email: pausch@stjarnhimlen.se or WWW: http://stjarnhimlen.se/

Break out of a frame

- 1. Fundamentals
- 2. Some useful functions
- 3. Rectangular and spherical coordinates
- 4. The time scale. A test date
- 5. The Sun's position
- 6. Sidereal time and hour angle. Altitude and azimuth
- 7. The Moon's position
- 8. The Moon's position with higher accuracy. Perturbations
- 9. The Moon's topocentric position
- 10. The orbital elements of the planets
- 11. The heliocentric positions of the planets
- 12. Higher accuracy - perturbations
- 13. Precession
- 14. Geocentric positions of the planets
- 15. The elongation and physical ephemerides of the planets
- 16. The positions of comets. Comet Encke and Levy

Computing rise and set times

Today it's not really that difficult to compute a planet's position from its orbital elements. The only thing you'll need is a computer and a suitable program. If you want to write such a program yourself, this text contains the formulae you need. The aim here is to obtain the planetary positions at any date during the 20'th and 21'st century with an error of one or at the most two arc minutes, and to compute the position of an asteroid or a comet from its orbital elements.

No programs are given, because different computers and calculators are programmed in different languages. It is easier to convert formulae to your favourite language than to translate a program from one programming language to another. Therefore formulae are presented instead. Each formula is accompanied with one numeric example - this will enable you to check your implementation of these formulae. Just remember that all numerical quantities contain rounding errors, therefore you may get slightly different results in your program compared to the numerical results presented here. When I checked these numerical values I used a HP-48SX pocket calculator, which uses 12 digits of accuracy.

The

The

The

The

The

The

a Mean distance, or semi-major axis e Eccentricity T Time at perihelionA 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. 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).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)

On microcomputers the situation is worse. Let's start with the programming language Basic: there we can count on having sin/cos/tan and atn (=arctan) in radians, but nothing more. arcsin and arccos are missing, and the trig functions don't work in degrees. However one can add one's own function library, like e.g. this:

95 rem Constants 100 pi = 3.14159265359 110 radeg = 180/pi 115 rem arcsin and arccos 120 def fnasin(x) = atn(x/sqr(1-x*x)) 130 def fnacos(x) = pi/2 - fnasin(x) 135 rem Trig. functions in degrees 140 def fnsind(x) = sin(x/radeg) 150 def fncosd(x) = cos(x/radeg) 160 def fntand(x) = tan(x/radeg) 170 def fnasind(x) = radeg*atn(x/sqr(1-x*x)) 180 def fnacosd(x) = 90 - fnasind(x) 190 def fnatand(x) = radeg*atn(x) 195 rem arctan in all four quadrants 200 def fnatan2(y,x) = atn(y/x) - pi*(x<0) 210 def fnatan2d(y,x) = radeg*atn(y/x) - 180*(x<0) 215 rem Normalize an angle between 0 and 360 degrees 217 rem Use Double Precision, if possible 220 def fnrev#(x#) = x# - int(x#/360#)*360# 225 rem Cube Root (needed for parabolic orbits) 230 def fncbrt(x) = exp( log(x)/3 )The code above follows the convensions of traditional Microsoft Basic (MBASIC/BASICA/GWBASIC). If you use some other Basic dialect, you may want to modify this code.

The code above gives you sin/cos/tan and their inverses, in radians and degrees. The functions fnatan2() and fnatan2d() may need some explanation: they take two arguments, x and y, and they compute arctan(y/x) but puts the angle in the correct quadrant from -180 to +180 degrees. This is really part of a conversion from rectangular to polar coordinates, where the angle is computed. The distance is then computed by:

sqrt( x*x + y*y )The function fnrev# normalizes an angle to between 0 and 360 degrees, by adding or subtracting even mutiples of 360 degrees until the final value falls between 0 and 360. The # sign means that the function and its argument use double precision. It is essential that this function uses more than single precision. More about this later.

