Department of Physics

User's Guide to the Night Sky

Solar and Sidereal Days

In 24 hours the Earth rotates ~361° and the Earth revolves ~1° around the Sun.

Let P_SID be the sidereal rotation period of the Earth (i.e. the sidereal day = 23.93445 h),
P_SOLAR be the length of the solar day (i.e. 24 h) and
P_YEAR be the length of the year (i.e. 365.256 d).

The angular rotation speed of the Earth is 360/P_SID [in ° per hour if P_SID is in hours].
The angular revolution speed of the Earth around the Sun is 360/P_YEAR [ in ° per day if P_YEAR is in days]

In the time period P_SOLAR the Earth rotates (360/P_SID) × P_SOLAR°
and the Earth revolves around the Sun (360/P_YEAR) × P_SOLAR°.

Now removing the complete turn of the Earth, i.e. 360°, we have
(360/P_SID) × P_SOLAR − 360° = (360/P_YEAR) × P_SOLAR

so 1/P_SID − 1/P_SOLAR = 1/P_YEAR

or 1/P_SID = 1/P_SOLAR + 1/P_YEAR

Hence if P_SOLAR is 24 h and P_YEAR is 365.256 × 24 h,
then P_SID is 23.93445 h ( = 23h 56m 04s).

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