"With them it is as though an artist would gather the hands,
feet, head and other members for his images from diverse models,
each part excellently drawn, but not related to a single body,
and since they in no way match each other, the result would be a
monster rather than a man."
Copernicus
This page is concerned with astronomical calculations. You can find out how to calculate a very rough position (given as the right ascension and declination) for the planet Venus, and for the planet Jupiter on any given day. The method will extend to any of the other planets, and is accurate to a few degrees either way in the sky - enough for binoculars! As an example of the method, I show calculations for the position of Jupiter and of Venus at 0 UT on April 20th/21st, 1997.
The method shown here is based on Section 55 of Peter Duffett- Smith's book Practical Astronomy with your Calculator (3rd Edition, 1990). You can plot your calculated position on a star chart - or convert to ALT and AZ coordinates and find the planets with a magnetic compass.
If you want accurate positions for the planets until 2006, I would suggest you look at the NASA Twelve Year Planetary Ephemeris (TYPE), provided by Fred Espenak. I have used this very useful resource to check the accuracy of the method described here.
The planets orbit the Sun following paths which are approximately elliptical. The planes of their orbits are tilted to the plane of the Ecliptic by varying small angles. Both these things complicate the calculation of the position of the planet. In the following calculations, we make two outrageous assumptions;
To find the position of the planet in its orbit today, we just need to know its position at some instant in time (known as an epoch) and how fast the planet is moving. This kind of information is listed for various dates in the Astronomical Ephemeris, and in Duffett-Smith's book. A set of 7 numbers define a proper eliptical orbit - and these are called the elements of the orbit. I list just the elements needed for this simplified calculation in the 'Elements' section below.
As we have assumed that the planes of the orbits are the same as the plane of the Ecliptic, a simple sum gives us the Heliocentric Longitude of the planet. The Heliocentric Latitude of the planet is assumed to be zero throughout. We also need to find the Heliocentric Longitude of the Earth, to allow the positions to be referred back to the Earth in the next step.
Having found the position of the planet and of the Earth as angles measured from the Sun, we need to refer that position to the Earth. This requires some formulas from trigonometry. We end up with the Geocentric Ecliptic Longitude of the planet - again the Geocentric Ecliptic Latitude of the planet is zero as a result of our assumptions.
Star maps and Planispheres are usually calibrated in Equatorial Coordinates, consisting of the right ascension and declination. We need to convert our Geocentric Ecliptic Longitude to RA and DEC. You can then either;
As the results of these calculations are of low accuracy, there is little point in correcting for light travel time or for planetrary aberration. See Duffett-Smith's book for more accurate methods and for corrections such as these.
We only need the Mean longitude at epoch (L) and the daily motion (n) for the planet and for Earth, and the semi major axis of the orbit (a) of the planet - which we regard as being the radius of the circular orbit. The values below are from the 1996 Astronomical Ephemeris;
Epoch JD 2450320.5 - 0h UT 25th August 1996 Planet n L a Mercury 4.092385 281.18017 0.3870975 Venus 1.602159 20.17002 0.7233235 Earth 0.9855931 333.58600 1.0000108 (taken as 1) Mars 0.5240218 73.77336 1.5237131 Jupiter 0.08310024 292.64251 5.202427 Saturn 0.03333857 8.91324 9.561943 Uranus 0.01162075 298.87491 19.30425 Neptune 0.005916098 297.58472 30.27750 n - daily motion in degrees per day L - Mean Longitude or Longitude at Epoch (degrees) a - Mean distance (au) - taken as radius of orbit here
I tried the calculations using the 1990.0 epoch elements given by Duffett-Smith in the current printing of Practical Astronomy with your Calculator, and the differences were very small compared to the error overall. The book can thus claim to be self contained, you won't need to go to a library to get new figures.
As we assume that all the orbits are circular, the planets travel round them at an even angular speed - we often see the 'mean daily motion' listed in degrees per day. To find the Heliocentric Longitude of the Earth a given number of days later than the epoch of the elements, we just multiply the number of days by the daily motion, and add on the Longitude it had at the epoch of the elements. Just as an (analogue) clock goes to zero when it reaches 12 - we have to subtract off multiples of 360 degrees to bring the Longitude back into the range 0 to 360.