One warning: some of these functions may divide by zero. If one tries to compute fnasin(1.0), it's computed as: atn(1/sqrt(1-1.0*1.0)) = atn(1/sqr(0)) = atn(1/0). This is not such a big problem, since in practice one rarely tries to compute the arc sine of exactly 1.0. Also, some dialects of Microsoft Basic then just print the warning message "Overflow", compute 1/0 as the highest possible floating-point number, and then continue the program. When computing the arctan of this very large number, one will get pi/2 (or 90 degrees), which is the correct result. However, if you have access to a more modern, structured, Basic, with the ability to define multi-line function, then by all means use this to write a better version of arcsin, which treats arcsin(1.0) as a special case.

There's a similar problem with the fncbrt() function - it only works for positive values. With a multi-line function definition it can be rewritten to work for negative values and zero as well, if one follows these simple rules: the Cube Root of zero is of course zero. The Cube Root of a negative number is computed by making the number positive, taking the Cube Root of that positive number, and then negating the result.

Two other popular programming language are C and C++. The standard library of these languages are better equipped with trigonometric functions. You'll find sin/cos/tan and their inverses, and even an atan2() function among the standard functions. All you need to do is to define some macros to get the trig functions in degrees (include

#define PI 3.14159265358979323846 #define RADEG (180.0/PI) #define DEGRAD (PI/180.0) #define sind(x) sin((x)*DEGRAD) #define cosd(x) cos((x)*DEGRAD) #define tand(x) tan((x)*DEGRAD) #define asind(x) (RADEG*asin(x)) #define acosd(x) (RADEG*acos(x)) #define atand(x) (RADEG*atan(x)) #define atan2d(y,x) (RADEG*atan2((y),(x))) double rev( double x ) { return x - floor(x/360.0)*360.0; } double cbrt( double x ) { if ( x > 0.0 ) return exp( log(x) / 3.0 ); else if ( x < 0.0 ) return -cbrt(-x); else /* x == 0.0 */ return 0.0; }In C++ the macros could preferably be defined as inline functions instead - this enables better type checking and also makes overloading of these function names possible.

The good ol' programming langauge FORTRAN is also well equipped with standard library trig functions: we find sin/cos/tan + inverses, and also an atan2 but only for radians. Therefore we need several function definitions to get the trig functions in degrees too. Below I give code only for SIND, ATAND, ATAN2D, plus REV and CBRT. The remaining functions COSD, TAND, ASIND and ACOSD are written in a similar way:

FUNCTION SIND(X) PARAMETER(RADEG=57.2957795130823) SIND = SIN( X * (1.0/RADEG) ) END FUNCTION ATAND(X) PARAMETER(RADEG=57.2957795130823) ATAND = RADEG * ATAN(X) END FUNCTION ATAN2D(Y,X) PARAMETER(RADEG=57.2957795130823) ATAN2D = RADEG * ATAN2(Y,X) END FUNCTION REV(X) REV = X - AINT(X/360.0)*360.0 IF (REV.LT.0.0) REV = REV + 360.0 END FUNCTION CBRT(X) IF (X.GE.0.0) THEN CBRT = X ** (1.0/3.0) ELSE CBRT = -((-X)**(1.0/3.0)) ENDIF ENDThe programming language Pascal is not as well equipped with trig functions. We have sin, cos, tan and arctan but nothing more. Therefore we need to write our own arcsin, arccos and arctan2, plus all the trig functions in degrees, and also the functions rev and cbrt. The trig functions in degrees are trivial when the others are defined, therefore I only define arcsin, arccos, arctan2, rev and cbrt:

const pi = 3.14159265358979323846; half_pi = 1.57079632679489661923; function arcsin( x : real ) : real; begin if x = 1.0 then arcsin := half_pi else if x = -1.0 then arcsin := -half_pi else arcsin := arctan( x / sqrt( 1.0 - x*x ) ) end; function arccos( x : real ) : real; begin arccos := half_pi - arcsin(x); end; function arctan2( y, x : real ) : real; begin if x = 0.0 then begin if y = 0.0 then (* Error! Give error message and stop program *) else if y > 0.0 then arctan2 := half_pi else arctan2 := -half_pi end else begin if x > 0.0 then arctan2 := arctan( y / x ) else if x < 0.0 then begin if y >= 0.0 then arctan2 := arctan( y / x ) + pi else arctan2 := arctan( y / x ) - pi end; end; end; function rev( x : real ) : real; var rv : real; begin rv := x - trunc(x/360.0)*360.0; if rv < 0.0 then rv := rv + 360.0; rev := rv; end; function cbrt( x : real ) : real; begin if x > 0.0 then cbrt := exp ( ln(x) / 3.0 ) else if x < 0.0 then cbrt := -cbrt(-x) else cbrt := 0.0 end;It's well worth the effort to ensure that all these functions are available. Then you don't need to worry about these details which really don't have much to do with the problem of computing a planetary position.

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) = 0.0000Rotate 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) = 0.3971The "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_degAlternatively:

Decl = atan2( 0.3971, sqrt( 0.8423 + 0.0000 ) ) = 23.40_degHere 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.

We call our day number d. It can be computed from a JD (Julian Day Number) or MJD (Modified Julian Day Number) like this:

d = JD - 2451543.5 = MJD - 51543.0We can also compute d directly from the calendar date like this:

d = 367*Y - (7*(Y + ((M+9)/12)))/4 + (275*M)/9 + D - 730530

Follow

Y is the year (all 4 digits!), M the month (1-12) and D the date. In this formula all divisions should be INTEGER divisions. Use "div" instead of "/" in Pascal, and "\" instead of "/" in Microsoft Basic. In C/C++ and FORTRAN it's sufficient to ensure that both operands to "/" are integers.

This formula yields d as an integer, which is valid at the start (at midnight), in UT or TDT, of that calendar date. If you want d for some other time, add UT/24.0 (here the division is a floating-point division!) to the d obtained above.

Example: compute d for 19 april 1990, at 0:00 UT :

We can look up, or compute the JD for this moment, and we'll get: JD = 2448000.5 which yields d = -3543.0

Or we can compute d directly from the calendar date:

d = 367*1990 - (7*(1990 + ((4+9)/12)))/4 + (275*4)/9 + 19 - 730530 d = 730330 - (7*(1990 + (13/12)))/4 + 1100/9 + 19 - 730530 d = 730330 - (7*(1990 + 1))/4 + 122 + 19 - 730530 d = 730330 - (7*1991)/4 + 122 + 19 - 730530 d = 730330 - 13937/4 + 122 + 19 - 730530 d = 730330 - 3484 + 122 + 19 - 730530 = -3543This moment, 1990 april 19, 0:00 UT/TDT, will be our test date for most numerical examples below. d is negative since our test date, 19 april 1990, is earlier than the "point of origin" of our day number, 31 dec 1999.

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 * dand the Sun's mean longitude, L:

L = w + MBy 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_degWe 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_degWe also compute:

L = w + M = 386.8388_deg = 26.8388_deg oblecl = 23.4406_degLet'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_degNow we compute the Sun's rectangular coordinates in the plane of the ecliptic, where the X axis points towards the perihelion:

x = cos(E) - e y = sin(E) * sqrt(1 - e*e)We plug in E and get:

x = -0.275370 y = +0.965834Convert to distance and true anomaly:

r = sqrt(x*x + y*y) v = arctan2( y, x )Numerically we get:

r = 1.004323 v = 105.9134_degNow we can compute the longitude of the Sun:

lon = v + w lon = 105.9134_deg + 282.7735_deg = 388.6869_deg = 28.6869_degWe'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 +0.000012The 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 = 0.0We plug in our longitude and r:

x = 0.881048 y = 0.482098 z = 0.0We 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 = 0.191778Convert to RA and Decl:

r = sqrt( xequat*xequat + yequat*yequat + zequat*zequat ) RA = atan2( yequat, xequat ) Decl = atan2( zequat, sqrt( xequat*xequat + yequat*yequat) ) 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"

SIDTIME = GMST0 + UT + LON/15where 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 + 12hWe've already computed L = 26.8388_deg, which yields:

GMST0 = 26.8388_deg/15 + 12h = 13.78925 hoursNow 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 21.3sTo 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 - RAHere 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_degIf 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) = +0.190953Now 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) = -0.308303Now 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_degLet'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).

