1. Fundamentals
A celestial body usually orbits the sun in an elliptical
orbit. Perturbations from other planets causes small deviations from this
elliptical orbit, but an unperturbed elliptical orbit can be used as a first
approximation, and sometimes as the final approximation. If the distance from
the Sun to the planet always is the same, then the planet follows a circular
orbit. No planet does this, but the orbits of Venus and Neptune are very close
to circles. Among the planets, Mercury and Pluto have orbits that deviate the
most from a circle, i.e. are the most eccentric. Many asteroids have even more
eccentric orbits, but the most eccentric orbits are to be found among the
comets. Halley's comet, for instance, is closer to the Sun than Venus at
perihelion, but farther away from the Sun than Neptune at aphelion. Some comets
have even more eccentric orbits that are best approximated as a parabola. These
orbits are not closed - a comet following a parabolic orbit passes the Sun only
once, never to return. In reality these orbits are extremely elongated ellipses
though, and these comets will eventually return, sometimes after many
millennia.
The perihelion and aphelion are the points in
the orbit when the planet is closest to and most distant from the Sun. A
parabolic orbit only has a perihemium of course.
The perigee and
apogee are points in the Moon's orbit (or the orbit of an artificial
Earth satellite) which are closest to and most distant from the
Earth.
The celestial sphere is an imaginary sphere around the
observer, at an arbitrary distance.
The celestial equator is the
Earth's equatorial plane projected on the celestial sphere.
The
ecliptic is the plane of the Earth's orbit. This is also the plane of the
Sun's yearly apparent motion. The ecliptic is inclined by approximately 23.4_deg
to the celestial equator. The ecliptic intersects the celestial equator at two
points: The Vernal Point (or "the first point of Aries"), and the Autumnal
Point. The Vernal Point is the point of origin for two different commonly used
celestial coordinates: equatorial coordinates and ecliptic
coordinates.
Right Ascension and Declination are equatorial
coordinates using the celestial equator as a fundamental plane. At the Vernal
Point both the Right Ascension and the Declination are zero. The Right Ascension
is usually measured in hours and minutes, where one revolution is 24 hours
(which means 1 hour equals 15 degrees). It's counted countersunwise along the
celestial equator. The Declination goes from +90 to -90 degrees, and it's
positive north of, and negative south of, the celestial
equator.
Longitude and Latitude are ecliptic coordinates,
which use the ecliptic as a fundamental plane. Both are measured in degrees, and
these coorinates too are both zero at the Vernal Point. The Longitude is counted
countersunwise along the ecliptic. The Latitude is positive north of the
ecliptic. Of course longitude and latitude are also used as terrestial
coordinates, to measure a position of a point on the surface of the
Earth.
Heliocentric, Geocentric, Topocentric. A
position relative to the Sun is heliocentric. If the position is relative to the
center of the Earth, then it's geocentric. A topocentric position is relative to
an observer on the surface of the Earth. Within the aim of our accuracy of 1-2
arc minutes, the difference between geocentric and topocentric position is
negligible for all celestial bodies except the Moon (and some occasional
asteroid which happens to pass very close to the Earth).
The orbital
elements consist of 6 quantities which completely define a circular,
elliptic, parabolic or hyperbolic orbit. Three of these quantities describe the
shape and size of the orbit, and the position of the planet in the
orbit:
a Mean distance, or semi-major axis
e Eccentricity
T Time at perihelion
A cirular orbit has zero eccentricity. An elliptical orbit has an
eccentricity between zero and one. A parabolic orbit has an eccentricity of
exactly one. Finally, a hyperbolic orbit has an eccentricity larger than one. A
parabolic orbit has an infinite semi-major axis, a, therefore one instead gives
the perihelion distance, q, for a parabolic orbit:
q Perihelion distance = a * (1 - e)
It is customary to give q instead of a also for hyperbolic orbit, and for
elliptical orbits with eccentricity close to one.
The three remaining
orbital elements define the orientation of the orbit in space:
i Inclination, i.e. the "tilt" of the orbit relative to the
ecliptic. The inclination varies from 0 to 180 degrees. If
the inclination is larger than 90 degrees, the planet is in
a retrogade orbit, i.e. it moves "backwards". The most
well-known celestial body with retrogade motion is Comet Halley.
N (usually written as "Capital Omega") Longitude of Ascending
Node. This is the angle, along the ecliptic, from the Vernal
Point to the Ascending Node, which is the intersection between
the orbit and the ecliptic, where the planet moves from south
of to north of the ecliptic, i.e. from negative to positive
latitudes.
w (usually written as "small Omega") The angle from the Ascending
node to the Perihelion, along the orbit.
These are the primary orbital elements. From these many secondary orbital
elements can be computed:
q Perihelion distance = a * (1 - e)
Q Aphelion distance = a * (1 + e)
P Orbital period = 365.256898326 * a**1.5/sqrt(1+m) days,
where m = the mass of the planet in solar masses (0 for
comets and asteroids). sqrt() is the square root function.
n Daily motion = 360_deg / P degrees/day
t Some epoch as a day count, e.g. Julian Day Number. The Time
at Perihelion, T, should then be expressed as the same day count.
t - T Time since Perihelion, usually in days
M Mean Anomaly = n * (t - T) = (t - T) * 360_deg / P
Mean Anomaly is 0 at perihelion and 180 degrees at aphelion
L Mean Longitude = M + w + N
E Eccentric anomaly, defined by Kepler's equation: M = E - e * sin(E)
An auxiliary angle to compute the position in an elliptic orbit
v True anomaly: the angle from perihelion to the planet, as seen
from the Sun
r Heliocentric distance: the planet's distance from the Sun.
This relation is valid for an elliptic orbit:
r * cos(v) = a * (cos(E) - e)
r * sin(v) = a * sqrt(1 - e*e) * sin(E)
x,y,z Rectangular coordinates. Used e.g. when a heliocentric
position (seen from the Sun) should be converted to a
corresponding geocentric position (seen from the Earth).
3. Rectangular and spherical coordinates
The position of a planet can be
given in one of several ways. Two different ways that we'll use are rectangular
and spherical coordinates.
Suppose a planet is situated at some RA, Decl
and r, where RA is the Right Ascension, Decl the declination, and r the distance
in some length unit. If r is unknown or irrelevant, set r = 1. Let's convert
this to rectangular coordinates, x,y,z:
x = r * cos(RA) * cos(Decl)
y = r * sin(RA) * cos(Decl)
z = r * sin(Decl)
(before we compute the sine/cosine of RA, we must first convert RA from
hours/minutes/seconds to hours + decimals. Then the hours are converted to
degrees by multiplying by 15)
If we know the rectangular coordinates, we
can convert to spherical coordinates by the formulae below:
r = sqrt( x*x + y*y + z*z )
RA = atan2( y, x )
Decl = asin( z / r ) = atan2( z, sqrt( x*x + y*y ) )
At the north and south celestial poles, both x and y are zero. Since
atan2(0,0) is undefined, the RA is undefined too at the celestial poles. The
simplest way to handle this is to assign RA some arbitrary value, e.g. zero.
Close to the celestial poles the formula asin(z/r) to compute the declination
becomes sensitive to round-off errors - here the formula atan2(z,sqrt(x*x+y*y))
is preferable.
