The position of the Sun in the sky as seen from a planet (such as the Earth) is determined by four things:
Below, I provide formulas that take all of these things into account, based on the rotation axes of the planets as defined by the IAU in "Report of the IAU/IAG Working Group on Cartographic Coordinates and Rotational Elements of the Planets and Satellites: 2000" by P.K. Seidelmann et al. I neglected small effects here and there, to keep the formulas simple. Remember that you can add multiples of 360 degrees to any angle without changing its direction.
When in this page an ecliptic or pole or coordinates such as ecliptic longitude or declination are mentioned, then these are appropriate for the planet where the observer is. Such coordinates are based on the orbit and equator of that planet and are not identical to the coordinates of the same names that we use on Earth. For example, the ecliptic of Mars, the ecliptic that is appropriate for Mars, is not the same as the ecliptic of Earth, and an earth-based atlas of the stars is of no use if you have a right ascension and declination calculated for the Sun as seen from Mars.
For examples, we'll calculate the position of the Sun for 1 April 2004 at 12:00 UTC, as seen from 52° north latitude and 5° east longitude (Netherlands) on Earth, and as seen from 14°36' south latitude and 184°36' west longitude (Gusev crater) on Mars.
For this kind of astronomical calculations, it is convenient to
express the date and time using an unending day numbering scheme.
Such a scheme is provided by the Julian Day Number J
.
The Julian Day Number
Calculation Page explains how you can calculate the Julian Day
Number for a date in the Gregorian calendar. For the calculations of
the position of the Sun you should express time as Universal Time
(UTC), and this includes the Julian Date. For example, JD 2453144.5
corresponds to 0:00 hours UTC on 19 May 2004, which is (for example)
equivalent to 1:00 hours Central European Time on 19 May 2004, or
20:00 hours (8 pm) Eastern Standard Time on 18 May 2004.
Only on Earth do the seasons repeat themselves after about one
calendar year (in the western - Gregorian - calendar), so, only for
the Earth, the day number d
in the calendar year can be
used instead of the Julian Day Number: d
= 1
corresponds to 0:00 UTC on January 1st, 2 to 0:00 UTC on January 2nd,
32 to 0:00 UTC on February 1st, and so on.
Because we see the Sun from the planet, we see the motion of the planet around the Sun reflected in the apparent motion of the Sun along the ecliptic, relative to the stars.
If the orbit of the planet were a perfect circle, then the planet as
seen from the Sun would move along its orbit at a fixed speed, and
then it would be simple to calculate its position (and also the
position of the Sun as seen from the planet). The position that the
planet would have relative to its perihelion if the orbit of the
planet were a circle is called the mean anomaly, indicated in the
formulas as M
.
You can calculate the mean anomaly of the planets (measured in
degrees) to reasonable accuracy using the following formula, for a
date given as a Julian Day Number (JD) J
:
(Eq. 1)
M = M₀ + M₁*(J − J_{2000})
(Eq. 2)
J_{2000} = 2451545
You should take M₀
(in degrees) and M₁
(in degrees per day) from the following table:
M₀ | M₁ | |
---|---|---|
Mercury | 174.7948 | 4.09233445 |
Venus | 50.4161 | 1.60213034 |
Earth | 357.5291 | 0.98560028 |
Mars | 19.3730 | 0.52402068 |
Jupiter | 20.0202 | 0.08308529 |
Saturn | 317.0207 | 0.03344414 |
Uranus | 141.0498 | 0.01172834 |
Neptune | 256.2250 | 0.00598103 |
Pluto | 14.882 | 0.00396 |
For the Earth, you can also use the following formula:
(Eq. 3) M = −3.59° + 0.98560° d
where d
is the time since 00:00 UTC at the beginning of
the most recent January 1st, measured in (whole and fractional) days.
The chosen date and time correspond to Julian Day Number 2453097, so
J = 2453097
, and also to d = 92.5
. For
the Earth, using equation 1 we find M_{earth} =
1887.1807° = 87.1807°
and with equation 3 we find
M_{earth} = 87.58°
. Because equation 1 is a bit
more accurate that equation 3, we'll use equation 1. For Mars, we find M_{mars} = 832.6531° = 112.6531°
.
The orbits of the planets are not perfect circles but rather ellipses, so the speed of the planet in its orbit varies, and therefore the apparent speed of the Sun along the ecliptic also varies throughout the planet's year.
The true anomaly (symbol ν
, nu) is the angular distance
of the planet from the perihelion of the planet, as seen from the Sun.
For a circular orbit, the mean anomaly and the true anomaly are the
same. The difference between the true anomaly and the mean anomaly is
called the Equation of Center, written here as C
:
(Eq. 4) ν = M + C
To calculate the Equation of Center or the true anomaly from the mean anomaly, you must first solve the Equation of Kepler. This equation cannot be solved in general, but you can find approximations to the solution that are more and more accurate. See the Calculation Page for the Equation of Kepler for more information about this. If the orbit looks more like a circle than like a parabola, then the following approximation is sufficiently accurate:
(Eq. 5) C ≈ C₁ sin M + C₂ sin(2 M) + C₃ sin(3 M) + C₄
sin(4 M) + C₅ sin(5 M) + C₆ sin(6 M)
You can find the coefficients C₁
through
C₆
in the following table. They depend on the
eccentricity e
of the planet's orbit. If a particular
coefficient is not mentioned, then it is equal to zero (to four
decimal places). The orbits of Mercury and Pluto deviate the most
from circularity, so they need the most coefficients. The column
E_{C}
shows the maximum error that you get if you use the
approximation with the coefficients from the table.