The heliocentric longitude of the Earth is given by;
hle - is the heliocentric longitude of the Earth ne - is the daily motion of the Earth D - is the number of days since the epoch of the elements Le - is the longitude of the Earth at the epoch of the elements 1) hle = ne * D + Le Subtract multiples of 360 to bring into range 0-360 degreesOn April 20th 1997,
ne = 0.9855931
D = 128 + 31 + 28 + 31 + 20 = 238 days since epoch of elements
Le = 333.58600
1) hle = 0.9855931 * 238 + 333.58600
= 568.1571578 - 360
= 208.1571578 degrees
This value is used to refer the positions of Venus and Jupiter
back to the Earth.
To find the geocentric longitude of Venus, we find the heliocentric longitude of the planet (step 2), and then change our origin of coordinates to the Earth (steps 3 to 5). The formula for referring to the Earth's position is different for inner and outer planets.
2) hlv = nv * D + Lv
Subtract multiples of 360 to bring into range 0-360 degrees
hlv - is the heliocentric longitude of Venus
nv - is the daily motion of Venus
D - is the number of days since the epoch of the elements
Lv - is the longitude of Venus at the epoch of the elements
On April 20th 1997,
nv = 1.602159
D = 128 + 31 + 28 + 31 + 20 = 238 days since epoch of elements
lv = 292.64251
2) hlv = 1.602159 * 238 + 20.17002
= 401.483862 - 360
= 41.483862 degrees
Recall that,
av - radius of Venus orbit
hle and hlv - the heliocentric longitudes of Earth and Venus
glv - the geocentric longitude of venus
3) calculate;
Y = av * sin(hle - hlv) (always in range -0.75 to 0.75)
X = 1 - av * cos(hle - hlv) (always in range 0.2 to 1.8)
4) Then find atan(Y/X). If Y is negative, then add 360 to the result
of atan(Y/X), X is always positive.
5) Then
glv = 180 + hle + atan(Y/X)
subtract or add multiples of 360 as needed to bring into
range 0-360
The geocentric latitude of venus is assumed to be zero.
Below are the figures for April 20th 1997,
av = 0.7233235
hle = 208.1571578
hlv = 41.483862
glv = the geocentric longitude of Venus
3) Y = 0.7233235 * sin(208.1571578 - 41.483862)
= 0.1667284448
X = 1 - 0.7233235 * cos(208.1571578 - 41.483862)
= 1.703845517
4) atan(Y/X) = 5.5588839611 degrees (no correction, as Y positive)
5) glv = 180 + 208.1571578 + 5.5588839611
= 393.74599741 - 360
= 33.74599741 degrees
You can skip to the section on transforming to RA and DEC to get the equatorial coordinates for Venus, or carry on to calculate the geocentric longitude for Jupiter.
To find the Geocentric longitude of Jupiter, we find the heliocentric longitude of the planet (step 2), and then change our origin of coordinates to the Earth (steps 3 to 5). The formula for referring to the Earth's position is different for inner and outer planets.
2) hlj = nj * D + Lj Subtract multiples of 360 to bring into range 0-360 degrees hlj - is the heliocentric longitude of Venus nj - is the daily motion of Venus D - is the number of days since the epoch of the elements Lj - is the longitude of Venus at the epoch of the elementsOn April 20th 1997,
nj = 0.08310024
D = 128 + 31 + 28 + 31 + 20 = 238 days since epoch of elements
lj = 292.64251
2) hlj = 0.08310024 * 238 + 292.64251
= 312.4203671 degrees
Recall that,
aj - radius of Jupiter's orbit
hle and hlj - the heliocentric longitudes of Earth and Jupiter
glj - the geocentric longitude of Jupiter
3) calculate;
Y = sin(hlj - hle) (always in range -1 to 1)
X = aj - cos(hlj - hle) (always in range 4.2 to 6.2)
4) Then find atan(Y/X).
If Y < 0 then add 360 to result - X is always positive
5) Then
glj = hlj + atan(Y/X)
subtract or add multiples of 360 as needed to bring into
range 0-360
The geocentric latitude of Jupiter is assumed to be zero.
For April 20th 1997, we have
aj = 5.20247
hle = 208.1571578
hlj = 312.4203671
glj = the geocentric longitude of Jupiter
3) Y = sin(312.4203671 - 208.1571578)
= 0.9691741343
X = 5.20247 - cos(312.4203671 - 208.1571578)
= 5.448846739
4) atan(Y/X) = 10.08559526 degrees (no correction, as Y positive)
5) glj = 312.4203671 + 10.08559526
= 322.5059624 degrees
In the (outrageous) approximation we are using, the Geocentric
Latitude of the planet is zero - because we assume that Jupiter's
orbit is in the plane of the Ecliptic.