Not only equatorial coordinates can be converted between
spherical and rectangular. These conversions can also be applied to ecliptic and
horizontal coordinates. Just exchange RA,Decl with long,lat (ecliptic
coordinates) or azimuth,altitude (horizontal coordinates).
A coordinate
system can be rotated. If a rectangular coordinate system is rotated around,
say, the X axis, one can easily compute the new x,y,z coordinates. As an
example, let's consider rotating an ecliptic x,y,z system to an equatorial x,y,z
system. This rotation is done around the X axis (which points to the Vernal
Point, the common point of origin in ecliptic and equatorial coordinates),
through an angle of oblecl (the obliquity of the ecliptic, which is
approximately 23.4 degrees):
xequat = xeclip
yequat = yeclip * cos(oblecl) - zeclip * sin(oblecl)
zequat = yeclip * sin(oblecl) + zeclip * cos(oblecl)
Now the x,y,z system is equatorial. It's easily rotated back to ecliptic
coordinates by simply switching sign on oblecl:
xeclip = xequat
yeclip = yequat * cos(-oblecl) - zequat * sin(-oblecl)
zeclip = yequat * sin(-oblecl) + zequat * cos(-oblecl)
When computing sin and cos of -oblecl, one can use the identities:
cos(-x) = cos(x), sin(-x) = -sin(x)
Now let's put this together to convert directly from spherical ecliptic
coordinates (long, lat) to spherical equatorial coordinates (RA, Decl). Since
the distance r is irrelevant in this case, let's set r=1 for
simplicity.
Example: At the Summer Solstice the Sun's ecliptic longitude
is 90 degrees. The Sun's ecliptic latitude is always very nearly zero. Suppose
the obliquity of the ecliptic is 23.4 degrees:
xeclip = cos(90_deg) * cos(0_deg) = 0.0000
yeclip = sin(90_deg) * cos(0_deg) = 1.0000
zeclip = sin(0_deg) = 0.0000
Rotate through oblecl = 23.4_deg:
xequat = 0.0000
yequat = 1.0000 * cos(23.4_deg) - 0.0000 * sin(23.4_deg)
zequat = 1.0000 * sin(23.4_deg) + 0.0000 * cos(23.4_deg)
Our equatorial rectangular coordinates become:
x = 0
y = cos(23.4_deg) = 0.9178
z = sin(23.4_deg) = 0.3971
The "distance", r, becomes: sqrt( 0.8423 + 0.1577 ) = 1.0000 i.e.
unchanged
RA = atan2( 0.9178, 0 ) = 90_deg
Decl = asin( 0.3971 / 1.0000 ) = 23.40_deg
Alternatively:
Decl = atan2( 0.3971, sqrt( 0.8423 + 0.0000 ) ) = 23.40_deg
Here we immediately see how simple it is to compute RA, thanks to the
atan2() function: no need to consider in which quadrant it falls, the atan2()
function handles this.
5. The Sun's position.
Today most people know that the Earth orbits the
Sun and not the other way around. But below we'll pretend as if it was the other
way around. These orbital elements are thus valid for the Sun's (apparent) orbit
around the Earth. All angular values are expressed in degrees:
w = 282.9404_deg + 4.70935E-5_deg * d (longitude of perihelion)
a = 1.000000 (mean distance, a.u.)
e = 0.016709 - 1.151E-9 * d (eccentricity)
M = 356.0470_deg + 0.9856002585_deg * d (mean anomaly)
We also need the obliquity of the ecliptic, oblecl:
oblecl = 23.4393_deg - 3.563E-7_deg * d
and the Sun's mean longitude, L:
L = w + M
By definition the Sun is (apparently) moving in the plane of the ecliptic.
The inclination, i, is therefore zero, and the longitude of the ascending node,
N, becomes undefined. For simplicity we'll assign the value zero to N, which
means that w, the angle between acending node and perihelion, becomes equal to
the longitude of the perihelion.
Now let's compute the Sun's position for
our test date 19 april 1990. Earlier we've computed d = -3543.0 which
yields:
w = 282.7735_deg
a = 1.000000
e = 0.016713
M = -3135.9347_deg
We immediately notice that the mean anomaly, M, will get a large negative
value. We use our function rev() to reduce this value to between 0 and 360
degrees. To do this, rev() will need to add 9*360 = 3240 degrees to this
angle:
M = 104.0653_deg
We also compute:
L = w + M = 386.8388_deg = 26.8388_deg
oblecl = 23.4406_deg
Let's go on computing an auxiliary angle, the eccentric anomaly. Since the
eccentricity of the Sun's (i.e. the Earth's) orbit is so small, 0.017, a first
approximation of E will be accurate enough. Below E and M are in
degrees:
E = M + (180/pi) * e * sin(M) * (1 + e * cos(M))
When we plug in M and e, we get:
E = 104.9904_deg
Now we compute the Sun's rectangular coordinates in the plane of the
ecliptic, where the X axis points towards the perihelion:
x = r * cos(v) = cos(E) - e
y = r * sin(v) = sin(E) * sqrt(1 - e*e)
We plug in E and get:
x = -0.275370
y = +0.965834
Convert to distance and true anomaly:
r = sqrt(x*x + y*y)
v = arctan2( y, x )
Numerically we get:
r = 1.004323
v = 105.9134_deg
Now we can compute the longitude of the Sun:
lon = v + w
lon = 105.9134_deg + 282.7735_deg = 388.6869_deg = 28.6869_deg
We're done!
How close did we get to the correct values? Let's
compare with the Astronomical Almanac:
Our results Astron. Almanac Difference
lon 28.6869_deg 28.6813_deg +0.0056_deg = 20"
r 1.004323 1.004311 +0.000012
The error in the Sun's longitude was 20 arc seconds, which is well below
our aim of an accuracy of one arc minute. The error in the distance was about
1/3 Earth radius. Not bad!
Finally we'll compute the Sun's ecliptic
rectangular coordinates, rotate these to equatorial coordinates, and then
compute the Sun's RA and Decl:
x = r * cos(lon)
y = r * sin(lon)
z = 0.0
We plug in our longitude:
x = 0.881048
y = 0.482098
z = 0.0
We use oblecl = 23.4406 degrees, and rotate these coordinates:
xequat = 0.881048
yequat = 0.482098 * cos(23.4406_deg) + 0.0 * sin(23.4406_deg)
zequat = 0.482098 * sin(23.4406_deg) + 0.0 * cos(23.4406_deg)
which yields:
xequat = 0.881048
yequat = 0.442312
zequat = 0.191778
Convert to RA and Decl:
r = 1.004323 (unchanged)
RA = 26.6580_deg = 26.6580/15 h = 1.77720 h = 1h 46m 37.9s
Decl = +11.0084_deg = +11_deg 0' 30"
The Astronomical Almanac says:
RA = 1h 46m 36.0s Decl = +11_deg 0' 22"
6. Sidereal time and hour angle. Altitude and azimuth
The Sidereal Time
tells the Right Ascension of the part of the sky that's precisely south, i.e. in
the meridian. Sidereal Time is a local time, which can be computed from:
SIDTIME = GMST0 + UT + LON/15
where SIDTIME, GMST0 and UT are given in hours + decimals. GMST0 is the
Sidereal Time at the Greenwich meridian at 00:00 right now, and UT is the same
as Greenwich time. LON is the terrestial longitude in degrees (western longitude
is negative, eastern positive). To "convert" the longitude from degrees to hours
we divide it by 15. If the Sidereal Time becomes negative, we add 24 hours, if
it exceeds 24 hours we subtract 24 hours.