C₁ | C₂ | C₃ | C₄ | C₅ | C₆ | E_{C} | |
---|---|---|---|---|---|---|---|
Mercury | 23.4400 | 2.9818 | 0.5255 | 0.1058 | 0.0241 | 0.0055 | 0.0026 |
Venus | 0.7758 | 0.0033 | 0.0000 | ||||
Earth | 1.9148 | 0.0200 | 0.0003 | 0.0000 | |||
Mars | 10.6912 | 0.6228 | 0.0503 | 0.0046 | 0.0005 | 0.0001 | |
Jupiter | 5.5549 | 0.1683 | 0.0071 | 0.0003 | 0.0001 | ||
Saturn | 6.3585 | 0.2204 | 0.0106 | 0.0006 | 0.0001 | ||
Uranus | 5.3042 | 0.1534 | 0.0062 | 0.0003 | 0.0001 | ||
Neptune | 1.0302 | 0.0058 | 0.0001 | ||||
Pluto | 28.3150 | 4.3408 | 0.9214 | 0.2235 | 0.0627 | 0.0174 | 0.0096 |
For the Earth, we find C_{earth} = 1.9148° * sin(87.1807°) +
0.0200° * sin(2 * 87.1807°) + 0.0003° * sin(3 * 87.1807°) =
1.9142°
and thence ν_{earth} = 89.0949°
. For
Mars we find C_{mars} = 10.6912° * sin(112.6531°) + 0.6228° *
sin(2 * 112.6531°) + 0.0503° * sin(3 * 112.6531°) + 0.0046° * sin(4
* 112.6531°) + 0.0005° * sin(5 * 112.6531°) = 9.4092°
and
ν_{mars} = 122.0623°
.
To find the position of the Sun in the sky we need to know what the
ecliptic longitude Π
is of the perihelion of the
planet, relative to the ecliptic and vernal equinox (the ascending
equinox) of the planet. The ecliptic of the planet is the plane of
the orbit of the planet, which makes an angle with the orbit
(ecliptic) of the Earth (and that angle is called the inclination of
the orbit). The vernal equinox of the planet is the point where the
Sun passes from south to north through the plane of the equator of the
planet. We also need to know the obliquity ε
of the
equator of the planet compared to the orbit of the planet. These two
values are listed in the following table for each planet, measured in
degrees.
Π | ε | |
---|---|---|
Mercury | 111.5943 | 0.02 |
Venus | 73.9519 | 2.64 |
Earth | 102.9372 | 23.45 |
Mars | 70.9812 | 25.19 |
Jupiter | 237.2074 | 3.12 |
Saturn | 99.4571 | 26.74 |
Uranus | 5.4639 | 82.22 |
Neptune | 182.1957 | 27.84 |
Pluto | 4.5433 | 57.46 |
The ecliptical longitude λ
(lambda) is the position
along the ecliptic, relative to the vernal equinox (so relative to the
stars). The mean longitude L
is the ecliptical
longitude that the planet would have if the orbit were a perfect
circle. The ecliptic longitude of the planet as seen from the Sun is
equal to
(Eq. 6) λ = ν + Π = M + Π + C = L + C
If you look at the Sun from the planet, then you're looking in exactly the opposite direction than if you look at the planet from the Sun, so those directions are 180° apart. So, the ecliptic longitude of the Sun, as seen from the planet, is equal to
(Eq. 7) λ_{sun} = ν + Π + 180° = M + Π + C + 180°
The value of λ_{sun}
determines when the (astronomical)
seasons begin: when λ_{sun} = 0°
, then spring begins in
the northern hemisphere, and autumn in the southern hemisphere. Each
next multiple of 90° brings the start of the next season.
The ecliptic latitude β_{sun}
(beta) of the Sun, the
perpendicular distance of the Sun from the ecliptic, is always so
small that we can ignore it here. With this, we now have the
ecliptic coordinates of the Sun.
As seen from Earth we find λ_{sun} = 372.0319° =
12.0321°
and as seen from Mars λ_{sun} = 373.0435° =
13.0435°
.
The equatorial coordinate system in the sky is tied to the rotation
axis of the planet. The equatorial coordinates are the right
ascension α
(alpha) and the declination
δ
(delta). The declination determines from which parts
of the planet the object can be visible, and the right ascension
determines (together with other things) when the object is visible.
With these formulas you can calculate the equatorial coordinates from the ecliptic coordinates:
(Eq. 8) sin α cos δ = sin λ cos ε cos β − sin
β sin ε
(Eq. 9) cos α cos δ = cos λ cos β
(Eq. 10) sin δ = sin β cos ε + cos β sin ε sin λ
For the Sun we have β_{sun} = 0
, so then
(Eq. 11) α_{sun} = arctan(sin λ_{sun} cos ε, cos
λ_{sun})
(Eq. 12) δ_{sun} = arcsin(sin λ_{sun} sin ε)
For future reference we define
(Eq. 13) α_{sun} = λ_{sun} + S
If ε
is close enough to 0° or 180°, and if we neglect
small terms, then we can approximate the relationship between the
right ascension α_{sun}
and the ecliptic longitude
λ_{sun}
of the Sun as seen from the planet by
(Eq. 14) arctan(tan(λ) cos(ε)) = λ − ((¼) ε² +
(1⁄24) ε⁴ + (17⁄2880) ε⁶) sin(2 λ) + ((1⁄32) ε⁴ + (1⁄96) ε⁶) sin(4
λ) − (1⁄192) ε⁶ sin(6 λ) + O(ε⁸).
