# Time period in which a planet rotates

I notice that Wikipedia give two contradictory terms for the time period in which a planet rotates in relation to an infinitely-distant object. In one place this is called a "sidereal day" and in another "stellar day". Which of these are correct? For that point, is this an IAU definition? I cannot find the definitions on the IAU website.

From the Sidereal Day page:

A sidereal day is the length of time it takes a planet to rotate from the perspective of a distant star.

From the Day page:

• stellar day - an entire rotation of a planet with respect to the distant stars
• sidereal day - a single rotation of a planet with respect to the vernal equinox
• This doesn't answer your question, but: for planets, and even most larger asteroids, the vernal equinox is nearly fixed with respect to the distant stars, so the two definitions would give nearly equal results. – barrycarter Oct 14 '14 at 15:48
• Both are correct. Asserting that one must be correct and the other be incorrect creates a false dilemma. – David Hammen Oct 16 '14 at 12:05
• @DavidHammen: Thank you. What would be the case for objects whose vernal equinox precesses at an exaggerated pace? – dotancohen Oct 16 '14 at 17:28
• @dotancohen -- Precession is very slow. It's not a factor in this regard. The biggest distinction between a sidereal vs solar day is the simple fact that a planet orbits the Sun. This distinction is small for the Earth (roughly a factor of 366/365). It's rather large for Venus. A Venusian sidereal day is about 243 Earth days long, but thanks to Venus's retrograde rotation, A Venusian solar day is "only" 116.75 Earth days long. – David Hammen Oct 16 '14 at 21:28

Effective 1985, UT1 is now computed using very long baseline interferometry of distant quasars, and so can be taken to be authoritative regarding the "fixed stars". The coordinated universal time (UTC) time approximates UT1 with atomic clocks. Anyway, UT1 derived the rotational period of the Earth as $p = 86164.09890369732\,\mathrm{s}$ of UT1 time and the mean sidereal day (in 2000) would be $86164.090530833\,\mathrm{s}$ of UT1 time, following the explanation:
The length of one sidereal day is defined by two successive transits of the mean equinox; while the Earth is rotating eastward, the mean equinox is moving westward due to precession. Therefore, one sidereal day is shorter than the Earth's rotational period by about $0.008\,\mathrm{s}$, the amount of precession in the right ascension in one day.