Now, how do we compute GMST0?
Simple - we add (or subtract) 180 degrees to (from) L, the Sun's mean longitude,
which we've already computed earlier. Then we normalise the result to between 0
and 360 degrees, by applying the rev() function. Finally we divide by 15 to
convert degrees to hours:
GMST0 = ( L + 180_deg ) / 15 = L/15 + 12h
We've already computed L = 26.8388_deg, which yields:
GMST0 = 26.8388_deg/15 + 12h = 13.78925 hours
Now let's compute the local Sidereal Time for the time meridian of Central
Europe (at 15 deg east longitude = +15 degrees long) on 19 april 1990 at 00:00
UT:
SIDTIME = GMST0 + UT + LON/15 = 13.78925h + 0 + 15_deg/15 = 14.78925 hours
SIDTIME = 14h 47m 21.3s
To compute the altitude and azimuth we also need to know the Hour Angle,
HA. The Hour Angle is zero when the clestial body is in the meridian i.e. in the
south (or, from the southern heimsphere, in the north) - this is the moment when
the celestial body is at its highest above the horizon.
The Hour Angle
increases with time (unless the object is moving faster than the Earth rotates;
this is the case for most artificial satellites). It is computed from:
HA = SIDTIME - RA
Here SIDTIME and RA must be expressed in the same unit, hours or degrees.
We choose hours:
HA = 14.78925h - 1.77720h = 13.01205h = 195.1808_deg
If the Hour Angle is 180_deg the celestial body can be seen (or not be
seen, if it's below the horizon) in the north (or in the south, from the
southern hemisphere). We get HA = 195_deg for the Sun, which seems OK since it's
around 01:00 local time.
Now we'll convert the Sun's HA = 195.1808_deg
and Decl = +11.0084_deg to a rectangular (x,y,z) coordinate system where the X
axis points to the celestial equator in the south, the Y axis to the horizon in
the west, and the Z axis to the north celestial pole: The distance, r, is here
irrelevant so we set r=1 for simplicity:
x = cos(HA) * cos(Decl) = -0.947346
y = sin(HA) * cos(Decl) = -0.257047
z = sin(Decl) = +0.190953
Now we'll rotate this x,y,z system along an axis going east-west, i.e. the
Y axis, in such a way that the Z axis will point to the zenith. At the North
Pole the angle of rotation will be zero since there the north celestial pole
already is in the zenith. At other latitudes the angle of rotation becomes
90_deg - latitude. This yields:
xhor = x * cos(90_deg - lat) - z * sin(90_deg - lat)
yhor = y
zhor = x * sin(90_deg - lat) + z * cos(90_deg - lat)
Since sin(90_deg-lat) = cos(lat) (and reverse) we'll get:
xhor = x * sin(lat) - z * cos(lat)
yhor = y
zhor = x * cos(lat) + z * sin(lat)
Finally we compute our azimuth and altitude:
azimuth = atan2( yhor, xhor ) + 180_deg
altitude = asin( zhor ) = atan2( zhor, sqrt(xhor*xhor+yhor*yhor) )
Why did we add 180_deg to the azimuth? To adapt to the most common way to
specify azimuth: from North (0_deg) through East (90_deg), South (180_deg), West
(270_deg) and back to North. If we didn't add 180_deg the azimuth would be
counted from South through West/etc instead. If you want to use that kind of
azimuth, then don't add 180_deg above.
We select some place in central
Scandinavia: the longitude is as before +15_deg (15_deg East), and the latitude
is +60_deg (60_deg N):
xhor = -0.947346 * sin(60_deg) - (+0.190953) * cos(60_deg) = -0.915902
yhor = -0.257047 = -0.257047
zhor = -0.947346 * cos(60_deg) + (+0.190953) * sin(60_deg) = -0.308303
Now we've computed the horizontal coordinates in rectangular form. To get
azimuth and altitude we convert to spherical coordinates (r=1):
azimuth = atan2(-0.257047,-0.915902) + 180_deg = 375.6767_deg = 15.6767_deg
altitude = asin( -0.308303 ) = -17.9570_deg
Let's round the final result to two decimals:
azimuth = 15.68_deg, altitude = -17.96_deg.
The Sun is thus 17.96_deg below the horizon at this moment and place. This
is very close to astronomical twilight (18_deg below the horizon).
7. The Moon's position
Let's continue by computing the position of the
Moon. The computatons will become more complicated, since the Moon doesn't move
in the plane of the ecliptic, but in a plane inclined somewhat more than 5
degrees to the ecliptic. Also, the Sun perturbs the Moon's motion significantly,
an effect we must account for.
The orbital elements of the Moon
are:
N = 125.1228_deg - 0.0529538083_deg * d (Long asc. node)
i = 5.1454_deg (Inclination)
w = 318.0634_deg + 0.1643573223_deg * d (Arg. of perigee)
a = 60.2666 (Mean distance)
e = 0.054900 (Eccentricity)
M = 115.3654_deg + 13.0649929509_deg * d (Mean anomaly)
The Moon's ascending node is moving in a retrogade ("backwards") direction
one revolution in about 18.6 years. The Moon's perigee (the point of the orbit
closest to the Earth) moves in a "forwards" direction one revolution in about
8.8 years. The Moon itself moves one revolution in aboout 27.5 days. The mean
distance, or semi-major axis, is expressed in Earth equatorial
radii).
Let's compute numerical values for the Moon's orbital elements on
our test date 19 april 1990 (d = -3543):
N = 312.7381_deg
i = 5.1454_deg
w = -264.2546_deg
a = 60.2666 (Earth equatorial radii)
e = 0.054900
M = -46173.9046_deg
Now the need for sufficient numerical accuracy becomes obvious. If we
would compute M with normal single precision, i.e. only 7 decimal digits of
accuracy, then the error in M would here be about 0.01 degrees. Had we selected
a date around 1901 or 2099 then the error in M would have been about 0.1
degrees, which is definitely worse than our aim of a maximum error of one or two
arc minutes. Therefore, when computing the Moon's mean anomaly, M, it's
important to use at least 9 or 10 digits of accuracy.
(This was a real
problem around 1980, when microcomputers were a novelty. Around then, pocket
calculators usually offered better precision than microcomputers. But these days
are long gone: nowadays microcomputers routinely offer double precision (14-16
digits of accuracy) support in hardware; all you need to do is to select a
compiler which really suports this.)
All angular elements should be
normalized to within 0-360 degrees, by calling the rev() function. We
get:
N = 312.7381_deg
i = 5.1454_deg
w = 95.7454_deg
a = 60.2666 (Earth equatorial radii)
e = 0.054900
M = 266.0954_deg
To normalize M we had to add 129*360 = 46440 degrees.
Next, we
compute E, the eccentric anomaly. We start with a first approximation (E0 and M
in degrees):
E0 = M + (180_deg/pi) * e * sin(M) * (1 + e * cos(M))
The eccentricity of the Moon's orbit is larger than of the Earth's orbit.
