I want to make a hypothetical planet that is 26% the mass of earth, with 64% the radius. Because I want a relatively short day to keep temperatures more stable, but I don't want a super fast rotation that would dramatically affect the weather, I'm wondering if I just kept the rotation speed the same as Earth, if the day would naturally be shorter by virtue of the planet being smaller. Not sure if this is a dumb and obvious question.

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    $\begingroup$ Increased rotation speed will increase the Coriolis force, but it's already so weak that you'd likely need orders of magnitude higher rotational speed for it to have a significant effect on weather. Wind speeds won't generally be any higher, as the atmosphere rotates at the same rate as the planet, whatever that rate is. You could even out temperature variation by having a surface with high thermal inertia - a watery planet will see less daily variation in surface temperature, since it takes a relatively large amount of energy to change the temperature of water. $\endgroup$ May 13, 2022 at 15:15
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    $\begingroup$ A smaller planet will have a smaller surface speed for the same rotational speed. If you want shorter days without changing the surface speed, you both have to make the planet smaller AND make it spin faster. $\endgroup$
    – Jemox
    May 13, 2022 at 16:13
  • $\begingroup$ @Echox - I don't get that...you said "without changing the surface speed" but also "make it spin faster." Doesn't that contradict itself? If you take a planet and you shrink it, but keep the speed of rotation the same, wouldn't the fact that it's smaller shorten the day? Point A on the planet would take less time to get back to the same position, moving at the same speed, if the total circumference distance it had to cover was less, right? $\endgroup$
    – Elhammo
    May 13, 2022 at 17:56
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    $\begingroup$ @Elhammo so the thing about rotating objects is that rotation speed is most preferably measured as an angular velocity, which is measured in a way that is independent of the velocity of a specific point on the surface of that object (which is what "surface speed" is). Earth's angular velocity is most easily written as "1 rotation per day". The surface speed of that for its current average radius is ~1000mph. If you shrunk its radius you could: a) keep angular velocity constant (surface velocity would shrink), or b) keep surface velocity constant (angular velocity would increase) $\endgroup$
    – Daevin
    May 13, 2022 at 18:16
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    $\begingroup$ @Elhammo The question confused me for a sec. 'speed of rotation' means 'the number of turns of the object divided by time'. By definition if you keep the same 'speed of rotation', then the length of a day would remain the same, one rotation per ~24 hours. I think you're talking about increasing the speed of rotation, but in a way that a point on the equator moves around at the same speed because the planet is smaller. You can't give a planet a value for this speed without mentioning the latitude. $\endgroup$ May 14, 2022 at 3:31

6 Answers 6


The length of the day is strictly a function of the time earth takes to rotate. If your planet takes 24 hours to rotate then regardless of its radius, it will have a 24 hour day.

You can develop an intuition for this yourself by imagining a coffee cup and a dinner plate. Mark a spot on each of them. Put the cup on the plate with the marked spots in the same place. Note how when you turn the plate the cup turns at the exact same rate. Note how it takes exactly 1 turn for the marked place on the cup and plate to end up back where they started. If that spot was high noon, and a rotation was 24 hours, then there will be 24 hours between high noon on both the cup and the plate despite their different radii.

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    $\begingroup$ Technically this is not exactly true; the length of the day is also affected by the time it takes the planet to orbit its sun. The effect of the orbit is usually very small compared to the effect of the rotation, for Earth it takes about 23 hours 56 minutes to complete a full rotation, but it takes another 4 minutes for the Earth to rotate a bit more in order to face the Sun at the same angle as it did to begin with, due to orbiting slightly. If the planet's orbital period is only a small multiple of its rotational period then the effect will be much more significant. $\endgroup$
    – kaya3
    May 13, 2022 at 13:11
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    $\begingroup$ In the extreme case, a planet that is not rotating at all would still have a day/night cycle, but the length of the cycle would be one orbital period (i.e. one year). $\endgroup$
    – kaya3
    May 13, 2022 at 13:15
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    $\begingroup$ @kaya3 You're not wrong. However none of that is dependent on the radius of the planet. A non rotating cup, and a non rotating plate, in the same orbit, will have exactly the same length of day in your more complex, and technically accurate model. $\endgroup$
    – sphennings
    May 13, 2022 at 15:23

Visual demonstration

The two planets, rotating at the same speed

Just above you have two planets, the biggy, earthy one and the little one you're making. I added Snakeybot the light green square and Doomy the dark blue one standing respectfully on the big and small planet to see the planets' rotation. As you can see, both falls into night and wake into the sun at the same time.

