For a given planet, if you know the sidereal rotation period and the sidereal revolution period, can you compute the length of one solar day?

For instance, for Earth, if you know one rotation is roughly 23h56m and one year is 365.25 days, can you recover that one solar day is 24 hours?

  • $\begingroup$ Try this answer to Synodic Day and Sidereal Day I think it should work. $\endgroup$
    – uhoh
    Commented Jul 7, 2021 at 14:11
  • $\begingroup$ Actually some of the days are shorter than 86400 seconds and others longer, because Earth's orbit is elliptical, and Earth's axis is inclined. 86400 is just the average day over a year. The longer and shorter days accumulate to time errors of +/- 15 minutes difference between mean solar time and apparent solar time. See en.wikipedia.org/wiki/Equation_of_time. $\endgroup$
    – stretch
    Commented Jul 7, 2021 at 16:43
  • $\begingroup$ Just multiply it by 366.25 (number of sidereal days in one year) to get the total number of hours in the year, and divide this by 365.25 (number of solar days in the year) to get the length of one solar day. Note that Earth makes exactly one revolution more with respect to stars in the year than with respect to Sun. You can do this reversely as well. $\endgroup$
    – User123
    Commented Jul 8, 2021 at 12:02

1 Answer 1


I'll take my own advice and "Try this answer to Synodic Day and Sidereal Day and see what happens.

As explained there, what we do in this case is subtract frequencies, just like radios heterodyne one frequency with another close by to downconvert.

$$f_{\text{diff}} = f_> - f_<$$

where the > and < indicate the larger and smaller frequencies.

The frequency of the Earth's orbit is very low, and subtracts from the Earth's natural 23h 56m 4.09s rotational frequency to give us a 24 hour day.

To convert a period into a frequency, we take the reciprocal, so it's:

$$\frac{1}{T_{syn}} \ = \ \frac{1}{T_<} - \frac{1}{T_>} \ = \ \frac{T_> - T_<}{T_> \ T_<}.$$

In this case the year will be the synodic period $T_{syn}$, and the sidereal period will be $T_<$, and we'll solve for the slightly longer, larger period solar day $T_>$.

$$\frac{1}{T_{solar}} \ = \ \frac{1}{T_{sid}} - \frac{1}{T_{year}} \ = \ \frac{T_{year} - T_{sid}}{T_{year} \ T_{sid}}.$$

sidereal day: 23 * 3600 + 56 * 60 + 4.09 =     86164.09 sec 
one year:     365.2564 * 24 * 3600    =     31558152.96 sec 

yields for the solar day: 86399.99 sec compare to 86400 = 24 * 3600 sec 

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