This means that our first approximation will have a bigger error - it'll be
close to the limit of what we can tolerate within our accuracy aim. If you want
to be careful, you should therefore use the iteration formula below: set E0 to
our first approximation, compute E1, then set E0 to E1 and compute a new E1,
until the difference between E0 and E1 becomes small enough, i.e. smaller than
about 0.005 degrees. Then use the last E1 as the final value. In the formula
below, E0, E1 and M are in degrees:
E1 = E0 - (E0 - (180_deg/pi) * e * sin(E0) - M) / (1 - e * cos(E0))
On our test date, the first approximation of E becomes: E=262.9689_deg The
iterations then yield: E = 262.9735_deg, 262.9735_deg ......
Now we've
computed E - the next step is to compute the Moon's distance and true anomaly.
First we compute rectangular (x,y) coordinates in the plane of the lunar
orbit:
x = r * cos(v) = a * (cos(E) - e)
y = r * sin(v) = a * sqrt(1 - e*e) * sin(E)
Our test date yields:
x = -10.68095
y = -59.72377
Then we convert this to distance and true anonaly:
r = sqrt( x*x + y*y ) = 60.67134 Earth radii
v = atan2( y, x ) = 259.8605_deg
Now we know the Moon's position in the plane of the lunar orbit. To
compute the Moon's position in ecliptic coordinates, we apply these
formulae:
xeclip = r * ( cos(N) * cos(v+w) - sin(N) * sin(v+w) * cos(i) )
yeclip = r * ( sin(N) * cos(v+w) + cos(N) * sin(v+w) * cos(i) )
zeclip = r * sin(v+w) * sin(i)
Our test date yields:
xeclip = +37.65311
yeclip = -47.57180
zeclip = -0.41687
Convert to ecliptic longitude, latitude, and distance:
long = 308.3616_deg
lat = -0.3937_deg
r = 60.6713
According to the Astronomical Almanac, the Moon's position at this moment
is 306.94_deg, and the latitude is -0.55_deg. This differs from our figures by
1.42_deg in longitude and 0.16_deg in latitude!!! This difference is much larger
than our aim of an error of max 1-2 arc minutes. Why is this so?
8. The Moon's position with higher accuracy. Perturbations
The big error
in our computed lunar position is because we completely ignored the
perturbations on the Moon. Below we'll compute the most important perturbation
terms, and then add these as corrections to our previous figures. This will cut
down the error to 1-2 arc minutes, or less.
First we need several
fundamental arguments:
Sun's mean longitude: Ls (already computed)
Moon's mean longitude: Lm = N + w + M (for the Moon)
Sun's mean anomaly: Ms (already computed)
Moon's mean anomaly: Mm (already computed)
Moon's mean elongation: D = Lm - Ls
Moon's argument of latitude: F = Lm - N
Let's plug in the figures for our test date:
Ms = 104.0653_deg
Mm = 266.0954_deg
Ls = 26.8388_deg
Lm = 312.7381_deg + 95.7454_deg + 266.0954_deg = 674.5789_deg
= 314.5789_deg
D = 314.5789_deg - 26.8388_deg = 287.7401_deg
F = 314.5789_deg - 312.7381_deg = 1.8408_deg
Now it's time to compute and add up the 12 largest perturbation terms in
longitude, the 5 largest in latitude, and the 2 largest in distance. These are
all the perturbation terms with an amplitude larger than 0.01_deg in longitude
resp latitude. In the lunar distance, only the perturbation terms larger than
0.1 Earth radii has been included:
Perturbations in longitude
(degrees):
-1.274_deg * sin(Mm - 2*D) (Evection)
+0.658_deg * sin(2*D) (Variation)
-0.186_deg * sin(Ms) (Yearly equation)
-0.059_deg * sin(2*Mm - 2*D)
-0.057_deg * sin(Mm - 2*D + Ms)
+0.053_deg * sin(Mm + 2*D)
+0.046_deg * sin(2*D - Ms)
+0.041_deg * sin(Mm - Ms)
-0.035_deg * sin(D) (Parallactic equation)
-0.031_deg * sin(Mm + Ms)
-0.015_deg * sin(2*F - 2*D)
+0.011_deg * sin(Mm - 4*D)
Perturbations in latitude (degrees):
-0.173_deg * sin(F - 2*D)
-0.055_deg * sin(Mm - F - 2*D)
-0.046_deg * sin(Mm + F - 2*D)
+0.033_deg * sin(F + 2*D)
+0.017_deg * sin(2*Mm + F)
Perturbations in lunar distance (Earth radii):
-0.58 * cos(Mm - 2*D)
-0.46 * cos(2*D)
Some of the largest perturbation terms in longitude even have received
individual names! The largest perturbation, the Evection, was discovered already
by Ptolemy (he made it one of the epicycles in his theory for the Moon's
motion). The two next largest perturbations, the Variation and the Yearly
equation, were discovered by Tycho Brahe.
If you don't need 1-2 arcmin
accuracy, you don't need to compute all these perturbation terms. If you only
include the two largest terms in longitude and the largest in latitude, the
error in longitude rarely becomes larger than 0.25_deg, and in latitude rarely
larger than 0.15_deg.
Let's compute these perturbation terms for our test
date:
longitude: -0.9847 - 0.3819 - 0.1804 + 0.0405 - 0.0244 + 0.0452 +
0.0428 + 0.0126 - 0.0333 - 0.0055 - 0.0079 - 0.0029
= -1.4132_deg
latitude: -0.0958 - 0.0414 - 0.0365 - 0.0200 + 0.0018 = -0.1919_deg
distance: -0.3680 + 0.3745 = +0.0066 Earth radii
Add this to the ecliptic positions we earlier computed:
long = 308.3616_deg - 1.4132_deg = 306.9484_deg
lat = -0.3937_deg - 0.1919_deg = -0.5856_deg
dist = 60.6713 + 0.0066 = 60.6779 Earth radii
Let's compare with the Astronomical Almanac:
longitude 306.94_deg, latitude -0.55_deg, distance 60.793 Earth radii
Now the agreement is much better, right?
Let's continue by
converting these ecliptic coordinates to Right Ascension and Declination. We do
as described earlier: convert the ecliptic longitude/latitude to rectangular
(x,y,z) coordinates, rotate this x,y,z, system through an angle corresponding to
the obliquity of the ecliptic, then convert back to spherical coordinates. The
Moon's distance doesn't matter here, and one can therefore set r=1.0. We
get:
RA = 309.5011_deg
Decl = -19.1032_deg
According to the Astronomical Almanac:
RA = 309.4881_deg
Decl = -19.0741_deg
9. The Moon's topocentric position.
The Moon's position, as computed
earlier, is geocentric, i.e. as seen by an imaginary observer at the center of
the Earth. Real observers dwell on the surface of the Earth, though, and they
will see a different position - the topocentric position. This position can
differ by more than one degree from the geocentric position. To compute the
topocentric positions, we must add a correction to the geocentric
position.
