John Pazmino 
 NYSkies AStronomy Inc
 2011 December 17
    The NYSkies Astronomy Seminar of 16 December 2011 discussed 
methods of determining the distances to celestial objects. One was the 
parallax or triangulation method. The technique involves measuring a star 
from different places in the Earth's orbit. The triangle of Earth at 
time 1, Earth at time 2, and star is solved for the Earth (either 
one)-star leg. The length of this side is enormous compared to the 
Earth's orbit, in the order at the very least of 100,000 Earth orbit 
    We mentioned that Kepler tried triangulation on the planets to 
discover that the planet orbits are ellipses, not off-centered 
circles. The question came up that as Earth moves from place 1 to 2 in 
her own orbit, the planet also moves. The two sightlines can not make 
a closed triangle. A star, while it has proper motion, is essentially 
stationary in space for the decade or so it takes to get sufficient 
observations for a good parallax. 
    How did Kepler get around the problem of the moving planet?
Kepler's work
    Kepler flourished in the early 1600s, a contemporary of Galileo 
and a worker at Tycho's observatory. When he took up the problem of 
planet motions he realized that it was futile to find parallax based 
on the Earth's own diameter. Tycho tried that for the supernova and 
comet of the 1570s and found them to be well beyond the Moon's 
    Tycho could measure no parallax on the planets, they, too, being 
indefinitely remote. At least he proved that both comet and supernova 
(stella nova) were in the celestial realm, not some luminous 
apparition among the clouds or under the Moon. 
    Kepler from his work with Tycho could mine observations of the 
planets to find clues to the planetary motions. He also had faith that 
Tycho was a careful and cautious observer whose data were of the 
highest precision up to his era. He was an adhaerent of the Copernicus 
solar system in which Earth orbits the Sun. He designed to try 
parallax based on the diameter of the Earth's orbit, which is 
immensely larger than the very Earth herself. 
    Kepler did not know the length of the orbit radius, in kilometers 
or what ever units were used in his time. He set the Earth orbit 
radius to unity as a supersize meterstick. We still do this today, 
calling this length the astronomical unit or AU. Since the Earth 
varies her distance from the Sun during the year, the radius is the 
mean distance, about 149,600,000 kilometers. 
Clever trick 
    For the stars, Kepler recognized that they were so very far away 
that they were essentially fixed in space, In fact, Kepler engaged the 
stars as the 'ground' against which to lay out the planet orbits. 
    His zeroth law, commonly cited as part of the first law, stated 
that the orbits of planets are planes fixed in space and passing thru 
the Sun. This is a truly profound statement. For the first time in 
human history Earth was replaced as the foundation of motion and 
motions were banked against the enclosing sphere of fixed stars. 
    His effort to find a frame or platform for citing absolute motion 
eventually became the basis for Einstein's theory of relativity quite 
300 years later. The separation of the two concepts, planes and 
ellipses, usually rolled into the first law brings out this feature of 
Kepler's thinking. 
    He also knew that planet orbits, under Copernicus, were stable and 
rigid in their planes. The planets traced the same path over and over 
again during each round of the Sun. 
    To get a sight on Mars from different places in Earth's orbit, 
Kepler had to be sure that Mars was in the same place in his own orbit 
at each sighting. He did this by picking Tycho's records of Mars 
spaced one Mars year apart. After one round of the Sun Mars stands at 
the same place in his orbit, ready for a new sighting from Earth. 
    The practical difficulty was that Tycho, while he made routine 
measures of Mars, did not purposely keep account of the period or year 
of Mars. Kepler had to do with records close to yearly intervals and 
interpolate among them. 
    With pairs of such observations in hand, Kepler plotted the path 
of Mars around the Sun and fitted a curve for it. The curve of best 
fit ended up being an ellipse and not a circle displaced one way or 
another from the Sun. The rest of the story is history. 
Replicating Kepler's method 
    Here we use a modern computer ephemeris or planetarium program to 
replicate Kepler's work with Mars. By now the 2-thous virtually all 
astronomy programs generate competent positions and motions of the 
planets. It really doesn't matter which you use. As long as it gives 
readouts of various parameters about the planet when selected from a 
chart of table, the program is good for this exercise. 
    The figures here are my own taken for the date of the Seminar, 16 
December 2011, and dates one and two Mars years before then. After 
working thru Kepler's method for these fates you may try for any other 
set of dates at one Mars year intervals. 
    The period of Mars's orbit was well known from both the Ptolemaeus 
and Copernicus models of the world, In the Copernicus scheme the orbit 
circulated around the Sun. In the Ptolemaeus system the orbit was the 
deferent around Earth. 