The model has some simplifications we have to remember : First I supposed the light was coming in straight lines. In real-life, you'd have a very, very little deviation angle. Not even big enough to see a single pixel move in my image. So we can move that away.

Then, there is an atmosphere. I'm not well-versed enough on this matter to have very accurate guesses, but it should be relatively minimal : sunlight can travel quite some distance under Earth atmosphere before it's dissipated, and the light deviation due to it traversing a medium should be constant, regardless of the height of the atmosphere. If the air was very dusty though, you'd get the feeling of earlier nights on the big planet with the big atmosphere especially near the poles, as light will get absorbed by more dust before reaching you.

Moreover, planets are not perfectly round. First they're more squashed on the poles, but more interestingly they have mountains. This has a slight impact on when you have sunsets or sunrises. If Snakeybot and Doomy the squares were mountains, you'll see that people standing on top of them will have a little more suntime, and people behind them will be in the dark earlier. On Earth you can lose a good 30 min to 1 hour of daylight by living in enclosed valleys. Here's what it looks like :

2 mountain examples

Sunlight will be less affected by green Snakeybot's presence than with blue Doomy, whether you stand on top of it (more sunlight) or behind it (less sunlight). This however make two important assumptions : the mountains are the same size, and they're like, gigantic (closer to 100-200km than 8km). The effect of correctly sized-mountains would be much smaller, or even non-existent if the mountains are scaled down, too!

What can we conclude?

Changing the size but keeping the turn rate will have no meaningful impact by itself. You'll do have to increase its rotation speed at some point to have shorter days.

Don't fall in despair though : Atmospheres are heavily influenced by many other factors. With some tweaking in landmasses, atmosphere composition, air temperature and so on, you could perhaps get shorter days with about the same weather behaviour.

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    $\begingroup$ One thing to note is that smaller planets and moons will tend to have taller mountains and deeper valleys, especially relative to their diameters. This is observable in our solar system and I imagine there are scaling laws that predict the extent of this. $\endgroup$
    – Eric Hirst
    May 13, 2022 at 18:21

Depends on the measurement of speed


V is the speed on the surface. It is defined as $\vec v=\vec \omega \times \vec r$, where r is the radius. Smaller earth has smaller surface velocity. Because the vector v and the rotational axis are orthogonal to one another at the equator, the formula falls into $| \vec v |= |\vec \omega| |\vec r|$ there. So the surface speed goes down, due to the radius going down.


$\omega$ is the rotational speed of a body, measured in rotations per time. It is the inverse of the rotational period $\tau=\frac 1 \omega$, the time it takes for one rotation of the body.

$\omega$ won't change if it is still one full revolution per 24 hours ($\tau=\pu{24 h}$). Unless you change $\omega$ and $\tau$ and thus the number of rotations in a given fixed time, you will keep the days to be always the same: $$\tau=\pu{1 day}=\pu{864000 s}=\frac {1}{\omega}$$ and $$\omega= 0,11574\times 10^{-6} \frac{1} {\pu s}=\frac 1 \tau$$


By shrinking Earth, you actively alter the Coriolis force on items, which is by the speed compared to earth's surface, which of course is different since we have a different speed of the surface: $$\vec {F_C}=-2m (\vec \omega \times \vec v')$$

  • $\begingroup$ I think this is the best answer given the current form of the question, which doesn't specify what is meant by "rotation speed". Also worth noting that since the question says "I don't want a super fast rotation that would dramatically affect the weather", the Coriolis effect has an effect on weather, but as long as the angular velocity 𝜔 is the same, the Coriolis force should be the same for a weather system moving at a given speed relative to the surface (same speed in the rotating frame of the surface). $\endgroup$
    – Hypnosifl
    May 14, 2022 at 22:24
  • $\begingroup$ @Hypnosifl actually... corriolis is going with $\omega \times v$, and since $v=\omega \times r$, it is basically, only at the equator, proportional to $\omega^2 r$ $\endgroup$
    – Trish
    May 14, 2022 at 22:43
  • $\begingroup$ The formula for the Coriolis force here seems to say it's proportional to 𝜔×𝑣' where 𝑣' is defined as the velocity of the object moving relative to the surface (say, a storm system), as measured in the rotating frame where the surface is at rest, whereas I thought in the equation 𝑣=𝜔×𝑟 which you gave for surface speed, 𝑣 referred to the velocity of a point on the surface as seen in the inertial rest frame of the rotating body's center. $\endgroup$
    – Hypnosifl
    May 14, 2022 at 23:23
  • $\begingroup$ @Hypnosifl both are right to some degree: the lowest air layer (moving with earh) is 𝑣=𝜔×𝑟, higher ones have different speeds, which makes it all so messy... $\endgroup$
    – Trish
    May 14, 2022 at 23:43
  • $\begingroup$ But if the lower layer of air is moving with the Earth, then in the rotating frame where Earth is at rest the lower layer is at rest too, and so the Coriolis force on it (which is defined relative to this frame) is zero, correct? And calculating the Coriolis force on higher levels of atmosphere should depend only on their velocity in this rotating frame, not on the velocity of the surface in an inertial frame. $\endgroup$
    – Hypnosifl
    May 15, 2022 at 0:11