Let's start by computing the Moon's parallax, i.e. the apparent
size of the (equatorial) radius of the Earth, as seen from the Moon:
mpar = asin( 1/r )
where r is the Moon's distance in Earth radii. It's simplest to apply the
correction in horizontal coordinates (azimuth and altitude): within our accuracy
aim of 1-2 arc minutes, no correction need to be applied to the azimuth. One
need only apply a correction to the altitude above the horizon:
alt_topoc = alt_geoc - mpar * cos(alt_geoc)
Sometimes one needs to correct for topocentric position directly in
equatorial coordinates though, e.g. if one wants to draw on a star map how the
Moon passes in front of the Pleiades, as seen from some specific location. Then
we need to know the Moon's geocentric Right Ascension and Declination (RA,
Decl), the Local Sidereal Time (LST), and our latitude (lat).
Our
astronomical latitude (lat) must first be converted to a geocentric latitude
(gclat) and distance from the center of the Earth (rho) in Earth equatorial
radii. If we only want an approximate topocentric position, it's simplest to
pretend that the Earth is a perfect sphere, and simply set:
gclat = lat, rho = 1.0
However, if we do wish to account for the flattening of the Earth, we
instead compute:
gclat = lat - 0.1924_deg * sin(2*lat)
rho = 0.99833 + 0.00167 * cos(2*lat)
Next we compute the Moon's geocentric Hour Angle (HA):
HA = LST - RA
We also need an auxiliary angle, g:
g = atan( tan(gclat) / cos(HA) )
Now we're ready to convert the geocentric Right Ascention and Declination
(RA, Decl) to their topocentric values (topRA, topDecl):
topRA = RA - mpar * rho * cos(gclat) * sin(HA) / cos(Decl)
topDecl = Decl - mpar * rho * sin(gclat) * sin(g - Decl) / sin(g)
Let's do this correction for our test date and for the geographical
position 15 deg E longitude (= +15_deg) and 60 deg N latitude (= +60_deg).
Earlier we computed the Local Sidereal Time as LST = SIDTIME = 14.78925 hours.
Multiply by 15 to get degrees: LST = 221.8388_deg
The Moon's Hour Angle
becomes:
HA = LST - RA = -87.6623_deg = 272.3377_deg
Our latitude +60_deg yields the following geocentric latitude and distance
from the Earth's center:
gclat = 59.83_deg rho = 0.9975
We've already computed the Moon's distance as 60.6779 Earth radii, which
means the Moon's parallax is:
mpar = 0.9443_deg
The auxiliary angle g becomes:
g = 88.642
And finally the Moon's topocentric position becomes:
topRA = 309.5011_deg - (-0.5006_deg) = 310.0017_deg
topDecl = -19.1032_deg - (+0.7758_deg) = -19.8790_deg
This correction to topocentric position can also be applied to the Sun and
the planets. But since they're much farther away, the correction becomes much
smaller. It's largest for Venus at inferior conjunction, when Venus' parallax is
somewhat larger than 32 arc seconds. Within our aim of obtaining a final
accuracy of 1-2 arc minutes, it might barely be justified to correct to
topocentric position when Venus is close to inferior conjunction, and perhaps
also when Mars is at a favourable opposition. But in all other cases this
correction can safely be ignored within our accuracy aim. We only need to worry
about the Moon in this case.
If you want to compute topocentric
coordinates for the planets anyway, you do it the same way as for the Moon, with
one exception: the parallax of the planet (ppar) is computed from this
formula:
ppar = (8.794/3600)_deg / r
where r is the distance of the planet from the Earth, in astronomical
units.
10. The orbital elements of the planets
To compute the positions of the
major planets, we first must compute their orbital
elements:
Mercury:
N = 48.3313_deg + 3.24587E-5_deg * d (Long of asc. node)
i = 7.0047_deg + 5.00E-8_deg * d (Inclination)
w = 29.1241_deg + 1.01444E-5_deg * d (Argument of perihelion)
a = 0.387098 (Semi-major axis)
e = 0.205635 + 5.59E-10 * d (Eccentricity)
M = 168.6562_deg + 4.0923344368_deg * d (Mean anonaly)
Venus:
N = 76.6799_deg + 2.46590E-5_deg * d
i = 3.3946_deg + 2.75E-8_deg * d
w = 54.8910_deg + 1.38374E-5_deg * d
a = 0.723330
e = 0.006773 - 1.302E-9 * d
M = 48.0052_deg + 1.6021302244_deg * d
Mars:
N = 49.5574_deg + 2.11081E-5_deg * d
i = 1.8497_deg - 1.78E-8_deg * d
w = 286.5016_deg + 2.92961E-5_deg * d
a = 1.523688
e = 0.093405 + 2.516E-9 * d
M = 18.6021_deg + 0.5240207766_deg * d
Jupiter:
N = 100.4542_deg + 2.76854E-5_deg * d
i = 1.3030_deg - 1.557E-7_deg * d
w = 273.8777_deg + 1.64505E-5_deg * d
a = 5.20256
e = 0.048498 + 4.469E-9 * d
M = 19.8950_deg + 0.0830853001_deg * d
Saturn:
N = 113.6634_deg + 2.38980E-5_deg * d
i = 2.4886_deg - 1.081E-7_deg * d
w = 339.3939_deg + 2.97661E-5_deg * d
a = 9.55475
e = 0.055546 - 9.499E-9 * d
M = 316.9670_deg + 0.0334442282_deg * d
Uranus:
N = 74.0005_deg + 1.3978E-5_deg * d
i = 0.7733_deg + 1.9E-8_deg * d
w = 96.6612_deg + 3.0565E-5_deg * d
a = 19.18171 - 1.55E-8 * d
e = 0.047318 + 7.45E-9 * d
M = 142.5905_deg + 0.011725806_deg * d
Neptune:
N = 131.7806_deg + 3.0173E-5_deg * d
i = 1.7700_deg - 2.55E-7_deg * d
w = 272.8461_deg - 6.027E-6_deg * d
a = 30.05826 + 3.313E-8 * d
e = 0.008606 + 2.15E-9 * d
M = 260.2471_deg + 0.005995147_deg * d
Let's compute all these elements for our test date, 19 april 1990 0h
UT:
N i w a e M
deg deg deg a.e. deg
Mercury 48.2163 7.0045 29.0882 0.387098 0.205633 69.5153
Venus 76.5925 3.3945 54.8420 0.723330 0.006778 131.6578
Mars 49.4826 1.8498 286.3978 1.523688 0.093396 321.9965
Jupiter 100.3561 1.3036 273.8194 5.20256 0.048482 85.5238
Saturn 113.5787 2.4890 339.2884 9.55475 0.055580 198.4741
Uranus 73.9510 0.7732 96.5529 19.18176 0.047292 101.0460
Neptune 131.6737 1.7709 272.8675 30.05814 0.008598 239.0063
11. The heliocentric positions of the planets
The heliocentric ecliptic
coordinates of the planets are computed in the same way as we computed the
geocentric ecliptic coordinates of the Moon: first we compute E, the eccentric
anomaly. Several planetary orbits have quite high eccentricities, which means we
must use the iteration formula to obtain an accurate value of E. When we know E,
we compute, as earlier, the distance r ("radius vector") and the true anomaly,
v. Then we compute ecliptic rectangular (x,y,z) coordinates as we did for the
Moon. Since the Moon orbits the Earth, this position of the Moon was geocentric.
