Sidereal and solar days

Builds on: Sections 1.1 and 1.3
Reading time: 35 min
Exercises: 5
Olympiad problems: 1

From Ankara at 22:00 in mid-January, the three stars of Orion's belt stand high in the south, 48° above the horizon. At the same time by the clock in mid-March they are 31° up in the south-west, and in mid-May the belt sets at about 20:50, before twilight has ended. The stars have not moved on the celestial sphere, so it is our clock that slips against them. A clock follows the Sun, and the Sun creeps eastward among the stars by about a degree a day as Earth goes round it. In this section we see why this makes a day measured by the Sun longer than a day measured by the stars, and how much earlier the stars rise every night as a result.

2.1.1Two kinds of day

Before you readMake a guess first

How long does Earth take to turn once on its axis, measured against the stars?

There are two natural ways to measure a day. The sidereal day is the time between two crossings of the meridian by the same star, which is the time Earth takes to turn once relative to the stars. The solar day is the time between two crossings of the meridian by the Sun, from one noon to the next. The length of the solar day changes a little through the year, for reasons explained in Section 2.2. Its average over the year is the mean solar day, and our clocks divide it into 24 hours of 3600 seconds, 86 400 seconds in all.

Figure 2.1 shows why the two days differ. At A it is noon for an observer on Earth, so the observer's meridian points at the Sun. As Earth turns, it also moves along its orbit, about 1° a day, because it goes 360° round the Sun in a year of 365 days. After one turn relative to the stars the meridian points in the same direction in space as it did at A, but Earth is now at a different place in its orbit, and the Sun is no longer in that direction. Earth has to turn a little further, through the angle it has moved round the Sun, before the Sun is back on the meridian. The solar day is therefore longer than the sidereal day, by the time Earth takes to turn about 1°, which is about 4 minutes.

Figure 2.1Why noon comes later than a full turninteractive

Days after A

Angle $θ$ moved round the Sun
Sidereal time at noon
Turns of Earth

The figure shows Earth's orbit round the Sun S, seen from above the north pole, with Earth drawn far larger than to scale. Earth moves round the orbit anticlockwise, as the small arrow shows. At A it is noon for the observer, and the meridian points at the Sun. B is Earth's position some days later. The dashed arrow at B points the same way in space as the noon arrow at A, which is where the meridian points after a whole number of turns relative to the stars. The heavy arrow points at the Sun, and the angle

θ

between them is the angle that Earth has moved round the Sun, the same

θ

as at S. The slider sets the number of days after A. After the angle is about 1°, too small to see. After it is 90°, and Earth has turned one quarter of a turn more relative to the stars than relative to the Sun. After it has turned one whole turn more.

Over a whole year the extra turns add up to exactly one. In a tropical year of 365.2422 mean solar days, the year of the seasons described in Section 2.4, Earth turns 365.2422 times relative to the Sun and 366.2422 times relative to the stars. The same count can be written as a relation between the two days and the year.

ResultSidereal and solar days(2.1)

\frac{1}{T_{sol}} = \frac{1}{T_{sid}} - \frac{1}{Y}

$T_{sol}$	mean solar day
$T_{sid}$	sidereal day, the time for one turn relative to the stars
$Y$	length of the year, in the same unit as the days

The relation holds for a planet that spins the same way as it goes round the Sun. For a planet that spins the opposite way, such as Venus, the minus sign becomes a plus.

Show the derivation

Relative to the stars, Earth turns at $360 ° / T_{sid}$ per unit of time. Seen from Earth, the Sun moves eastward among the stars at $360 ° / Y$ , in the same direction as Earth turns. Relative to the Sun, Earth therefore turns at the difference of the two rates, and the time for one turn relative to the Sun is the solar day,

\frac{360 °}{T_{sol}} = \frac{360 °}{T_{sid}} - \frac{360 °}{Y} .

Dividing by 360° gives (2.1). If the planet spins the opposite way, the Sun moves against the direction of the spin, the two rates add, and the minus sign becomes a plus.

Quick checkUse the result above

How many sidereal days are there in a tropical year of 365.2422 mean solar days?

sidereal days

Solving (2.1) for the sidereal day gives $T_{sid} = 24 h \times 365.2422 / 366.2422$ , which is 23 h 56 min 4.09 s. The ratio of the two days is used constantly, so it gets its own result.

ResultRate of sidereal time(2.2)

\begin{aligned} k = \frac{T_{sol}}{T_{sid}} & = 1.002738 \\ T_{sid} & = 23 h 56 min 4.09 s \end{aligned}

$k$	number of sidereal days, or sidereal hours, in one mean solar day, or one ordinary hour

A clock that keeps sidereal time gains

(k - 1) \times 24 h = 3 min 56.56 s

on an ordinary clock every day.