(Eq. 15) α_{sun} = λ_{sun} + S ≈ λ_{sun} + A₂
sin(2 λ_{sun}) + A₄ sin(4 λ_{sun}) + A₆ sin(6 λ_{sun})
and the relationship between the declination δ_{sun}
and the ecliptic length by
(Eq. 16) arcsin(sin(λ) sin(ε)) = (ε − (1⁄6) ε³ + (1⁄120) ε⁵) sin(λ)
+ ((1⁄6) ε³ − (1⁄12) ε⁵) sin(3 λ) + (3⁄40) ε⁵ sin(5 λ) + O(ε⁷)
(Eq. 17) δ_{sun} ≈ D₁ sin(λ_{sun}) + D₃
sin(λ_{sun})³ + D₅ sin(λ_{sun})⁵
with A₂
, A₄
, A₆
,
D₁
, D₃
, and D₅
(measured
in degrees) from the following table. The columns E_{A}
and E_{D}
show the greatest errors you make for
α_{sun}
and δ_{sun}
when you use these
approximations.
A₂ | A₄ | A₆ | E_{A} | D₁ | D₃ | D₅ | E_{D} | |
---|---|---|---|---|---|---|---|---|
Mercury | −0.0000 | 0.0000 | 0.0000 | 0.0000 | ||||
Venus | −0.0305 | 0.0001 | 2.6427 | 0.0009 | 0.0036 | |||
Earth | −2.4680 | 0.0530 | -0.0014 | 0.0003 | 22.8008 | 0.5999 | 0.0493 | 0.0003 |
Mars | −2.8605 | 0.0712 | -0.0022 | 0.0004 | 24.3870 | 0.7331 | 0.0706 | 0.0011 |
Jupiter | −0.0424 | 0.0001 | 3.1151 | 0.0015 | 0.0034 | |||
Saturn | −3.2364 | 0.0911 | -0.0031 | 0.0009 | 25.7790 | 0.8649 | 0.0951 | 0.0010 |
Uranus | -42.5725 | 12.8039 | -2.6057 | 17.6902 | 56.9067 | -0.8355 | 26.1482 | 3.34 |
Neptune | −3.5195 | 0.1077 | -0.0039 | 0.0163 | 26.7577 | 0.9662 | 0.1164 | 0.060 |
Pluto | -17.1633 | 2.4178 | -0.3035 | 0.5052 | 48.3114 | 4.7880 | 4.3582 | 0.19 |
The approximations for Uranus aren't very good, because Uranus lies almost on its side: You'll do better to use the full formulas for Uranus.
As seen from Earth, and using equation 11, we find:
α_{sun} = arctan(sin(12.0321°) * cos(23.45°),
cos(12.0321°)) = 11.0639°
, and with equation 12
we find δ_{sun} = arcsin(sin(12.0321°) * sin(23.45°)) =
4.7585°
. As seen from Mars, we find α_{sun} =
arctan(sin(13.0435°) * cos(25.19°), cos(13.0435°)) =
11.8398°
and δ_{sun} = arcsin(sin(13.0435°) *
sin(25.19°)) = 5.5123°
.
Using equations 15 and 17 we find
for the Earth α_{sun} = 12.0321° − 2.468° * sin(2 * 12.0321°)
+ 0.053° * sin(4 * 12.0321°) − 0.0014° * sin(6 * 12.0321°) =
11.0639°
and δ_{sun} = 22.8008° * sin(12.0321°) +
0.5999° * sin(12.0321°)³ + 0.0493° * sin(12.0321°)⁵ =
4.7585°
and for Mars α_{mars} = 13.0435° − 2.8605° *
sin(2 * 13.0435°) + 0.0712° * sin(4 * 13.0435°) − 0.0022° * sin(6 *
13.0435°) = 11.8397°
and δ_{mars} = 24.387° *
sin(13.0435°) + 0.7331° * sin(13.0435°)³ + 0.0706° * sin(13.0435°)⁵
= 5.5123°
, so the approximations yield practically the same
results in these cases as the full equations 11 and
12.
Where a celestial body is in your sky depends on your geographical
coordinates (latitude φ
[phi] north, longitude
l_{w}
west), on the position of the body between the
stars (its equatorial coordinates α
and
δ
), and on the rotation angle of the planet at your
location, relative to the stars. That latter angle is expressed in
the sidereal time θ
(theta). The sidereal time is the
right ascension that is on the celestial meridian at that moment. The
sidereal time is equal to
(Eq. 18) θ = θ₀ + θ₁ * (J − J_{2000}) − l_{w}
with θ₀
and θ₁
from the next table.