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_degNow 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_degTo 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 = a * (cos(E) - e) y = a * sqrt(1 - e*e) * sin(E)Our test date yields:

x = -10.68095 y = -59.72377Then we convert this to distance and true anonaly:

r = sqrt( x*x + y*y ) = 60.67134 Earth radii v = atan2( y, x ) = 259.8605_degNow 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 = -0.41687Convert to ecliptic longitude, latitude, and distance:

long = atan2( yeclip, xeclip ) lat = atan2( zeclip, sqrt( xeclip*xeclip + yeclip*yeclip ) ) r = sqrt( xeclip*xeclip + yeclip*yeclip + zeclip*zeclip ) long = 308.3616_deg lat = -0.3937_deg r = 60.6713According 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?

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 - NLet'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_degNow 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 radiiAdd 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 r = 60.6713 + 0.0066 = 60.6779 Earth radiiLet's compare with the Astronomical Almanac:

longitude 306.94_deg, latitude -0.55_deg, distance 60.793 Earth radiiNow the agreement is much better, right?

Let's continue by converting these ecliptic coordinates to Right Ascension and Declination. We do as described earlier for the Sun: 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 start by computing the obliquity of the ecliptic, or to simply re-use the value of oblecl we've already computed for the Sun's position:

oblecl = 23.4393_deg - 3.563E-7_deg * d = 23.4406_degNext, we compute the Moon's rectangular ecliptic coordinates and rotate them to get rectangular equatorial coordinates and then RA and Decl:

xeclip = r * cos(long) * cos(lat) yeclip = r * sin(long) * cos(lat) zeclip = r * sin(lat) xequat = xeclip yequat = yeclip * cos(oblecl) - zeclip * sin(oblecl) zequat = yeclip * sin(oblecl) + zeclip * cos(oblecl) RA = atan2( yequat, xequat ) Decl = atan2( zequat, sqrt( xequat*xequat + yequat*yequat ) )To simplify we set r = 1.0 and then we get:

xeclip = cos(306.9484_deg) * cos(-0.5856_deg) = 0.601064 yeclip = sin(306.9484_deg) * cos(-0.5856_deg) = -0.799135 zeclip = sin(-0.5856_deg) = -0.010220 xequat = 0.601064 = 0.601064 yequat = -0.799135 * cos(23.4406_deg) - -0.010220 * sin(23.4406_deg) = -0.729119 zequat = -0.799135 * sin(23.4406_deg) + -0.010220 * cos(23.4406_deg) = -0.327271 RA = atan2( -0.729119, 0.601064 ) = -50.4988_deg = 309.5011_deg Decl = atan2( -0.327271, sqrt( 0.601064*0.601064 + -0.729119*-0.729119) ) = -19.1032_degThe Moon's position according to the Astronomical Almanac is:

RA = 309.4881_deg Decl = -19.0741_deg

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 = 1.0However, 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 - RAWe 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_degOur latitude +60_deg yields the following geocentric latitude and distance from the Earth's center:

gclat = 59.83_deg rho = 0.9975We've already computed the Moon's distance as 60.6779 Earth radii, which means the Moon's parallax is:

mpar = 0.9443_degThe auxiliary angle g becomes:

g = 88.642And 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_degThis 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 / rwhere r is the distance of the planet from the Earth, in astronomical units.