In the previous sections we found the Geocentric Longitude (glp) and Latitude of the planets Venus and Jupiter. In the (radical) approximation we are using, the Geocentric Latitude for both planets is zero - as the orbits are assumed to be in the plane of the Ecliptic. We now want to find the Equatorial coordinates of the planets - the right ascension (RA) and declination (DEC). We use the formulas below, and use the value 23.43928 degrees for the inclination of the Equator to the Ecliptic (E).
E - is the angle of the Equator to the Ecliptic, and
glp - is the Geocentric Longitude of the planet,
glv for Venus and glj for Jupiter
6) Y = sin(glp)*sin(E)
X = cos(glp)
ra' = arctan(Y/X)
If X < 0 then ra = ra' + 180
If Y < 0 and X >> 0 then ra = ra' + 360
else ra = ra'
DEC = ARCSIN(sin(glp)*sin(E))
E = 23.43928 glv = 33.74599741
Substituting these values into the formulas above gives;
6) Y = sin(33.74599741) * cos(23.43928)
= 0.5096724726
X = 0.8315084215
ra' = arctan(Y/X) = arctan(0.6129492612)
= 31.50618405 degs
no correction as X and Y both positive
ra = 31.50618405/15
= 2.10041227 hrs
DEC = ARCSIN(sin(33.74599741) * sin(23.43928))
= 12.76603228
= 12.77 degrees
TYPE gives RA = 2.16 hrs and DEC = 12.21 degrees
for Venus on this day.
E = 23.43928 glj = 322.5059624
Substituting these values into the formulas above gives;
6) Y = sin(322.5059624) * cos(23.43928)
= - 0.5584519886
X = cos(322.5059624)
= 0.7934166858
ra' = arctan(Y/X) = arctan(-0.7038571265)
= -35.14007177 degs
add 360 to ra', as Y is negative and X positive
ra = -35.14007177 + 360 = 324.8599282
= 21.65732855 hrs
DEC = ARCSIN(sin(322.5059624) * sin(23.43928))
= -14.01160621 degrees
= -14.01
TYPE gives RA = 21.38 hrs and DEC = - 15.94 degrees for Jupiter on this day.
The very crude method outlined here will give you positions for Venus within 0.1 hours of RA and around 3 degrees of DEC. For Jupiter, the RA is only good for 0.4 hours (nearly 6 degrees on the sky) and about 3 degrees in DEC. This level of accuracy is sufficient to put the planet in the right constellation - and within the correct binocular field (7.3 degrees) in most cases.
I wrote a simple BASIC program to repeat the estimated method for every second day in 1997 - and I compared my results with data transferred from the NASA Twelve Year Planetary Ephemeris, using a spreadsheet. The error in each measurement is defined as
error = TYPE value - estimated value
I was expecting to see a double periodicity in the error, one for the Earth's departure from circular motion, and another corresponding to the planet's eccentricity. The graphs for Venus in 1997 confirmed this general picture.
The main features of the error in the RA and DEC for Venus in 1997 are;
The main features of the error in the RA and DEC for Jupiter in 1997 are;
Plotting the errors over 4 years (1997 to 2000 inclusive) shows more of the picture;
The main features of the error in the RA and DEC for Jupiter over the period 1997 to 2000 inclusive are;
Some simple periodic adjustment term may work for Jupiter over the whole 12 year TYPE period - but then you might as well try some interpolation scheme using the TYPE values themselves.
A copy of Practical Astronomy with your Calculator by Peter Duffett-Smith provides details on more accurate orbital calculations which are feasible on a programmable calculator. You don't always want to boot up a PC to work out a rough azimuth for a comet - or to see how far below the horizon the Sun will be at a certain time. Here is the full reference;
Practical Astronomy with your calculator
by Peter Duffett-Smith
Cambridge University Press
ISBN 0-521-35699-7 (3rd Edition)
Cost about UK Pounds 10.
A short review by Sam Wormley of Practical Astronomy with your calculator is available.
Peter Duffett-Smith has a home page about his books, and his photograph reveals him to be totally different to my expectation - which just goes to show something or other!
[Root]
Last Modified 26th April 1997
Keith Burnett