θ₀ | θ₁ | |
---|---|---|
Mercury | 13.5964 | 6.1385025 |
Venus | 215.2995 | −1.4813688 |
Earth | 280.1600 | 360.9856235 |
Mars | 313.4803 | 350.89198226 |
Jupiter | 146.0727 | 870.5366420 |
Saturn | 174.3479 | 810.7939024 |
Uranus | 17.9705 | -501.1600928 |
Neptune | 52.3996 | 536.3128492 |
Pluto | 56.3183 | −56.3623195 |
For the Netherlands on Earth we find θ_{earth} = 560529.8477° −
(−5°) = 14.8477°
(after subtracting multiples of 360°) and for
Gusev on Mars θ_{mars} = 544897.8367° − 184.6° =
33.2367°
.
The position of a celestial body in the sky is specified by its
altitude h
above the horizon, and its azimuth
A
. The altitude is 0° at the horizon, +90° in the
zenith (straight over your head), and −90° in the nadir (straight
down). The azimuth is the direction along the horizon, which we
measure from south to west. South has azimuth 0°, west +90°, north
+180°, and east +270° (or −90°, that's the same thing). The altitude
and azimuth are the horizontal coordinates. To calculate the
horizontal coordinates from the equatorial coordinates, you can use
the following formulas:
(Eq. 19) sin A cos h = sin H cos δ
(Eq. 20) cos A cos h = cos H sin φ cos δ − sin δ cos φ
(Eq. 21) sin h = sin φ sin δ + cos φ cos δ cos H
(Eq. 22) H = θ − α
(Eq. 23) A = arctan(sin H, cos H sin φ − tan δ cos φ)
The H
is the hour angle, which indicates how long ago
(measured in sidereal time) the celestial body passed through the
celestial meridian.
For the Netherlands on Earth we find H = 3.7838°
so
A = arctan(sin(3.7838°), cos(3.7838°) * sin(52°) −
tan(4.7585°) * cos(52°)) = 5.1302°
and h =
arcsin(sin(52°) * sin(4.7585°) + cos(52°) * cos(4.7585°) *
cos(3.7838°)) = 42.6542°
. For Gusev on Mars we find H
= 21.3969°
and A = arctan(sin(21.3969°),
cos(21.3969°) * sin(−14.6°) − tan(5.5123°) * cos(−14.6°)) =
131.9648°
and h = arcsin(sin(−14.6°) * sin(5.5123°)
+ cos(−14.6°) * cos(5.5123°) * cos(21.3969°)) = 60.7657°
.
At 12:00 UTC on 1 April 2004, the Sun as seen from the Netherlands
stands about 5° west of south at 43° above the horizon, and as seen
from the Gusev crater on Mars the Sun then stands about 3° south by
northwest at 61° above the horizon.
The transit of a celestial body is the moment at which the body passes
through the celestial meridian. The transit of the Sun is noon, the
middle of the day, at 12 hours solar time. The hour angle H =
H_{target}
of the Sun is then equal to 0. We have
(Eq. 24) θ = α_{sun} + H_{target} mod 360°
We keep H
in the formula for later convenience. Using
this and previous equations, you can find the time of transit by
making a guess for a value of J
for which equation
24 holds, calculating θ
and
α_{sun}
for it, and then checking whether they satisfy
equation 24. If they do not, then you have to adjust
J
. All in all, it is a big search operation for the
correct value of J
. Moreover, equation 24
does not help you to understand which things are most important to the
time of solar transit. You do get such insight if you approximate the
solution by neglecting smaller terms. The answer may then not be as
accurate as the correct solution, but does indicate clearly what the
solution looks like, is usually easier to calculate, and provides an
excellent starting guess in the search for the real value of
J
.
With equations 7, 5, 15, and 18, and after omitting smaller terms, we find
(Eq. 25) J_{transit} ≈ J_{2000} + (H_{target} + M₀ + Π + 180° −
θ₀ + l_{w} + C₁ sin M + A₂ sin(2 L_{sun}))/(θ₁ − M₁) mod 360°/(θ₁ −
M₁) = J_{2000} + J₀ + (H_{target} + l_{w}) * J₃/360° + J₁ sin M + J₂
sin(2 L_{sun}) mod J₃
where
(Eq. 26) L_{sun} = M + Π + 180°
(Eq. 27) J₀ = (M₀ + Π + 180° − θ₀) J₃/360° mod J₃
(Eq. 28) J₁ = C₁ J₃/360°
(Eq. 29) J₂ = A₂ J₃/360°
(Eq. 30) J₃ = 360°/(θ₁ − M₁)
If you already have λ_{Sun}
, then you can use it instead
of L_{Sun}
in equation 25: that is a bit more
accurate. J₀
provides the date and time of a transit
of the Sun. J₁
shows by how much the time of transit
can vary because of the eccentricity e
of the orbit.
J₂
indicates how much the time of transit can change
because of the obliquity ε
of the ecliptic.
J₃
is the average length of the solar day (from one
transit to the next). All values are measured in Earth days of 24
hours.
J₀ | J₁ | J₂ | J₃ | |
---|---|---|---|---|
Mercury | 45.3495 | 11.4556 | 175.9386 | |
Venus | 87.8650 | -0.2516 | 0.0099 | −116.7505 |
Earth | 0.0009 | 0.0053 | -0.0069 | 1.0000000 |
Mars | 0.9044 | 0.0305 | -0.0082 | 1.027491 |
Jupiter | 0.3345 | 0.0064 | 0.4135775 | |
Saturn | 0.0766 | 0.0078 | -0.0040 | 0.4440276 |
Uranus | 0.1027 | -0.0106 | 0.0849 | -0.7183165 |
Neptune | 0.3841 | 0.0019 | -0.0066 | 0.6712575 |
Pluto | 3.8479 | -0.5023 | 0.3045 | −6.386797 |
To find the date and time of a solar transit near Julian Date
J
, you now proceed as follows:
(Eq. 31) n_{(*)} = (J − J_{2000} − J₀)/J₃ − (H_{target} + l_{w})/360°
and then take for n
the whole number nearest to
n_{(*)}
.