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 * dMars:

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 * dJupiter:

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 * dSaturn:

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 * dUranus:

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 * dNeptune:

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 * dLet'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

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_degMercury's heliocentric ecliptic rectangular coordinates become:

x = -0.367821 y = +0.061084 z = +0.038699Convert to spherical coordinates:

lon = 170.5709_deg lat = +5.9255_deg r = 0.374862The Astronomical Almanac gives these figures:

lon = 170.5701_deg lat = +5.9258_deg r = 0.374856The 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 30.19284For e.g. Saturn, the Astronomical Almanac says:

lon = 289.3864_deg lat = +0.1816_deg r = 10.01850The 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.

First we need three fundamental arguments:

Jupiters mean anomaly: Mj Saturn mean anomaly: Ms Uranus mean anomaly: MuThen 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_degPerturbations in Saturn's longitude:

-0.1560_deg + 0.0206_deg + 0.0850_deg - 0.0070_deg - 0.0124_deg = - 0.0699_degPerturbations 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_degPerturbations 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

We will simplify the precession correction further by doing it in eliptic coordinates: the correction is simply done by adding

3.82394E-5_deg * ( 365.2422 * ( epoch - 2000.0 ) - d )to the ecliptic longitude. We ignore precession in ecliptic latitude. "epoch" is the epoch we wish to precess to, and "d" is the "day number" we used when computing our planetary positions.

Example: if we wish to precess computations done at our test date 19 April 1990, when d = -3543, we add the quantity below (degrees) to the ecliptic longitude:

3.82394E-5_deg * ( 365.2422 * ( 2000.0 - 2000.0 ) - (-3543) ) = = 0.1355_degSo we simply add 0.1355_deg to our ecliptic longitude to get the position at 2000.0.

Let's do this for Mercury on our test date - we add the x, y and z coordinates separately:

xsun = +0.881048 ysun = +0.482098 zsun = 0.0 xplan = -0.367821 yplan = +0.061084 zplan = +0.038699 ----------------------------------------------------------------- xgeoc = +0.513227 ygeoc = +0.543182 zgeoc = +0.038699Now we have rectangular geocentric coordinates of Mercury. If we wish, we can convert this to spherical coordinates - then we get geocentric ecliptic longitude and latitude. This is useful if we want to precess the position to some other epoch: we then simply add the approproate precessional value to the longitude. Then we can convert back to rectangular coordinates.

But for the moment we want the "epoch of the day": let's rotate the x,y,z, coordinates around the X axis, as described earlier. Then we'll get equatorial rectangular geocentric (whew!) coordinates:

xequat = +0.513227 yequat = +0.482961 zequat = 0.251582We can convert these coordinates to spherical coordinates, and then we'll (finally!) get geocentric Right Ascension, Declination and distance for Mercury:

RA = 43.2598_deg Decl = +19.6460_deg R = 0.748296Note that the distance now is different. This is quite natural since the distance now is from the Earth and not, as earlier, from the Sun.

Let's conclude by checking the values given by the Astronomical Almanac:

RA = 43.2535_deg Decl = +19.6458_deg R = 0.748262

Let's start by computing the apparent diameter of the planet:

d = d0 / RR 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" polThe Sun's apparent diameter at 1 astronomical unit is 1919.26". The Moon's apparent diameter is:

d = 1873.7" * 60 / rwhere 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 about 20 degrees, the planet is hard to observe, and if it's smaller than about 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 is the "haversine" function. 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 - elongFinally 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))**2Here 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 * dHere 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.

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) / kIf 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_degThe 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_degLet'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 = -67.54502and for comet Levy:

D = -3354.3046 d - D = -63.6954We'll continue by computing the Mean Anomaly for comet Encke:

M = -20.2751_deg = 339.7249_degThe 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 = 1.3885Now 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 = 1.431947Then 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 = 1.432059The 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_degTo 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_degand for comet Levy we get:

N = 138.6637_deg - (-0.5676_deg) = 139.2313_degUsing 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 0.449919These 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*FVHere 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.