Do you mean speed of rotation of a point on the surface relative to the center, or rotations relative to the sun?


If a big planet and a little planet are rotating such that it is noon at the X at the same time for each, you can say they are rotating at the same rate. It takes the same time for each planet to complete one full rotation relative to the sun.

On these two planets rotating at the same rate relative to the sun, the X is moving faster on the big planet relative to its center. The X has to complete a large circumference circle in the same time that the small planet X completes the small circumference circle.

You can have small planets with day length the same as larger planets.

All that said and drawings drawn I am not sure speed of rotation has a direct relationship with weather.

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    $\begingroup$ I'm talking about rotation speed. I guess I was assuming that it was the rotation speed of the planet that would affect weather (I don't want super intense winds) more so than the day/night cycle, so I wanted to keep that more similar to Earth. Not sure if that's a correct assumption, though. I also like the idea of having a less than 24 hour day to stabilize day/night temperature changes (which might be more extreme on a planet with less atmosphere). $\endgroup$
    – Elhammo
    May 12, 2022 at 23:24
  • $\begingroup$ If it were "relative to the sun", it would have been called orbiting, not rotation. $\endgroup$
    – iBug
    May 13, 2022 at 18:55

If you spin the ball to do 25 rotations per second, then it will do 25 rotations per second, and the length of each rotation will be 0.04 seconds. Nothing fancy about that. And the good news is that any speed here is permissible. Just spin it as fast or as slow as you want, and in the vacuum of space it will keep spinning. So you can really set the day to be as long or as short as you want it to be.

Mind you though that faster spinning speeds result in a larger centrifugal force and thus the apparent surface gravity will get smaller. For Earth this centrifugal force is fairly small, but for your planet it might be a bit more noticeable. Best double check.

Now, orbital mechanics are different and the mass of the planet affects how far from the center (star) and how fast it will be going (well, if you want it to stick around that is). This would determine the length of your year as well as how hot the planet will be (the latter also depends on the size and type of the star). This is the real thing you should be worried about. This is what makes many such quaint hypothetical planets nonviable. Unfortunately I don't know enough about the topic to help you here.


As mentioned in other answers, the length of the rotation would not change.

However, if you mean "daylight hours", that's a bit different. The "lit" side of the planet depends on the surface exposed to the sun/star. For a basic model, we can model the sun/star as a point light source. Then we draw tangential lines to the surface. enter image description here Basically, the smaller your planet, the more of it will be illuminated at any given time. If you planet is reasonably distant from the star, then the change is negligible, but it is there - you may have a couple seconds more daylight.

  • $\begingroup$ that was my first idea, planet with radius half of radius of it's orbit will have length of day quarter of axial rotation period, obviously shorter $\endgroup$ May 13, 2022 at 12:52
  • $\begingroup$ If the only difference is mere seconds on a 24h cycles, I doubt that's worth changing the whole diameter and mass of the planet x). $\endgroup$ May 13, 2022 at 13:08
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    $\begingroup$ This isn't really an accurate model, on earth the sun is far too big and close to be treated as a point source. It's also much bigger than the earth, so you have the diagram sort of backwards - it should be the larger sun on the left shining onto a smaller planet on the right. You do get the same effect that a smaller planet gets slightly more illumination as the sun wraps "around" the horizon - the edge of the sun is still visible even after its center has dipped below the horizon. $\endgroup$ May 13, 2022 at 13:56
  • $\begingroup$ @Nuclear Hoagie Very valid. Considering the horizon is a lot closer, whilst the "day" would be the same length, Summer and Winter daylight hours will be affected. On the whole the daylight hours would increase due to first light and twilight bleed. But rise and set, yes, impacted. $\endgroup$
    – mckenzm
    May 15, 2022 at 19:00

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