(Eq. 32) J_{(*)} = J_{2000} + J₀ + (H_{target} +
l_{w}) * J₃/360° + J₃ * n
This J_{(*)}
is a reasonable estimate for the date and
time of the transit near J
, except that the
J₁
and J₂
corrections are not in it
yet.
M
and L_{sun}
for
J_{(*)}
and then get a better estimate for the date and
time of the solar transit from
(Eq. 33) J_{transit} ≈ J_{(*)} + J₁ sin M + J₂ sin(2 L_{sun})
H
for J_{transit}
and take J_{transit}
+ (H_{target} − H)/360° * J₃
as improved value of
J_{transit}
. You can repeat this until
J_{transit}
no longer changes.J_{transit}
is a Julian Date, which counts days and
which is equal to a whole number at 12:00 UTC (i.e., noon, UTC). So,
a Julian Date like 2453096.9898 that ends in .9898 is .0102 days
before the next whole number, i.e., 0.0102 days = 0.0102*24 = 0.245
hours = 0.0102*24*60 = about 15 minutes before 12:00 UTC, i.e., about
11:45 UTC.For our example, we looked near J = 2453097
. Which
solar transit is closest to that in the Netherlands and in Gusev
crater on Mars? For the Netherlands (l_{w} = −5°
) we
find n_{(*)} = (2453097 − 2451545 − 0.0009)/1 − (−5°)/360° =
1552.013
, so n = 1552
, so J_{(*)} =
2451545 + 0.0009 + (−5°) * 1⁄360° + 1 * 1552 = 2453096.9870
(with H_{target} = 0
. For the value of M
for J_{(*)}
we take the value we found for
J
, because the difference is negligible, so M =
87.1807°
and L_{sun} = 87.1807° + 102.9372° + 180° =
370.1179° = 10.1179°
. With that, we find J_{transit} =
2453096.9869 + 0.0053 * sin(87.1807°) − 0.0069 * sin(2 * 10.1179°) =
2453096.9898
. If you use the repetition method to increase
the accuracy, then you get J_{transit} = 2453096.9895
.
The solar transit at 5° east longitude happens on 1 April 2004 around
11:45 UTC.
For the Gusev crater on Mars (l_{w} = 184,6°
) we find
n_{(*)} = (2453097 − 2451545 − 0.9044)/1.02749 − 184.6°/360° =
1509.084
, so n = 1509
, so J_{(*)} =
2451545 + 0.9044 + 184.6° * 1.02749⁄360° + 1509 * 1.02749 =
2453096.9137
. For M
we take the value that we
found earlier, so M = 112.6531°
and L_{sun} =
112.6531° + 70.9812° + 180° = 363.6343° = 3.6343°
. With that,
we find J_{transit} = 2453096.9137 + 0.0305 * sin(112.6531°) −
0.0082 * sin(2 * 3.6343°) = 2453096.9408
. Using the
repetition method, this becomes J_{transit} =
2453096.9392
. The solar transit in Gusev crater happens on 1
April 2004 around 10:32 UTC.
Above, we calculated when the Sun is highest in the sky, expressed in the Earthly time scale of the Julian Date. With this, you still don't know at what time that is on your clock, and that time scale is definitely not very handy if you are on another planet where the day has a different length from the day on Earth that the Julian Date is based on.
If you are not interested in transforming to times and dates for other planets, then you just want to know at what time the Sun is highest in the sky on your planet according to your clock, which is tied to your solar time and the season on your planet. Solar time is the time determined by the Sun. Three different kinds of solar time are relevant:
The answer to the question at what time the Sun is highest in the sky depends a lot on the time scale that you use. In true solar time, the answer is "always at exactly 12 o'clock", but true solar time is otherwise not very convenient at all. You can't use it if the Sun does not shine or if it is hidden behind clouds, and a mechanical clock or watch would have to be a lot more complicated if it had to show true solar time. In mean solar time, the answer is "on average at 12 o'clock", but it may differ from 12 o'clock from day to day. In official clock time, the answer is "on average at a fixed time", but that fixed time depends on your location, though it is usually near 12 o'clock.
So how large is the difference between the mean solar time and the true solar time? This is called the Equation of Time. The Equation of Time is how much you have to add to true solar time ("sundial time") to get mean solar time ("local time").