The planets though orbit the Sun, which means we get heliocentric positions
instead. Also the semi-major axis, a, and the distance, r, which was given in
Earth radii for the Moon, are given in astronomical units for the planets, where
one astronomical unit is 149.6 million kilometers.
Let's do this for
Mercury on our test date: the first approximation of E is 81.3464_deg, and
following iterations yield 81.1572_deg, 81.1572_deg .... From this we
find:
r = 0.374862
v = 93.0727_deg
Mercury's heliocentric ecliptic rectangular coordinates become:
x = -0.367821
y = +0.061084
z = +0.038699
Convert to spherical coordinates:
lon = 170.5709_deg
lat = +5.9255_deg
r = 0.374862
The Astronomical Almanac gives these figures:
lon = 170.5701_deg
lat = +5.9258_deg
r = 0.374856
The agreement is almost perfect! The discrepancy is only a few arc
seconds. This is because it's quite easy to get an accurate position for
Mercury: it's close to the Sun where the Sun's gravitational field is strongest,
and therefore perturbations from the other planets will be smallest for
Mercury.
If we compute the heliocentric longitude, latitude and distance
for the other planets from their orbital elements, we get:
Heliocentric
longitude latitude distance
lon lat r
Mercury 170.5709_deg +5.9255_deg 0.374862
Venus 263.6570_deg -0.4180_deg 0.726607
Mars 290.6297_deg -1.6203_deg 1.417194
Jupiter 105.2543_deg +0.1113_deg 5.19508
Saturn 289.4523_deg +0.1792_deg 10.06118
Uranus 276.7999_deg -0.3003_deg 19.39628
Neptune 282.7192_deg +0.8575_deg 30.19284
For e.g. Saturn, the Astronomical Almanac says:
lon = 289.3864_deg
lat = +0.1816_deg
r = 10.01850
The difference is here much larger! For Mercury our discrepancy was only a
few arc seconds, but for Saturn it's up to four arc minutes! And still we've
been lucky, since sometimes the discrepancy can be up to one full degree for
Saturn. This is the planet that's perturbed most severely, mostly by
Jupiter.
12. Higher accuracy - perturbations
To reach our aim of a final accuracy
of 1-2 arc minutes, we must compute Jupiter's and Saturn's perturbations on each
other, and their perturbations on Uranus. The perturbations on, and by, other
planets can be ignored, with our aim for 1-2 arcmin accuracy.
First we
need three fundamental arguments:
Jupiters mean anomaly: Mj
Saturn mean anomaly: Ms
Uranus mean anomaly: Mu
Then these terms should be added to Jupiter's heliocentric
longitude:
-0.332_deg * sin(2*Mj - 5*Ms - 67.6_deg)
-0.056_deg * sin(2*Mj - 2*Ms + 21_deg)
+0.042_deg * sin(3*Mj - 5*Ms + 21_deg)
-0.036_deg * sin(Mj - 2*Ms)
+0.022_deg * cos(Mj - Ms)
+0.023_deg * sin(2*Mj - 3*Ms + 52_deg)
-0.016_deg * sin(Mj - 5*Ms - 69_deg)
For Saturn we must correct both the longitude and latitude. Add this to
Saturn's heliocentric longitude:
+0.812_deg * sin(2*Mj - 5*Ms - 67.6_deg)
-0.229_deg * cos(2*Mj - 4*Ms - 2_deg)
+0.119_deg * sin(Mj - 2*Ms - 3_deg)
+0.046_deg * sin(2*Mj - 6*Ms - 69_deg)
+0.014_deg * sin(Mj - 3*Ms + 32_deg)
and to Saturn's heliocentric latitude these terms should be added:
-0.020_deg * cos(2*Mj - 4*Ms - 2_deg)
+0.018_deg * sin(2*Mj - 6*Ms - 49_deg)
Finally, add this to Uranus heliocentric longitude:
+0.040_deg * sin(Ms - 2*Mu + 6_deg)
+0.035_deg * sin(Ms - 3*Mu + 33_deg)
-0.015_deg * sin(Mj - Mu + 20_deg)
The perturbation terms above are all terms having an amplitude of 0.01
degrees or more. We ignore all perturbations in the distances of the planets,
since a modest perturbation in distance won't affect the apparent position very
much.
The largest perturbation term, "the grand Jupiter-Saturn term" is
the perturbation in longitude with the largest amplitude in both Jupiter and
Saturn. Its period is 918 years, and its amplitude is a large part of a degree
for both Jupiter and Saturn. There is also a "grand Saturn-Uranus term", which
has a period of 560 years and an amplitude of 0.035 degrees for Uranus, but less
than 0.01 degrees for Saturn. Other included terms have periods between 14 and
100 years. Finally we should mention the "grand Uranus-Neptune term", which has
a period if 4200 years and an amplitude of almost one degree. It's not included
in our perturbation terms here, instead its effects have been included in the
orbital elements for Uranus and Neptune. This is why the mean distances of
Uranus and Neptune are varying.
If we compute the perturbations for our
test date, we get:
Mj = 85.5238_deg Ms = 198.4741_deg Mu = 101.0460:
Perturbations in Jupiter's longitude:
+ 0.0637_deg - 0.0236_deg + 0.0038_deg - 0.0270_deg - 0.0086_deg
- 0.0049_deg - 0.0155_deg = -0.0120_deg
Jupiter's heliocentric longitude, with perturbations: 105.2423_deg
The Astronomical Almanac says: 105.2603_deg
Perturbations in Saturn's longitude:
-0.1560_deg + 0.0206_deg + 0.0850_deg - 0.0070_deg - 0.0124_deg
= - 0.0699_deg
Perturbations in Saturn's latitude:
+0.0018_deg + 0.0035_deg = +0.0053_deg
Saturns position, with perturbations: 289.3824_deg +0.1845_deg
The Astronomical Almanac says: 289.3864_deg +0.1816_deg
Perturbations in Uranus' longitude:
+0.0017_deg - 0.0332_deg - 0.0012_deg = -0.0327_deg
Uranus heliocentric longitude, with perturbations: 276.7672_deg
The Astronomical Almanac says: 276.7706_deg
15. The elongation and physical ephemerides of the planets
When we
finally have completed our computation of the heliocentric and geocentric
coordinates of the planets, it could also be interesting to know what the planet
will look like. How large will it appear? What are its phase and magnitude
(brightness)? These computations are much simpler than the computations of the
positions.
Let's start by computing the apparent diameter of the
planet:
d = d0 / R
R is the planet's geocentric distance in astronomical units, and d is the
planet's apparent diameter at a distance of 1 astronomical unit. d0 is of course
different for each planet. The values below are given in seconds of arc. Some
planets have different equatorial and polar diameters:
Mercury 6.74"
Venus 16.92"
Earth 17.59" equ 17.53" pol
Mars 9.36" equ 9.28" pol
Jupiter 196.94" equ 185.08" pol
Saturn 165.6" equ 150.8" pol
Uranus 65.8" equ 62.1" pol
Neptune 62.2" equ 60.9" pol
The Sun's apparent diameter at 1 astronomical unit is 1919.26". The Moon's
apparent diameter is:
d = 1873.7" * 60 / r
where r is the Moon's distance in Earth radii.
Two other quantities
we'd like to know are the phase angle and the elongation.
