I am designing a habitable planet with a diameter of 12.879 km. orbiting around a K5V orange dwarf star. Due to a number of planetery collisions during its creation, the planet ended up being tidally locked to it's parent star. Assuming that the star has the following stats:

Mass: 0.70 M

Radius: 0.665 R

Luminosity: 0.153 L

Surface gravity (log g): 4.40 cgs

Temperature: 4,526 K

Rotation: 35.37 d

Age: 6.1 Gyr

How how fast does the planet need to orbit around the star and around its axis in order to be tidally locked and hiw far should it be?

  • 4
    $\begingroup$ The question provides lots of irrelevant data and omits one of the two necessary pieces of data. The only values that are relevant are the mass of the star, which is given as 0.7 M (where M is, we can guess, the head of the British Secret Intelligence Service, currently Richard Moore CMG); and the radius of the orbit of the planet, which is not given. All the other numbers sprinkled in the question are utterly useless. $\endgroup$
    – AlexP
    Sep 11 at 19:43
  • 1
    $\begingroup$ The question could be reduced to "what's the range of orbital periods in the habitable zone of a star with X mass, Y luminosity?".. There's a calculator which might help for some of that. $\endgroup$ Sep 11 at 21:26

2 Answers 2


If the luminosity of a K5V star with 0.70 the mass of the Sun is 0.153 the luminosity of the Sun, it is easy to calculte the distance where the planet will receive exactly as much radiation from the star as Earth gets from the Sun. I call that the Earth Equivalent Distance or EED.

The EED should equal the square root of the star's luminosity relative to the luminosity of the Sun. If the star's luminosity is 0.153 times the luminosity of the Sun, the EED of the star will be at the square root of 0.153, or 0.391152144, times 1 Astronomical Unit, or AU. Thus the EED will be at about 0.391152144 AU.

One AU is defined as 149,597,870,700 meters, which is 149,597,870.7 kilometers, or 92,955,807.273 miles. So if the planet orbits at the EED of the star, it will have the semi-major axis of its orbit at a distance of 58,515,527,860 meters, or 58,515,527.86 kilometers, or 36,359,863.31 miles.

I note that the table in this article:


says that a K5V star with a mass of 0.70 the mass of the Sun, will have a luminosity 0.17 that of the Sun.

If the star has 0.17 the luminosity of the Sun, the EED will be at 0.412310562 AU. That corresponds to a distance of 61,680,782,230 meters, or 61,680,782.23 kilometers, or 38,326,661.14 miles.

You really should find out which is the more accurate luminosity.

The star has 0.70 times the mass of the Sun. The Sun has a mass of 1.9885 times 10 to the 24th power kilograms, or 1,988,500,000,000,000,000,000,000 kilograms.

I looked for online planetary orbital period calculators.

According to this calculator,


the orbital period at 58,515,527.86 kilometers should be 81.84786 Earth days, and the orbital period at 61,680,782.23 kilometers should be 115.497 days.

According to this calculator,


the orbital period at a distance of 0.391152144 AU would be 106.75 days, and at a distance of 0.412310562 AU would be 115 days 13 hours, or 115.541666666 days.

The second figure is fairly close to that from the first calculator, but the first one is quite different.

The answer by user177107 to this question:


Has a table with characterstics of different types of stars. It says a K5V star has a mass of 0.68 Sun, a radius of 0.698 Sun, a surface temperature of 4410 K, and a luminosity of 0.165 Sun. It says that the orbital distance to receive the same amount of radiation as Earth gets from the Sun is 0.406 AU, and the orbital period at that distance is 114.84 days with an orbital speed of 38.518 kilometers per second.

You don't have to put your planet at the exact EED distance from its. The planet can be closer or farther from the star, as long as it is within the star's circumstellar habitable zone.


Unfortunately, estimates of the circumstellar habitable zone of the Sun to use as a basis for calculating the inner and outer edges of the habitable zonezones of other stars vary greatly. If a science fiction writer is certain they will never need to have another habitable planet in the same solar system they can put their only habitable planet somewhere between 0.99 and 1.01 of the EED of the star to play it safe. But if they want to have several habitable planets in different orbits around the star they need to check various estimates of the Sun's habitable zone to see if a wide habitable zone seems plausible to them.

  • $\begingroup$ I did some research on the sites you suggested and I will set my star's luminosity at 0.17. However, even though this answers my question about where to place the planet and how long it will take to fullfil one orbit, I also need to know how fast the planet revolves around itself. If the planet spins too slowly it may not have sufficient winds due to Corriolis effect to transfer heat around the planet and provide a suitable biosphere. $\endgroup$ Sep 12 at 9:16
  • 1
    $\begingroup$ @ΓΙΑΝΝΗΣ ΜΙΧΑΗΛΙΔΗΣ A tidally locked planet rotates one time for every orbit around its star. If you go with 0.17 as the luminosity of the star, that should be an orbital & rotation period of about 115 days. If the planet has ad iameter of 12,879 km, it has a circumference of 40,460.5718 kilometers at the equator, So the equator will turn about 350.181 km. per Earth day and 14.59 kilometers per Earth hour. Continued. $\endgroup$ Sep 13 at 1:59
  • $\begingroup$ @ΓΙΑΝΝΗΣ ΜΙΧΑΗΛΙΔΗΣ Many astrbiolgists feared than a tidally looked planet would have its water and air freeze out on the dark side., but some modern studies suggest that a decent atmosphere should transfer heat to the dark side. en.wikipedia.org/wiki/Planetary_habitability#Size But I don'tknow how fast the tidally locked planets in those studies would be spinning or what effect Corriolis effects would have. Continued. $\endgroup$ Sep 13 at 2:03
  • $\begingroup$ @ΓΙΑΝΝΗΣ ΜΙΧΑΗΛΙΔΗΣ According to rand.org/content/dam/rand/pubs/commercial_books/2007/… page 71, a star can a habitable zone without tidally locked planets down to a mass of 0.88 Sun, and the entire habitable zone will be tidally locked at a mass of 0.72 Sun. So all planets in the habitable zone of your K5V star should be tidally locked, unless modern calculatins put the limit different fromDole's calculations. YOu should see what rotatinperiod wer used in this study: ui.adsabs.harvard.edu/abs/1997Icar..129..450J/abstract $\endgroup$ Sep 13 at 2:14

A tidally locked planet rotates once each year. It doesn't matter what type the star is or the statistics of your planet. The only way to keep the same face of the planet facing the star is for the planet to rotate once each year or orbit of the planet.

Whether or not the statistics you give can result in a tidally locked planet is a different question.

I answered the one question in your title. You are allowed to ask one and only one question per post. Asking more than one question is a reason to close your question (see Vote to Close reason Needs More Focus).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .