So, I'm trying to figure out a world's cycle (length of a year, day, etc.). The planet, Vixeruka, is about 3 times larger than Earth is, making it a super-Earth, and it's the second planet from its sun, a K4V class star, so it's a main sequence star. The planet has rings around it and two moons. It's located in the habitable zone as well. Pretty much dab smack in the middle of the Goldilocks Zone to be specific.

So my best guess is that it takes the planet to go around its sun in about say 14 months, and that a day is about 28 hours long, and there's about 8 days in a week. But I'm not sure if I've got my math right.

My questions are how do I properly figure out the the period of rotation of the planet and the period of revolution, and what would those statistics be?

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    $\begingroup$ Don't use hard science tag with science based tag. These two are exclusive. $\endgroup$
    – Mołot
    Dec 16, 2017 at 23:56
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    $\begingroup$ "About 8 days in a week." Weeks are human constructions; you can have weeks of five, six, seven, eight, nine, ten or more days, as it pleases you. For example, the Romans used an eight day "week"; the seven day week of oriental origin was introduced towards the end of the 1st century BCE and by the beginning of the 4th century CE had completely replaced the indigenous eight day cycle. And the period of rotation of a planet has nothing to do with its period of revolution. $\endgroup$
    – AlexP
    Dec 16, 2017 at 23:57
  • $\begingroup$ I see. Didn't know about the science based tag restrictions. My bad! Still, I need to figure out both the period of rotation of the planet and the period of revolution. $\endgroup$
    – SCPilot
    Dec 17, 2017 at 0:12
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    $\begingroup$ Your question can't be answered with any precision. Revolution period is determined by star mass and orbital distance (major axis of elliptical orbit, to be precise) so we can make some guesses based on star class and derived Goldilock zone, but that would be very crude. Rotation speed is anyone guess, depending on details of planet formation. As @AlexP correctly stated "week" is a purely conventional construct, based on lunar cycle which is not synchronous with rotation (and You didn't mention a main moon, just rings). $\endgroup$
    – ZioByte
    Dec 17, 2017 at 1:15
  • $\begingroup$ @ZioByte You can figure out everything you need from the question. $\endgroup$
    – kingledion
    Dec 17, 2017 at 3:06

2 Answers 2


How long in time and space is my orbit?

Kepler's laws govern orbits, and they are relatively straightforward to plug in and calculate yourself. There are a variety of important and less important definitions and rules involved in Kepler's laws, but the heart of the matter can be expressed in one equation.

$$T^2 = \frac{4\pi^2}{G(M+m)}a^3$$

$T$ is the orbital period of the planet; $G = 6.67\times10^{-11}$ m$^3$kg$^{-1}$s$^{-2}$ is the gravitational constant; $M$ is the mass of the star; $m$ is the mass of the planet; and $a$ is the semi-major axis of the orbit. Orbits are elliptical, but if you pretend that your orbit is a circle, then $a$ becomes the distance between the planet and the Sun.

For an example in motion; for Earth, $a=1.50\times10^{11}$ m; $M=1.99\times10^{30}$ kg; $m=5.97\times10^{24}$ kg. Plugging those numbers in $$\begin{align}T^2 &= \frac{4\pi^2}{6.67\times10^{-11}(1.99\times10^{30}+5.97\times10^{24})}\left(1.50\times10^{11}\right)^3\\ T&=3.17\times10^{7}\text{ sec} = 366 \text{ days} \end{align}$$

Close enough!

How far is my planet from the sun?

The Earth is in the habitable zone of Sol. If we had a stonger greenhouse effect, by having more carbon dioxide or water vapor in the atmosphere, we could be farther away at the same temperature. If we had a higher albedo, due to more cloud cover, snow, or sand, we could be closer at the same temperature. There is a pretty varied zone in which Earth could exist.

But if we orbited a star that was not Sol, we might not be able to stay in the same place. A way to figure out how far away the orbit would need to be would be to use the inverse square law based on the difference between luminosity of your star and luminosity of the sun.

$$\frac{L_{sol}}{L_{other}} = \left(\frac{r_{sol-E}}{r_{other}}\right)$$

For example, our sun has 1 unit of luminosity, and the distance from the Earth to the sun is 1 AU. A K4V star is interesting, because there is no spectral standard star for K4V. However, 61 Cygni is the standard for K5V; and Epsilon Indi is also K5V and one of the closest sun-like stars. These two have luminosities of 0.15 and 0.22 that of Sol, respectively.

Lets say your sun is 0.2 times the luminosity of Sol. Then plugging in

$$\begin{align} \frac{1}{0.2} &=\left(\frac{1}{r_{other}}\right)^2\\r_{other}&=\sqrt{0.2}=0.45 \end{align}$$

Your plant must be about 0.45 AU from the star to be in the habitable zone of its star. Going back to the first equation, we can use a mass of 0.75 Sol to get

$$\begin{align} T^2 &= \frac{4\pi^2}{G(1.5\times10^{30})}6.8\times10^{10}\\ T&= 129 \text{ days} \end{align}$$

What is my rotation period?

This one doesn't really matter. Earth's day is 24h; Mars's is 25h; Venus's is 2784 hours. Up to you!

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    $\begingroup$ 7 hours and no one else upvoted it? Shame :( $\endgroup$
    – Mołot
    Dec 17, 2017 at 10:43

If the star is like our Sun, the planetary year would not be very different than Earth's, because the Goldilocks range would put your planet in a similar orbit as Earth's. Even at 3x Earth Mass, the total mass of your planet / star system wouldn't be much different than the Earth/Sun. 3x Earth masses is 0.00000901 solar masses.

In a binary system the separation between the objects in the system depends on the mass of the system and the period of the orbit (Kepler's Third Law). The period of an orbit equation computes the separation given the other two values.

$$ T = \sqrt{\frac{4 \pi ^2 \cdot R^3}{M \cdot G}}$$


If you want a more exotic year, use the equation in Planetary Separation Calculator to give you the planetary radius you want. I used a star with 1.9 solar masses and a period (planetary year) of 460 days, and it gave me a radius of 1.444 au.


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