The phase angle
tells us the phase: if it's zero the planet appears "full", if it's 90 degrees
it appears "half", and if it's 180 degrees it appears "new". Only the Moon and
the inferior planets (i.e. Mercury and Venus) can have phase angles exceeding
about 50 degrees.
The elongation is the apparent angular distance of the
planet from the Sun. If the elongation is smaller than ca 20 degrees, the planet
is hard to observe, and if it's smaller than ca 10 degrees it's usually not
possible to observe the planet.
To compute phase angle and elongation we
need to know the planet's heliocentric distance, r, its geocentric distance, R,
and the distance to the Sun, s. Now we can compute the phase angle, FV, and the
elongation, elong:
elong = acos( ( s*s + R*R - r*r ) / (2*s*R) )
FV = acos( ( r*r + R*R - s*s ) / (2*r*R) )
When we know the phase angle, we can easily compute the phase:
phase = ( 1 + cos(FV) ) / 2 = hav(180_deg - FV)
hav(FV) is the "haversine" of the phase angle. The "haversine" (or "half
versine") is an old and now obsolete trigonometric function; it's defined
as:
hav(x) = ( 1 - cos(x) ) / 2 = sin^2 (x/2)
As usual we must use a different procedure for the Moon. Since the Moon is
so close to the Earth, the procedure above would introduce too big errors.
Instead we use the Moon's ecliptic longitude and latitude, mlon and mlat, and
the Sun's ecliptic longitude, mlon, to compute first the elongation, then the
phase angle, of the Moon:
elong = acos( cos(slon - mlon) * cos(mlat) )
FV = 180_deg - elong
Finally we'll compute the magnitude (or brightness) of the planets. Here
we need to use a formula that's different for each planet. The phase angle, FV,
is in degrees:
Mercury: -0.36 + 5*log10(r*R) + 0.027 * FV + 2.2E-13 * FV**6
Venus: -4.34 + 5*log10(r*R) + 0.013 * FV + 4.2E-7 * FV**3
Mars: -1.51 + 5*log10(r*R) + 0.016 * FV
Jupiter: -9.25 + 5*log10(r*R) + 0.014 * FV
Saturn: -9.0 + 5*log10(r*R) + 0.044 * FV + ring_magn
Uranus: -7.15 + 5*log10(r*R) + 0.001 * FV
Neptune: -6.90 + 5*log10(r*R) + 0.001 * FV
** is the power operator, thus FV**6 is the phase angle (in degrees)
raised to the sixth power. If FV is 150 degrees, then FV**6 becomes ca 1.14E+13,
which is a quite large number.
Saturn needs special treatment due to its
rings: when Saturn's rings are "open" then Saturn will appear much brighter than
when we view Saturn's rings edgewise. We'll compute ring_mang like this:
ring_magn = -2.6 * sin(abs(B)) + 1.2 * (sin(B))**2
Here B is the tilt of Saturn's rings which we also need to compute. Then
we start with Saturn's geocentric ecliptic longitude and latitude (los, las)
which we've already computed. We also need the tilt of the rings to the
ecliptic, ir, and the "ascending node" of the plane of the rings, Nr:
ir = 28.06_deg
Nr = 169.51_deg + 3.82E-5_deg * d
Here d is our "day number" which we've used so many times before. For our
test date d = -3543. Now let's compute the tilt of the rings:
B = asin( sin(las) * cos(ir) - cos(las) * sin(ir) * sin(los-Nr) )
This concludes our computation of the magnitudes of the planets.
16. The positions of comets. Comet Encke and Levy.
If you want to
compute the position of a comet or an asteroid, you must have access to orbital
elements that still are valid. One set of orbital elements isn't valid forever.
For instance if you try to use the 1986 orbital elements of comet Halley to
compute its appearance in either 1910 or 2061, you'll get very large errors in
your computed positions - sometimes the errors will be 90 degrees or
more.
Comets will usually have a new set of orbital elements computed for
each perihelion. The comets are perturbed most severely when they're close to
aphelion, far away from the gravity of the Sun but maybe much closer to Jupiter,
Saturn, Uranus or Neptune. When the comet is passing through the inner solar
system, the perturbations are usually so small that the same set of orbital
elements can be used for the entire apparition.
Orbital elements for an
asteroid should preferably not be more than about one year old. If your accuracy
requirements are lower, you can of course use older elements. If you use orbital
elements that are five years old for a main-belt asteroid, then your computed
positions can be several degrees in error. If the orbital elements are less than
one year old, the errors usually stay below approximately one arc minute, for a
main-belt asteroid.
If you have access to valid orbital elements for a
comet or an asteroid, proceed as below to compute its position at some
date:
1. If necessary, precess the angular elements N,w,i to the epoch of
today. The simples way to do this is to add the precession angle to N, the
longitude of the ascending node. This method is approximate, but it's good
enough for our accuracy aim of 1-2 arc minutes.
2. Compute the day number
for the time or perihelion, call it D. Then compute the number of days since
perihelion, d - D (before perihelion this number is of course
negative).
3. If the orbit is elliptical, compute the Mean Anomaly, M.
Then compute r, the heliocentric distance, and v, the true anomaly.
4. If
the orbit is a parabola, or close to a parabola (the eccentricity is 1.0 or
nearly 1.0), then the algorithms for elliptical orbits will break down. Then use
another algorithm, presented below, to compute r, the heliocentric distance, and
v, the true anomaly, for near-parabolic orbits.
5. When you know r and v,
proceed as with the planets: compute first the heliocentric, then the
geocentric, position.
6. If needed, precess the final position to the
desired epoch, e.g. 2000.0
A quantity we'll encounter here is Gauss'
gravitational constant, k. This constant links the Sun's mass with our time unit
(the day) and the length unit (the astronomical unit). The EXACT value of Gauss'
gravitational constant k is:
k = 0.01720209895 (exactly!)
If the orbit is elliptical, and if the perihelion distance, q, is given
instead of the mean distance, a, we start by computing the mean distance a from
the perihelion distance q and the eccentricity e:
a = q / (1 - e)
Now we compute the Mean Anomaly, M:
M = (180_deg/pi) * (d - D) * k / (a ** 1.5)
a ** 1.5 is most easily computed as: sqrt(a*a*a)
Now we know the Mean Anomaly, M. We proceed as for a planetary orbit by
computing E, the eccentric anomaly. Since comet and asteroid orbits often have
high eccentricities, we must use the iteration formula given earlier, and be
sure to iterate until we get convergence for the value of E.
The orbital
period for a comet or an asteroid in elliptic orbit is (P in days):
P = 2 * pi * (a ** 1.5) / k
If the comet's orbit is a parabola, the algorithm for elliptic orbits will
break down: the semi-major axis and the orbital period will be infinite, and the
Mean Anomaly will be zero. Then we must proceed in a different way. For a
parabolic orbit we start by computing the quantities a, b and w (where a is not
at all related to a for an elliptic orbit):
a = 1.5 * (d - D) * k / sqrt(2 * q*q*q)
b = sqrt( 1 + a*a )
w = cbrt(b + a) - cbrt(b - a)
cbrt is the Cubic Root function. Finally we compute the true anomaly, v,
and the heliocentric distiance, r:
v = 2 * atan(w)
r = q * ( 1 + w*w )
From here we can proceed as usual.
