Astronomy Answers
Sidereal Time

Where a celestial body is in your sky depends on the rotation angle of the planet at your location, relative to the stars. That latter angle is expressed by the sidereal time θ (theta). The sidereal time is the right ascension that is on the celestial meridian at that moment. If the sidereal time is again the same (at the same location), then the stars are again in the same directions in the sky. We measure the sidereal time in degrees here. It can also be measured in hours. You can transform between degrees and hours using 1 hour = 15°.

The most important symbols that we use in this page are:

∆J

How much time has passed since 1 January 2000, 00:00 UTC (Julian Day number 2451544.5), measured in (whole and fractional) days.

L

The heliocentric mean longitude of the Earth, measured in degrees.

lw

The west longitude of the observation location, measured in degrees. West longitude is equal to negative east longitude.

t

The clock time, in the local time zone, measured in hours.

tz

UTC minus the clock time, measured in hours. In other words: how many hours you have to add to the clock time to get UTC.

θ

The sidereal time, measured in degrees. Divide by 15 to get sidereal time measured in hours.

1. From Clock Time to Sidereal Time

The mean sidereal time (relative to the mean equinox of the date) at a particular moment as seen from a particular location is equal to

(Eq. 1) θ ≡ L₀ + L₁ ∆J + L₂ ∆J² + L₃ ∆J³ − lw mod 360°

(Eq. 2) L₀ = 99.967794687°

(Eq. 3) L₁ = 360.98564736628603°

(Eq. 4) L₂ = 2.907879 × 10-13°

(Eq. 5) L₃ = −5.302 × 10-22°

The mod 360° (modulus 360°) part means that you can add (or subtract) multiples of 360° to the result; usually we do that until the result ends up between 0 and 360°. This also means that a given sidereal time keeps coming back (just like a given clock time).

∆J must take into account not just the date but also the time in UTC (Universal Time). If ∆Jd is number of whole days since 0:00 local time at the beginning of 1 January 2000, then we have

(Eq. 6) ∆J ≡ ∆Jd + (t + tz)/24

If you combine equations 1 and 6 then you get

(Eq. 7) θ ≡ θ₀ + θ₁ t mod 360°

where

(Eq. 8) θ₀ ≡ L₀ + L₁ ∆Jd + θp mod 360°

(Eq. 9) θp ≡ L₂ ∆Jd² + L₃ ∆Jd³ − lw + θ₁ tz

(Eq. 10) θ₁ ≡ M₀ + M₁ ∆Jd + M₂ ∆Jd²

(Eq. 11) M₀ = 15.04106864026192°

(Eq. 12) M₁ = 2.423233 × 10-14°

(Eq. 13) M₂ = −6.628 × 10-23°

Because ∆Jd is always a whole number, and because we calculateθp modulus 360°, we can use L₁ mod 360°, i.e., 0.98564736628603° instead of 360.98564736628603°. This keeps the numbers in the calculations smaller.

θp and θ₁ change with time only very slowly. In practice, you need to calculate them only once for a given problem, except if you need very high accuracy or if you want to calculate sidereal times for a very long period of time (for example, of at least a century).

2. From Sidereal Time to Clock Time

In the opposite direction, you find the local time t from

(Eq. 14) t = (θ − θ₀)/θ₁ = (θ − θp − L₀ − L₁ ∆Jd)/θ₁ mod ts

(Eq. 15) ts ≡ 360°/θ₁

ts says how many clock-hours correspond to 24 sidereal hours. θp and θ₁ depend a bit on the date (∆Jd), so ts depends (a little) on the date, too. For the most accurate calculations you should calculate θp and ts anew for each day. For quicker calculations, you can calculate θp and ts for a date in the middle of the desired interval, use those to calculate t (from formula 14) between 0 and ts for that middle date, and then add multiples of ts to it to find estimates for nearby clock times. Those estimates will be pretty good, as long as the interval is not too long (e.g., not longer than a few years).

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Last updated: 2012-01-13