Calendars and Julian dates

Builds on: Sections 2.1 and 2.3
Reading time: 35 min
Exercises: 5
Olympiad problems: 0

In Rome, Thursday 4 October 1582 was followed by Friday 15 October. Pope Gregory XIII had removed ten days from the calendar, because over the centuries the old calendar had slipped against the seasons. The March equinox, which the church used to fix the date of Easter, was falling on 11 March instead of on or about 21 March, the date the church had fixed for it. In this section we look at the different kinds of year, at how calendars keep in step with the seasons, and at the continuous count of days that astronomers use instead of dates.

2.4.1Kinds of year

Before you readMake a guess first

Which is longer, the year of the seasons, from one March equinox to the next, or the time Earth takes to go once round the Sun relative to the stars?

A year can be measured in several ways, and the results differ by minutes. The tropical year is the time between two March equinoxes, so it is the year of the seasons, and calendars try to follow it. The sidereal year is the time Earth takes to go once round the Sun relative to the stars. The anomalistic year is the time from one perihelion, Earth's closest approach to the Sun, to the next. The three differ because the equinoxes drift westward round the ecliptic (Section 2.7) and the perihelion drifts slowly eastward. Astronomers also use the Julian year, exactly 365.25 days, as a unit, for example in the light year and in formulas that count time from the year 2000.

ResultLengths of the year(2.7)

\begin{aligned} tropical year & = 365.2422 d \\ sidereal year & = 365.2564 d \\ anomalistic year & = 365.2596 d \\ Julian year & = 365.25 d \end{aligned}

The days are mean solar days of 86 400 s. The tropical year is 365 d 5 h 48 min 45 s and the sidereal year 365 d 6 h 9 min 10 s.

2.4.2Calendars

A calendar must count whole days, but the tropical year is not a whole number of days, so a calendar adds an extra day, a leap day, from time to time. The Julian calendar, introduced by Julius Caesar in 45 BC, adds a leap day every fourth year. Its average year is 365.25 days, which is 0.0078 days, or 11 minutes, longer than the tropical year. The error is small, but it adds up to one day every 128 years, and by 1582 the equinox fell on 11 March. The reform removed ten days to bring it back to about 21 March.

The Gregorian calendar keeps the leap day every fourth year but drops it in years that end in 00, unless the year can be divided by 400. So 1900 was not a leap year and 2000 was. That removes three leap days every 400 years, leaving 97, and the average year becomes $365 + 97 / 400 = 365.2425$ days. This is 0.0003 days longer than the tropical year, so the Gregorian calendar slips by one day in about 3200 years. Countries adopted it at different times. Britain and its colonies changed in 1752, removing 11 days, and Russia in 1918 and Greece in 1923, removing 13. Some Orthodox churches still use the Julian calendar, which is now 13 days behind the Gregorian one.

Figure 2.5The March equinox in two calendarsinteractive

Year

March equinox, Julian calendar
March equinox, Gregorian calendar
Gap between the calendars

The figure shows the date of the March equinox in every year from AD 1 to 2400, worked out from the motion of the Sun. In the Julian calendar the date falls steadily, by about three days every 400 years. In the Gregorian calendar, counted back before 1582 with the same leap-year rule, it stays between 19 and 22 March. Each curve is about a day wide because the leap days pull the date back by about a day every fourth year, and the steps in the Gregorian curve come from the century years that have no leap day. The slider picks a year, and the table gives the date in each calendar. At the Council of , which set the rule for Easter, the equinox fell on 20 March. By it fell on 11 March in the Julian calendar, nine days earlier, and it falls on 7 March in the Julian calendar and 20 March in the Gregorian one.

Quick checkUse the result above

Is 2100 a leap year in the Gregorian calendar?

2.4.3Julian dates

To find the time between two events, astronomers avoid calendars altogether. The Julian date, JD, counts days and fractions of a day continuously from noon UT on 1 January 4713 BC in the Julian calendar, a date chosen so far back that all of recorded history comes after it. The day starts at noon, so that a whole night of observations in Europe falls on one day number. The Julian date at noon UT on 1 January 2000 is exactly 2 451 545.0, and this instant, called J2000.0, is the starting point of many formulas, including those for sidereal time in Section 2.5. Julian dates have nothing to do with the Julian calendar apart from the name.

