitational locking on Mercury (being closest to the sun).

Among the remaining seven planets, there is a rough correlation between size and length of day, in that the planets form four groups where the length of day is similar in each group and gets shorter as the planets get larger.

  1. Jupiter – 9.9 Earth hours

    Saturn – 10 hours 39 minutes and 24 seconds

  2. Uranus – 17 hours 14 minutes and 24 seconds

    Neptune – 16 hours 6 minutes and 36 seconds

  3. Earth – 23 hours and 56 minutes

    Mars – 24 hours 39 minutes and 35 seconds

  4. Pluto – 6.39 Earth days

However, there is ultimately too little data to really make any kind of definitive statements about the conditions causing planetary days to vary. Ultimately, this would give a world designer considerable leeway to use whatever value they wish.

However, if you dug a little deeper, you would notice that well studied moons of the Solar System all have synchronous periods of rotation - that is their rotational period is the same as their orbital period around their primary. This is caused by the same gravitational/tidal locking that affect Mercury and Venus. Titan, for example has an orbital period of just under 16 days, which is close to the value that you specified.

  • $\begingroup$ > all have synchronous periods of rotation - that is their day is the > same as their orbital period around their primary. It's because most satellite in the solar system are tidal locked. Mercury isn't but it would make sense to say that it's closeness to the Sun is slowing the planet's rotation. $\endgroup$
    – Vincent
    Commented Sep 29, 2014 at 14:38
  • $\begingroup$ Mercury is not in 1:1 lock like the Moon, yet it is tide-locked in the sense that the sun stands nearly still in its sky at perihelion when the solar tide is strongest. $\endgroup$ Commented Oct 19, 2019 at 6:57
  • $\begingroup$ Remember that Earth's rotation was originally about 11 hours. The effort of shoving our very large Moon that far out has used up about 60% of the rotational momentum of Earth. (via tidal friction) $\endgroup$
    – PcMan
    Commented Dec 2, 2020 at 6:51
  • $\begingroup$ @PcMan I've read suggestions that early on earth's rotational period could have been as short as 5 hours. $\endgroup$ Commented Nov 27, 2023 at 16:00

You pretty much have free range here. The speed a planet rotate depends on the angular momentum it has after forming. This is influenced by a host of factors, from the composition of the planet, to its distance from the star and the gravity it experiences as well as any impacts it experiences such as comets hitting it. See this post on the naked scientist Since these factors happened in the distant past, you can make up pretty much day length without affecting your current world.

However, I would be conscious of the effect of the increased day length. If the day lasts two weeks of earth time then a point on the planets surface would be light for a week then dark for a week. The temperature swing would be enormous. During the day portion, constant sunlight drive the temperature ridiculously high, and it would be difficult for any creatures to sleep or force them to go a long time without sleep if they have any sort of day/night cycle. During the night, constant darkness would send the temperature plummiting. It would be very difficult to survive these wild swings. Plants especially could suffer from prolonged absence from the sun. Just something you need to consider after you choose your day length.

  • $\begingroup$ Conditions on a world with week long days and nights would be extreme but not as extreme as you suggest. Remember that the Arctic and Antarctic experience 6 month days and nights. The Earth’s atmosphere is quite good at moving heat around. $\endgroup$
    – Slarty
    Commented Nov 19, 2019 at 8:38

I assume you mean the solar day (determined e.g. as the average time between two sunrises). Although you didn't say it, I also assume that you want a planet in the habitable zone (otherwise you've got much more freedom in your choices).


You want a planet orbiting a small star, ideally a red dwarf, on the inner edge of the habitable zone, and in addition a large moon orbiting the planet in an orbit with period longer than the planet's day.

General considerations

The two fundamental determining factors are the rotation of the planet (obviously) and the revolution of the planet around the central star. Basically the solar day is the sidereal day (rotation period of the planet) times one plus or minus (sign depending on the rotational direction of the planet relative to its revolution) the inverse of the number of days in the year.

The revolution period (year)

The effect of the revolution period (the length of a year) is low if there are many days per year, but the slower the rotation/the faster the revolution, the larger is the effect. The most extreme case is when the planet is tidally locked to the star, so that the sidereal day is exactly the same length as a year; then the solar day is infinite (the sun always sits on the same place).

The revolution period is, of course, determined by the stellar mass and the planet's distance to the star. The condition centripetal force = gravitational force reads

G m M/r^2 = m omega^2 r

(G is the gravitational constant, M the star mass, m the planet mass, r the distance, and omega the angular velocity) and thus for the revolution time (length of year), T = 2 pi/omega, you get

T = 2 pi r^(3/2) / (G M)^(1/2) ~ r^(3/2)/M^(1/2)

(here ~ means "proportional to").

More massive stars are generally brighter and therefore habitable planets will be further out; the distance is proportional to the square root of the stellar luminosity (brightness), because the light intensity falls off with the square. The mass-luminosity relation says that the luminosity of a main sequence star (the most common type of star, and especially the type you'd expect life-bearing planets around) scales roughly with the mass to the power of 7/2, so the distance of a habitable planet should be proportional to the stellar mass to the power of 7/4. Inserting this in the above equation, you get

T ~ M^(5/4)

so the year's length grows with the mass of the central star, although the dependency is not too strong. To have a maximal effect on the day length, you want the year as short as possible, that is you want to have a small star (ideally a red dwarf), and you want your planet on the inner edge of the habitable zone. As it turns out, small stars are also the most likely to have life-bearing planets, since they have the longest life time. The search for habitable planets indeed concentrates on red dwarfs.

The rotation period (sidereal day)

The by far biggest effect on the day length has, of course, the rotational speed of the planet. The rotational speed is normally determined by the initial angular momentum when the planet formed, but may also be altered subsequently by large collisions (as for example the collision of the earth with another planet which gave us our moon).

After that, the rotational speed is reduced by tidal forces due to the inhomogeneity of the gravitational field of other bodies.

All planets observe tidal forces from the central star; however the tidal forces fall off with the third power of the distance from the star, while they are proportional to its mass. Inserting the previous estimation of the planet's distance as function of the stellar mass, you get

F_tidal,star ~ M^(-7/4)

To have the strongest effect, you want the star's tidal force to be as large as possible (OK, maybe not too large, or you'll end up with a tidally locked planet), which due to the negative exponent in the mass again means you need a small star mass, and of course again you want to be on the inner edge of the habitable zone.

However, the star is not the only source of tidal forces; indeed, on earth the biggest tidal forces don't come from the sun, but from the moon. So to further slow down the planet's rotation, you want a large moon. Of course the moon should be sufficiently far away that its revolution period is slower than the planet's day, as if it is faster, you'll get the opposite effect: The moon will spin the planet up.

And of course, how much the planet was spun down by tidal forces depends very much on the age of the planet. An old planet will be slower than a newly formed one. Of course an old planet implies an old star, that is a long-lived one, so we are again back at the requirement at a small star.

  • $\begingroup$ If the moon is closer than the geostationary orbit, not only will it speed up the planet's rotation, it will also decay the moon's orbit; you don't want to be planning to settle on such a planet for very long. $\endgroup$ Commented Nov 27, 2023 at 16:51

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