We found earlier (equation 24) for the moment at which the
Sun is highest in the sky (transit) that θ = α_{Sun} mod
360°
. For the right ascension α_{Sun}
of the
Sun we found α_{Sun} = L + C + S
. For the sidereal time
θ
measured in degrees we construct a different formula:
(Eq. 34) θ = L + t + 180° mod 360°
Here t
is the mean solar time measured in degrees (0° =
midnight, 180° = noon). The explanation for this equation is as
follows: The difference between the sidereal time and the solar time
reflects the motion of the planet around the Sun so it increases by
360° in a planet year. The sidereal time is tied to the regular
rotation of the planet around its axis, which does not notice the
varying speed at which the planet orbits around the Sun (associated
with eccentricity e
) or the position of the Sun above
or below the celestial equatior (associated with the obliquity
ε
of the ecliptic). Therefore, the sidereal time is
tied to the mean longitude L
of the Sun, which
increases at a fixed rate by 360° during a planet year, and which is
independent of e
and ε
. During the
descending equinox (when λ_{Sun} = L = 180°
) the
sidereal time is equal to the mean solar time, so we need the extra
180° in the formula. During a planet day, θ
increases
by 360° (the planet turns once around its axis) plus or minus a small
amount because the sidereal time runs a little faster or slower than
the mean solar time, but that difference is already caught in
L
.
If we now set θ
equal to α_{Sun}
then we
find
(Eq. 35) L + t + 180° = θ = α_{Sun} = L + C + S mod 360°
or
(Eq. 36) t = 180° + C + S mod 24
The Equation of Time ∆t
is equal to the difference
between t
and 180° (because 180° corresponds to
12:00:00 local mean solar time):
(Eq. 37) ∆t = C + S
If you divide the Equation of Time measured in degrees by 15 then you get the Equation of Time measured in hours.
C
depends on eccentricity e
of the
planet's orbit and on the mean anomaly M
of the planet,
i.e., on where the planet is compared to its perihelion.
S
depends on the obliquity ε
of the
ecliptic of the planet and on the longitude λ_{Sun}
of
the Sun, i.e., on the season. A first-order approximation is:
(Eq. 38) ∆t ≈ C₁ sin M + A₂ sin(2 λ_{Sun})
where we've omitted many smaller terms. The following table shows the
amplitudes, i.e., the largest contribution that the C
and S
terms give to the Equation of Time for each
planet, measured in planet minutes.
C | S | |
---|---|---|
Mercury | 94.5 | 0 |
Venus | 3.1 | 0.1 |
Earth | 7.7 | 9.9 |
Mars | 42.8 | 11.4 |
Jupiter | 22.2 | 0.2 |
Saturn | 25.4 | 13.0 |
Uranus | 21.2 | 178.1 |
Neptune | 4.1 | 14.1 |
Pluto | 114.6 | 69.3 |
For the Earth, the contributions of the orbit (C
) and
the season (S
) are about equally large; for Uranus and
Neptune the influence of the season is much greater than that of the
orbit; and for the other planets the influence of the orbit is much
greater than the influence of the season.
For the hour angle that corresponds to h = 0
we find
(Eq. 39) H = arccos(−tan δ tan φ)
There are two solutions: for sunset H
holds, and for
sunrise −H
. By using formulas found earlier and by
omitting small terms, we find
(Eq. 40) H ≈ 90° + H₁ sin λ_{sun} tan φ + H₃ (sin
λ_{sun})³ tan φ * (3 + (tan φ)²) + H₅ (sin λ_{sun})⁵ tan φ * (15 +
10 (tan φ)² + 3 (tan φ)⁴)
with
(Eq. 41) H₁ = ε − (1⁄6) ε³ + (1⁄120) ε⁵
(Eq. 42) H₃ = (1⁄6) ε³ − (1⁄12) ε⁵
(Eq. 43) H₅ = (1⁄40) ε⁵
where you must measure ε
in radians instead of in
degrees. You transform from degrees to radians by multiplying the
number of degrees by π/180 ≈ 0.017453292
.
The values of H₁
, H₃
, and
H₅
are listed for all planets in the following table,
measured in degrees. For Uranus and Pluto it is doubtful that the
approximations yield reasonable results.
H₁ | H₃ | H₅ | |
---|---|---|---|
Mercury | 0.018 | ||
Venus | 2.643 | 0.001 | |
Earth | 22.801 | 0.600 | 0.016 |
Mars | 24.387 | 0.733 | 0.024 |
Jupiter | 3.115 | 0.002 | |
Saturn | 25.779 | 0.865 | 0.032 |
Uranus | 56.907 | -0.836 | 8.716 |
Neptune | 26.758 | 0.966 | 0.039 |
Pluto | 48.311 | 4.788 | 1.453 |
Sunrise is the moment at which the top of the solar disk touches the horizon in the morning. Sunset is the same, but at night. To calculate the times of sunrise and sunset, you must take into account (besides the effects mentioned earlier) also the following things:
h = 0
), then half of the Sun is
still above the horizon. The size of the Sun must be taken into
account.To compensate for these two effects, you can set h
equal to the h₀
from the following table (measured in
degrees), instead of 0°. The refraction has been added in to the
value for the Earth, but not for the other planets, because they have
either no atmosphere of any consequence, or else they have one that is
so dense that you cannot see the Sun at all from the surface. The
table also gives the average diameter d_{Sun}
of the
solar disk (in degrees).
h₀ | d_{Sun} | |
---|---|---|
Mercury | -0.69 | 1.38 |
Venus | -0.37 | 0.74 |
Earth | -0.83 | 0.53 |
Mars | -0.17 | 0.35 |
Jupiter | -0.05 | 0.10 |
Saturn | -0.03 | 0.06 |
Uranus | -0.01 | 0.03 |
Neptune | -0.01 | 0.02 |
Pluto | -0.01 | 0.01 |
With equation 21 you can then find the hour angle
(Eq. 44) H = arccos((sin h₀ − sin φ sin δ)/(cos φ
cos δ))
Here again, H
holds for sunset and −H
for sunrise. Set H_{target}
equal to the value of
H
for sunset or −H
for sunrise, and then
you can use equations 25ff to calculate the times of sunset
and sunrise, just like for the solar transit.