    The modern value is 1.0807 Earth year or 686.9 Earth day. This 
figure may differ slightly among references because an orbit is not a 
trolley track in space. It wiggles and wanders from gravity tugs of 
other planets. 
Earth's orbit
    Some treatments of Kepler's planetary work state that Kepler 
assumed Earth had a circular centered orbit. This makes the maths 
easier but there was no reason to make this assumption. The Earth's 
orbit in Copernicus was well established with varying distance from 
the Sun during the year. Kepler used the instant distance for each of 
Tycho's records. 
    Tycho, by the way, favored a mixed Copernicus-Ptolemaeus world 
where the planets orbited the Sun but the Sun then orbited Earth. For 
the predicting the places and movement of the planets in the sky this 
Tycho system was just as good as the other two. Kepler went with the 
pure Copernicus model. 
    Kepler was blessed by a fantastic historical accident. In 1603-
1604 logarithms were invented! Both the common series by Briggs and 
the natural one by Napier were published almost simultaneously. 
Logarithms made calculations immensely easier by reducing operations 
by one level. Multiply-divide were handled by addition-subtraction; 
power-root, multiply-divide. 
    Logarithm tables were published for the trig functions, making 
them as easy to manipulate as regular numbers. Kepler himself noted 
that if it wasn't for logarithms, he probably would have given up with 
his planetary studies. 
    Until the introduction of electronic calculettes in the 1970s ws a 
astronomers had to be fluent in logarithmic computations. Many 
astronomy equations were written from start in logarithms. The 
calculette by 1980 become so pervasive and cheap that school began 
dropping tuition in logarithms. Probably now in the 2-thous, 
logarithms are a quaint footnote in school maths books. 
    For this article you should have a scientific or engineering 
calculette, not arithmetic one. Excellent models are at sale for a 
couple tens of dollars that will satisfy every astronomy situation. 
    Parallax works with a triangle set up on a baseline with apex at 
the target body. Sightlines to the target are made at each end of the 
baseline with appropriate angles taken. By trig formulae the length of 
the sightlines are worked up for the distance to the target. 
    The work here employs the ecliptic coordinates, these being vastly 
simpler to describe the motion and position of planets than the 
equatorial right ascension and declination. To further ease the maths, 
I neglect the latitude component and let Mars run in the ecliptic in 
pure longitudinal motion. This does let small errors creep in but the 
principles are just as valid as with both component factored in. 
Observing Mars
    From an ephemeris program we collect the following 'observations' 
of Mars spaced one Mars year apart. 
    parameter     | 16 Dec 2011 | 28 Jan 2010 | 12 Mar 2008 
    long of Mars  | 165.3 deg   | 130.4 deg   |  92.6 deg 
    long of Earth |  83.9 deg   | 128.2 deg   | 172.0 deg 
    dist of Earth | 0.984 AU    | 0.985 AU    | 0.994 AU 
    for pair 2011 Dec 16-2010 Jan 28 
    angle at Sun | abs(83.6 deg-128.2 deg) = 44.3 deg 
    Your peculiar program may offer the longitude of Sun, not Earth, 
if it confines to only an Earth viewpoint. The longitude of Earth as 
seen from Sun is merely 180 degrees added or subtracted from the solar 
longitude. Do which ever throws the value within 0-360 degrees. 
    For added realism, you can pick off the longitudes of Sun and Mars 
from a computer planetarium by scaling off of the background stars and 
the ecliptic coordinate grid. The accuracy will deteriorate somewhat 
from that of a numerical output of an ephemeris program but the 
principles are just as valid. 
Sun-Earth triangle 
    We take the three pairs of observations to get sightlines to Mars: 
2008-2010, 2010-2011, and 2088-2011. The details are here laid out for 
the 2010-2011 pair and you can use them as the model to work out the 
other two pairs. 
    We need the baseline length and orientation for the 2010 and 2011 
dates. The sketch below shows the geometry. 
                             / E11 
                          /     | 
                        /       | 
                    E10        | 83.9d 
                       \       | 0.984AU 
                   128.2d\    | 
                   0.985AU \  | 
    You really better make sketches of the planetary situation to keep 
the angles and distances right way round. This is specially true when 
working with trigonometric formulae. Pass up on this procedure will, 
WILL, throw your results wildly wrong. 
    The baseline is E11 to E10 and is found by the Law of Cosines We 
first need the angle at the Sun, which is the difference in the Earth 
longitude for the two dates, 128.2 - 83.6 = 44.3 deg. The absolute 
value is used, the signum being discarded. 
    c^2 = a^2 + b^2 - (2*a*b*cos C) 
    (E11_E10)^2 = (S_E11)^2+(S_E10)^2-2*(S_E11)*(S_E10)*(cos S) 
                = (0.984)^2+(0.985^2)-(2*0.984*0.985*cos 44.3) 
                = 0.551 AU2 
    (E11_E10) = 0.742 AU 
    One common mistake is to miss taking the square root of the c^2! 