Finally we have the case that's
most common for newly discovered comets: the orbit isn't an exact parabola, but
very nearly so. It's eccentricity is slightly below, or slightly above, one. The
algorithm presented here can be used for eccentricities between about 0.98 and
1.02. If the eccentricity is smaller than 0.98 the elliptic algorithm should be
used instead. No known comet has an eccentricity exceeding 1.02.
As for
the purely parabolic orbit, we start by computing the time since perihelion in
days, d - D, and the perihelion distance, q. We also need to know the
eccentricity, e. Then we can proceed as:
a = 0.75 * (d - D) * k * sqrt( (1 + e) / (q*q*q) )
b = sqrt( 1 + a*a )
W = cbrt(b + a) - cbrt(b - a)
f = (1 - e) / (1 + e)
a1 = (2/3) + (2/5) * W*W
a2 = (7/5) + (33/35) * W*W + (37/175) * W**4
a3 = W*W * ( (432/175) + (956/1125) * W*W + (84/1575) * W**4 )
C = W*W / (1 + W*W)
g = f * C*C
w = W * ( 1 + f * C * ( a1 + a2*g + a3*g*g ) )
v = 2 * atan(w)
r = q * ( 1 + w*w ) / ( 1 + w*w * f )
This algorithm yields the true anomaly, v, and the heliocentric distance,
r, for a nearly-parabolic orbit.
Now it's time for a practical example.
Let's select two of the comets that were seen in the autumn of 1990: Comet
Encke, a well-known periodic comet, and comet Levy, which was easily seen
towards a dark sky in the autumn of 1990. When passing the inner solar system,
the orbit of comet Levy was slightly hyperbolic.
According the the
Handbook of the British Astronomical Association the orbital elements for comet
Encke in 1990 are:
T = 1990 Oct 28.54502 TDT
e = 0.8502196
q = 0.3308858
w = 186.24444_deg
N = 334.04096_deg 1950.0
i = 11.93911_deg
The orbital elements for comet Levy are (BAA Circular 704):
T = 1990 Oct 24.6954 ET
e = 1.000270
q = 0.93858
w = 242.6797_deg
N = 138.6637_deg 1950.0
i = 131.5856_deg
Let's also choose another test date, when both these comets were visible:
1990 Aug 22, 0t UT, which yields a "day number" d = -3418.0
Now we
compute the day numbers at perihelion for these two comets. We get for comet
Encke:
D = -3350.45498 d - D = -67.54502
and for comet Levy:
D = -3354.3046 d - D = -63.6954
We'll continue by computing the Mean Anomaly for comet Encke:
M = -20.2751_deg = 339.7249_deg
The first approximation plus successive approximation for the Eccentric
anomaly, E, becomes (degrees):
E = 309.3811 293.5105 295.8474 295.9061 295.9061_deg ....
Here we clearly see the great need for iteration: the initial
approximation differs from the final value by 14 degrees. Finally we compute the
true anomaly, v, and heliocentric distance, r, for comet Encke:
v = 228.8837_deg
r = 1.3885
Now it's time for comet Levy: we'll compute the true anomaly, v, and the
heliocentric distance, r, for Levy in two different ways. First we'll pretend
that the orbit of Levy is an exact parabola. We get:
a = -1.2780823 b = 1.6228045 w = -0.7250189
v = -71.8856_deg
r = 1.431947
Then we repeat the computation but accounts for the fact that Levy's orbit
deviates slightly from a parabola. We get:
a = -1.2781686 b = 1.6228724 W = -0.7250566
c = 2.9022000 f = -1.3498E-4 g = -1.60258E-5
a1= 0.8769495 a2= 1.9540987 a3= 1.5403455
w = -0.7250270
v = -71.8863_deg
r = 1.432059
The difference is small in this case - only 0.0007 degrees or 2.5 arc
seconds in true anomaly, and 0.000112 a.u. in heliocentric distance. Here it
would have been sufficient to treat Levy's orbit as an exact
parabola.
Now we know the true anomaly, v, and the heliocentric distance,
r, for both Encke and Levy. Next we proceed by precessing N, the longitude of
the ascending node, from 1950.0 to the "epoch of the day". Let's compute the
precession angle from 1950.0 to 1990 Aug 22:
prec = 3.82394E-5_deg * ( 365.2422 * ( 1950.0 - 2000.0 ) - (-3418) )
prec = -0.5676_deg
To precess from 1990 Aug 22 to 1950.0, we should add this angle to N. But
now we want to do the opposite: precess from 1950.0 to 1990 Aug 22, therefore we
must instead subtract this angle:
For comet Encke we get:
N = 334.04096_deg - (-0.5676_deg) = 334.60856_deg
and for comet Levy we get:
N = 138.6637_deg - (-0.5676_deg) = 139.2313_deg
Using this modified value for N we proceed just like for the planets. I
won't repeat the details, but merely state some intermediate and final
results:
Sun's position: x = -0.863890 y = +0.526123
Heliocentric: Encke Levy
x +1.195087 +1.169908
y +0.666455 -0.807922
z +0.235663 +0.171375
Geoc., eclipt.: Encke Levy
x +0.331197 +0.306018
y +1.192579 -0.281799
z +0.235663 +0.171375
Geoc., equat.: Encke Levy
x +0.331197 +0.306018
y +1.000414 -0.326716
z +0.690619 +0.045133
RA 71.6824_deg 313.1264_deg
Decl +33.2390_deg +5.7572_deg
R 1.259950 0.449919
These positions are for the "epoch of the day". If you want positions for
some standard epoch, e.g. 2000.0, these positions must be precessed to that
epoch.
Finally some notes about computing the magnitude of a comet. To
accurately predict a comet's magnitude is usually hard and sometimes impossible.
It's fairly common that a magnitude prediction is off by 1-2 magnitudes or even
more. For comet Levy the magnitude formula looked like this:
m = 4.0 * 5*log10(R) + 10*log10(r)
where R is the geocentric distance and r the heliocentric distance. The
general case is:
m = G * 5*log10(R) + H*log10(r)
where H usually is around 10. If H is unknown, it's usually assumed to be
10. Each comet has it's own G and H.
Some comets have a different
magnitude formula. One good example is comet Encke, where the magnitude formula
looks like this:
m1 = 10.8 + 5*log10(R) + 3.55 * ( r**1.8 - 1 )
"m1" refers to the total magnitude of the comet. There is another cometary
magnitude, "m2", which refers to the magnitude of the nucleus of the comet. The
magnitude formula for Encke's m2 magnitude looks like this:
m2 = 14.5 + 5*log10(R) + 5*log10(r) + 0.03*FV
Here FV is the phase angle. This kind of magnitude formula looks very much
like the magnitude formula of asteroids, for a very good reason: when a comet is
far away form the Sun, no gases are evaporated from the surface of the comet.
Then the comet has no tail (of course) and no coma, only a nucleus. Which means
the comet then behaves much like an asteroid.
During the last few years
it's become increasingly obvious that comets and asteroids often are similar
kinds of solar-system objects. The asteroid (2060) Chiron has displayed cometary
activity and is now also considered a comet. And in some cases comets that have
"disappeared" have been re-discovered as asteroids! Apparently they "ran out of
gas" and what remains of the former comet is only rock, i.e. an
asteroid.