We found earlier that as seen from the Earth δ_{Sun} =
4.7585°
and α_{Sun} = 11.0639°
. According to
the above table, we have for the Earth h₀ = −0.83°
.
With φ = 52°
we then find from equation 44
that H = arccos((sin(−0.83°) − sin(52°)
sin(4.7585°))/(cos(52°) cos(4.7585°))) = 97.4785°
. As seen
from Mars we found δ_{sun} = 5.5123°
and α_{sun}
= 11.8398°
. With h₀ = −0.17°
for Mars and
φ = −14.6°
for Gusev when then find H =
arccos((sin(−0.17°) − sin(−14.6°) sin(5.5123°))/(cos(−14.6°)
cos(5.5123°))) = 88.7361°
.
For our example, we looked near J = 2453097
. Which
solar transit is closest to that in the Netherlands and in Gusev
crater on Mars? For the Netherlands (l_{w} = −5°
) we
find n_{(*)} = (2453097 − 2451545 − 0.0009)/1 − (−5°)/360° −
97.4785°/360° = 1551.7422
, so n = 1552
, so
J_{(*)} = 2451545 + 0.0009 + (97.4785° + (−5°)) * 1⁄360° + 1 *
1552 = 2453097.2578
. For the value of M
for
J_{(*)}
we take the value we found for J
,
because the difference is small, so M = 87.1807°
and
L_{sun} = 87.1807° + 102.9372° + 180° = 370.1179° =
10.1179°
. With that, we find J_{set} = 2453097.2578
+ 0.0053 * sin(87.1807°) − 0.0069 * sin(2 * 10.1179°) =
2453097.2607
. Or we can estimate that the sunset is
J₃ H/360°
days later than the solar transit we
calculated earlier, and thenwe find J_{set} = 2453096.9895 +
1 * 97.4785°/360° = 2453097.2603
. If you use the repetition
method to increase the accuracy, then you get J_{set} =
2453097.2606
. The sunset at 5° east longitude happens on 1
April 2004 around 18:15 UTC.
For the Gusev crater on Mars (l_{w} = 184.6°
) we find
n_{(*)} = (2453097 − 2451545 − 0.9044)/1.02749 − (88.7361° +
184.6°)/360° = 1508.8375
, so n = 1509
, so
J_{(*)} = 2451545 + 0.9044 + (88.7361° + 184.6°) * 1.02749⁄360° +
1509 * 1.02749 = 2453097.1670
. For M
we take
the value that we found earlier, so M = 112.6531°
and
L_{sun} = 112.6531° + 70.9812° + 180° = 363.6343° =
3.6343°
. With that, we find J_{transit} = 2453097.1670 +
0.0305 * sin(112.6531°) − 0.0082 * sin(2 * 3.6343°) =
2453097.1941
. Using the repetition method, this becomes
J_{transit} = 2453097.1921
. The solar transit in Gusev
crater happens on 1 April 2004 around 16:37 UTC.
If you watch sunrise or sunset from great altitude, then there are two more corrections to be made:
The correction ∆H
to equation 39 that is
necessary because of the large Sun and the atmosphere is approximately
equal to
(Eq. 45) ∆H ≈ −h₀/√(cos(φ)² − sin(δ_{sun})²)
For the Netherlands and Belgium (with φ
about 50°) this
is about 5 to 6 minutes.
To calculate the duration of sunrise or sunset you must calculate the
moment that the top of the solar disk touches the horizon, as
described above (with h₀
), and also the moment that
the bottom of the solar disk touches the horizon. To calculate that
last one, you should use h₀ + d₀
instead of
h₀
.
For places on Earth, we can combine various approximations into the following (with results measured in hours UTC):
(Eq. 46) t_{transit} ≈ 12h00m + l_{w}/15° + 24 * (J₀ + J₁ sin M + J₂ sin
2 L_{Sun}) = 12h01m + l_{w}/15° + 7.6m sin M − 9.9m sin 2 L_{Sun}
(Eq. 47) t_{rise} ≈ t_{transit} − H/15° ≈ 6h00m + l_{w}/15° + 24 * (J₀ + J₁
sin M + J₂ sin 2 L_{Sun}) − (H₁ tan φ sin L_{Sun} + H₃ tan φ (3
+ (tan φ)²) (sin L_{Sun})³)/15° = 6h01m + l_{w}/15° + 7.6m sin M −
9.9m sin 2 L_{Sun} − (1h31m tan φ sin L_{Sun} + 2.2m tan φ (3 +
(tan φ)²) (sin L_{Sun})³ + ∆H/15°)
(Eq. 48) t_{set} ≈ t_{transit} + H/15° ≈ 18h00m + l_{w}/15° + 24 * (J₀ +
J₁ sin M + J₂ sin 2 L_{Sun}) + (H₁ tan φ sin L_{Sun} + H₃ tan φ
(3 + (tan φ)²) (sin L_{Sun})³)/15° = 18h01m + l_{w}/15° + 7.6m sin M
− 9.9m sin 2 L_{Sun} + (1h31m tan φ sin L_{Sun} + 2.2m tan φ (3 +
(tan φ)²) (sin L_{Sun})³ + ∆H/15°)
For a spot at φ = 50°
and l_{w} = −5°
(in
the middle of the Netherlands) we find, approximately, measured in
Central European Time (CET, equal to UTC plus one hour):
(Eq. 49) t_{rise} ≈ 6h34m − 1h48,6m sin L − 13.2m sin(L)³ + 7.6m sin
M − 9.9m sin(2 L)
(Eq. 50) t_{transit} ≈ 12h40m + 7.6m sin M − 9.9m sin(2 L)
(Eq. 51) t_{set} ≈ 18h46m + 1h48.6m sin L + 13.2m sin(L)³ + 7.6m
sin M − 9.9m sin(2 L)
These formulas, though approximations, are reasonably accurate. A comparison, for all days of the year 2000, between the results of the full formulas and the results of the last approximate formulas shows differences of at most 3.5 minutes for the times of sunrise and sunset, and less than 1 minute for the time of transit.