NoteStrictly, the sidereal day of (2.2) is measured relative to the March equinox point, as sidereal time is. That point drifts slowly westward among the stars (Section 2.7), so a turn relative to the distant stars takes about 0.008 s longer. For olympiad problems the difference never matters.

2.1.2Stars rise earlier every night

Every star crosses the meridian once per sidereal day, which by the clock is 3 min 55.9 s less than 24 hours. So each star crosses the meridian, rises and sets about 3 min 56 s earlier each day by the clock, about 2 hours earlier each month, and round the whole clock in a year. The stars that are high in the south at midnight change with the seasons for the same reason. At midnight the meridian points away from the Sun, and as the Sun moves round the sky in a year, so does the part of the sky opposite it.

Example 2.1Sirius two months later

At Greenwich, Sirius crosses the meridian at 23:05 by the clock on 15 January 2026. At what time does it cross on 15 March?

Solution

From 15 January to 15 March is $16 + 28 + 15 = 59$ days. Each day Sirius crosses 3 min 55.9 s, or 3.932 min, earlier, so in 59 days it moves earlier by

59 \times 3.932 min = 232.0 min = 3 h 52.0 min .

AnswerOn 15 March Sirius crosses the meridian at about

23 : 05 - 3 : 52 = 19 : 13

, early in the evening.

2.1.3Days on other planets

Before you readMake a guess first

Venus turns once relative to the stars in 243 days and goes round the Sun in 225 days. It spins the opposite way to its orbit. How long is a day on Venus, from one noon to the next?

(2.1) is not special to Earth. The same reasoning gives the length of a solar day on any planet, and the same counting gives the month of the phases of the Moon, as the exercises show.

Example 2.2A day on Venus

Venus turns once relative to the stars in 243.02 days, in the opposite direction to its motion round the Sun, and goes round the Sun in 224.70 days. How long is its solar day?

Solution

Venus spins against the direction of its orbit, so the rates add, and by (2.1) with the sign changed,

\frac{1}{T_{sol}} = \frac{1}{243.02} + \frac{1}{224.70},

so $T_{sol} = 116.75$ days.

AnswerFrom one noon to the next on Venus takes 116.75 of our days, about half of Venus's year, so the Sun rises in the west and sets in the east about twice in each Venus year.

Example 2.3A day on Marsfinish it

Mars turns once relative to the stars in 24 h 37 min 22.7 s, the same way as it goes round the Sun, and its year is 686.98 of our days. How long is a solar day on Mars, in hours?

The start

In hours, the sidereal day is $24 + 37 / 60 + 22.7 / 3600 = 24.6230$ h and the year is $686.98 \times 24 = 16 487.5$ h. Mars spins the same way as it orbits, so by (2.1)

\frac{1}{T_{sol}} = \frac{1}{24.6230} - \frac{1}{16 487.5} .

Your stepNow find the solar day of Mars in hours.

hours

Common errorsSection 2.1

For Earth the sidereal day is the shorter of the two. A student who swaps them finds the stars rising later each night instead of earlier.
A star crosses the meridian 3 min 55.9 s earlier each day by the clock, while sidereal time at a fixed clock time is 3 min 56.6 s later each day. The two numbers differ because one is measured in ordinary minutes and the other in sidereal minutes, and over many days the difference adds up.
The minus sign in (2.1) is only for a planet that spins the same way as it orbits. For Venus, which spins the other way, it is a plus.
The year in (2.1) has to be in the same unit as the two days, so a year in days must not be mixed with days in hours.

ExercisesSection 2.15 questions

2.1.1
How many sidereal days are there in 30 mean solar days?
sidereal days
2.1.2
For an observer on Earth, which is longer?
2.1.3
A star rises at 21:00 tonight. How many minutes earlier will it rise 30 days from now?
minutes
2.1.4
The Moon goes round Earth once relative to the stars in 27.322 days, the same way as Earth goes round the Sun, and the sidereal year is 365.256 days. How long is the month of the Moon's phases, from one full Moon to the next?
days
2.1.5
Mercury turns once relative to the stars in 58.646 days, the same way as it goes round the Sun, and its year is 87.969 days. How long is its solar day?
days

Olympiad problemsPast IOAA problems, restated1 problem

2012If Earth turned the other way

Adapted from IOAA 2012 (Rio de Janeiro), theory problem 2. We have restated it in our own words and added the length of the year. The solution was last checked on 17 October 2026.

Find the length of the sidereal day on Earth, taking the year as 365.2422 mean solar days of 24 h. How long would the solar and sidereal days be, in our hours, minutes and seconds, if Earth turned in the opposite direction at the same speed?

I solved it without the solution

AnswerThe sidereal day is 23 h 56 min 4.09 s. If Earth turned the other way at the same speed, the sidereal day would be unchanged and the solar day would be 23 h 52 min 9.5 s, shorter than the sidereal day instead of longer.