Let there be a planet for which the duration of a solar day is equal to a year around the sun such that:

1 "day" = 1 "year" = 100 Earth days (2400 hours or 8640000 seconds)

Then, would we say that the rotation of the planet is 0 or equal to 100 days?


A planet that is not rotating with respect to the stars is not rotating. In your case this would give it a solar day length of 100 Earth-days. Such planets are not likely to exist in reality.

This is in contrast to the actual case when a body is tidally locked, such as our moon. The moon rotates once per (lunar) month at exactly the same rate as it orbits the Earth.

There are different ways of defining "length of a day". There is the (average) time from midday to midday (this is 24 hours, or one solar-day) or there is the time it takes for the stars to return to the same position that they were the previous night. This is the "sidereal day" and is about 23hours 56 minutes. The Earth takes 23hours 56 minutes to rotate (and not 24 hours)

For your planet, the sidereal day length is undefined and the solar day is 100 Earth-days. We would say that the planet is not rotating.

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  • $\begingroup$ Great, thanks a lot! However, my assumption for the question was that the duration of solar day is 100 earth-days and the duration of a sidereal day is not defined. In such a case, would we say that the planet does not rotate? Or that it rotates in 100 earth-days? $\endgroup$ – Aryaman Bansal Mar 8 at 10:04
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    $\begingroup$ To which kind of lunar month do you refer? (humor mostly) $\endgroup$ – uhoh Mar 8 at 10:28
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    $\begingroup$ @AryamanBansal I've edited to describe the (practically unlikely) sitution that you describe. $\endgroup$ – James K Mar 8 at 11:03

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