From the preceding formulas we can deduce how the times of sunrise,
transit, and sunset depend on the location (near any chosen spot).
The times of sunrise, transit, and sunset (measured in a fixed time
zone) get earlier by 4 minutes for each degree that you go to the
east. This corresponds to 2.16/cos φ
seconds per
kilometer. The time of transit (i.e., high noon) doed not depend on
the geographical latitude. In addition, sunrise gets earlier and
sunset later by a number of minutes per degree that you go north that
is approximately equal to 1.59 sin(L)/cos(φ)²
and
that is equivalent to about 0.86 sin(L)/cos(φ)²
seconds per kilometer. The shift of the times of sunrise and sunset
with latitude is greatest at the beginning of summer and winter. At
the beginning of spring and autumn it is 0, so then the times of
sunrise and sunset do not depend on the latitude (just like the time
of transit).
The planetarium programs Redshift 3 and Redshift 5 and the current procedure (AA) yield the following values for the azimuth and altitude of the Sun in the sky at JD 2451545.0 (1-1-2000, 12:00 UTC) as seen from the location with latitude and longitude equal to zero on each of the planets:
Redshift 5 | Redshift 3 | AA | ||||
---|---|---|---|---|---|---|
A | h | A | h | A | h | |
Mercury | 89:59:36 | −4:28:58 | 90:00 | −4:21 | 270:00:00 | −4:29:32 |
Venus | 263:39:25 | -70:00:13 | 263:39 | -70:00 | 83:40:01 | -69:59:25 |
Earth | 178:03:33 | +66:57:05 | 177:37 | +66:57 | 357:20:14 | +66:55:48 |
Mars | 233:17:03 | +44:46:23 | 233:21 | +44:41 | 53:08:10 | +44:58:23 |
Jupiter | 275:36:38 | -56:47:58 | 273:19 | +22:27 | 93:19:59 | +23:07:15 |
Saturn | 115:08:60 | +33:21:18 | 115:08 | +33:18 | 294:58:39 | +32:59:52 |
Uranus | 223:07:12 | +45:26:37 | 223:09 | +45:25 | 44:13:28 | +45:46:26 |
Neptune | 217:28:20 | -54:09:28 | 218:03 | -54:32 | 37:50:23 | -54:44:57 |
Pluto | 125:32:00 | -43:59:37 | 125:32 | -44:00 | 305:34:32 | -43:59:11 |
And here are similar results for JD 2453097.0 (1-4-2004, 12:00 UTC):
Redshift 5 | Redshift 3 | AA | ||||
---|---|---|---|---|---|---|
A | h | A | h | A | h | |
Mercury | 89:52:29 | -87:19:22 | 89:01 | -87:06 | 270:00:00 | -87:19:36 |
Venus | 266:46:35 | +35:02:25 | 266:47 | +35:03 | 86:46:32 | +35:02:39 |
Earth | 11:08:57 | +85:07:29 | 11:10 | +85:07 | 194:18:16 | +85:05:20 |
Mars | 77:36:54 | -63:14:20 | 77:46 | -62:54 | 257:35:00 | -63:27:54 |
Jupiter | 92:10:30 | +46:10:15 | 91:36 | +20:10 | 271:36:17 | +20:04:19 |
Saturn | 231:05:23 | +47:32:41 | 231:24 | +47:11 | 50:41:50 | +47:56:45 |
Uranus | 143:35:09 | -72:11:32 | 144:19 | -72:22 | 321:08:05 | -72:43:02 |
Neptune | 173:45:46 | -61:31:02 | 172:54 | -61:57 | 353:01:20 | -61:59:07 |
Pluto | 135:36:03 | -41:02:35 | 135:37 | -41:04 | 315:36:48 | -41:01:38 |
If you take into account that the Redshift programs measure azimuth from the north point but I measure it from the south point (corresponding to a difference of 180°), then the results fit to within about a degree, except that the position from Jupiter as calculated by Redshift 5 is very different from that calculated by either Redshift 3 or myself, whereas the results from Redshift 3 and myself fit nicely. I therefore suspect that Redshift 5 is in error in that case.
http://aa.quae.nl/en/reken/zonpositie.html;
Last updated: 2012-10-18