ResultJulian date of a calendar date(2.8)

\begin{aligned} JD & = ⌊ 365.25 (Y + 4716) ⌋ + ⌊ 30.6001 (M + 1) ⌋ \\ + D + B - 1524.5 \end{aligned}

$Y, M, D$	year, month and day, with the time of day as a fraction of $D$ counted from 0h UT
$B$	for a Gregorian date, $B = 2 - A + ⌊ A / 4 ⌋$ with $A = ⌊ Y / 100 ⌋$ , and for a Julian date, $B = 0$
$⌊ x ⌋$	the whole number part of $x$

For a date in January or February,

Y

is taken as the year before and

M

as 13 or 14, which puts the leap day at the end of the counting year.

Example 2.8The Julian date of a theory paper

The theory paper of IOAA 2026 was sat on 28 September 2026. Find the Julian date at 0h UT on that day, and the number of days since J2000.0.

Solution

September is after February, so $Y = 2026$ and $M = 9$ . Then $A = 20$ and $B = 2 - 20 + 5 = - 13$ , and

\begin{aligned} JD & = ⌊ 365.25 \times 6742 ⌋ + ⌊ 30.6001 \times 10 ⌋ \\ + 28 - 13 - 1524.5 \\ = 2 462 515 + 306 \\ + 28 - 13 - 1524.5 \\ = 2 461 311.5 . \end{aligned}

The number of days since J2000.0 is $2 461 311.5 - 2 451 545.0 = 9766.5$ .

AnswerThe Julian date is 2 461 311.5, which is 9766.5 days after J2000.0. The half day appears because Julian dates start at noon.

Example 2.9The day of the weekfinish it

The day of the week follows from the Julian date at 0h UT. Dividing $JD + 1.5$ by 7, the remainder is 0 for a Sunday, 1 for a Monday, and so on. On which day of the week was 28 September 2026?

The start

From Example 2.8, $JD + 1.5 = 2 461 313$ .

Your stepNow find the remainder when 2 461 313 is divided by 7. The box below takes the remainder, and the solution names the day.

Two other forms are common. The modified Julian date, $MJD = JD - 2 400 000.5$ , has fewer digits and starts at midnight, so 28 September 2026 at 0h UT is MJD 61 311. The heliocentric Julian date corrects for the time light takes to cross Earth's orbit. Light from a star reaches Earth and the Sun at different moments, up to 8.3 minutes apart, which matters when timing a variable star or an eclipse to the minute (Chapter 8). The heliocentric Julian date is the time at which the light would have reached the Sun.

ResultHeliocentric Julian date(2.9)

HJD = JD - \frac{r}{c} \cos ψ

$r$	distance from Earth to the Sun
$c$	speed of light, so that $r / c$ is 499 s, or 8.32 min, for $r = 1$ au
$ψ$	angle between the star and the Sun, as seen from Earth

When the star is near the Sun in the sky, Earth is further from the star than the Sun is, and the light reaches the Sun first, so the correction is subtracted.

Common errorsSection 2.4

A Julian date starts at noon, not midnight. Midnight at the start of a date has a Julian date ending in .5.
The Julian date and the Julian calendar are different things that share a name.
In the formula for the Julian date, January and February belong to the previous year, with months 13 and 14. Forgetting this puts the leap day in the wrong place and gives an answer one to three days out.
Years that end in 00 are leap years in the Gregorian calendar only if they can be divided by 400.

ExercisesSection 2.45 questions

2.4.1
How many leap years are there from 1901 to 2000, including both?
leap years
2.4.2
What is the Julian date at 0h UT on 1 January 2000?
2.4.3
The first artificial satellite was launched on 4 October 1957, and the first people landed on the Moon on 20 July 1969. How many days apart are these dates?
days
2.4.4
By how many minutes is the average year of the Julian calendar longer than the tropical year?
minutes
2.4.5
A variable star lies near the ecliptic, opposite the Sun in the sky. How does its heliocentric Julian date compare with the Julian date of the same observation?