You'll end up with way too short a distance to Mars, one that may 
escape notice unless you're familiar with Mars apparitions. 
    The angles at E11 and E10 are the same by the symmetry of the 
Earth orbit and are (180-44.3)/2 = 67.8d. This ia not exactly true but 
since the two legs from Sun to Earth are so nearly the same, the error 
is very small. 
    The results so far are gathered here for easy reference
    E11_Sun_E10 triangle 
    E11_E10   | 0.742AU 
    S angle   | 44.3 deg 
    E11 angle | 67.8 deg 
    E10 angle | 67.8 deg 
Mars-Earth triangle 
    We now need the base and angles for the Earth-Mars triangle. The 
baseline length we already got from above as 0.742 AU. The two angles 
at the ends of the baseline are found by studying the sketch below. 
                     M- - - -         [98.6d] 
                      \       - - - - E11 
                       \            /  | 
                       \          /    |83.9d  
                  130.4d\       /      |     
                        \     / 0.742AU| 
                         \  / 
                    [182.2d] \ 
    The angles between Sun-Earth-Mars at the two dates are the 
differences between the Sun and Mars longitude, rectified to the 
proper quadrant. I mark them in bumpers in the diagram. They are 
nearly the elongations of Mars from Sun in the Earth's sky. 
    Here is where a sketch saves your life. The orientation of the 
lines in the figure are based on ecliptic longitude. By adding or 
subtracting them against 180 or 360 degrees you obtain the large 
interior angles. You must pay attention to the sense of the angles! 
    We need the angles along the baseline, so we subtract out the 67.8 
deg from each of the elongation angles. At E10 we have 182.2d-67.8d = 
114.4d and at E11, 98.6d-67.8d = 30.8d. The third angle, at Mars, is 
180d-(114.4d+30.8d) = 34.9d. Here are the new parameters 
    Earth-Mars triangle 
    dist E08_E10  | 0.742 AU      
    angle at E10  | 114.4 deg 
    angle at E11  |  30.8 deg 
    angle at Mars |  34.9 day 
    With all three angles and one side, we employ the Law of Sines to 
get the length of each Earth-Mars sightline. 
    a/sin A = b/sin B = c/sin C
    (E11_M)/(sin E10) = (E10_M)/sin E11) = (E11_E10)/sin M 
    (E11_M) = (E11_E10)*(sin E10)/(sin M) 
            = (0.742AU)*(sin 114.4d)/(sin 34.9d) 
            = 1.181 AU 
This compares favorably with the computer ephemeris value of 1.183 AU 
for 16 December 2011. Kepler had no such modern planetary tools. Doing 
the same calc for the side E10_M yields 0.662 AU, also in good 
agreement with an ephemeris. 
    Note well that by picking dates when Mars returns to the same 
place in his own orbit we pinned Mars in space relative to Earth by a 
triangulation just as if Mars was standing still. 
    This process Kepler repeated for many pairs of observations spaced 
at multiples of a Mars year to build up a plot of Mars around the Sun. 
Distance from Sun
    The distance of Mars from Sun comes directa mente from the 
triangle Sun-Earth-Mars for either date. We have for the 2011 case the 
diagram below 
                M\\\\\\\\ 1.181AU 
                  \      \\\\\\\E11 [98.6d] 
                    \            | 
                      \          | 
                        \       | 0.984AU 
                          \     | 
                            \  | 
    We apply the Law od Cosines having two sides and the included 
angle. The included angle is almost the elongation of Mars from Sun, 
98.6d in bumpers in the diagram. 
    c^2 = a^2 + b^2 - (2*a*b*cos C) 
    (S_M)^2 = (E11_S)^2+(E11_M)^2-(2*(E11_S)*(E11_M)*cos E11
            = (0.984)^2+(1.181(^2-(2*(0.984)*(1.181)*cos(98.6)) 
            = 2.711 AU2 
    (S_M) = 1.646 AU 
    Doing this with the 2010 date and parameters also yields 1.646 AU. 
The two distances apply to the same Sun-Mars line and should be very 
nearly equal. An ephemeris gives 1.647 AU for Mars on both dates. 
Graphical solution
    If the maths are out of reach, you can work the problem entirely 
by graphical methods. You must have steady hand and the proper tools. 
Have a T-square, drawing board, protractor, triangles and ruler, and 
sharp pencils. 
    Be diligent in your linework, taking care to keep the lines crisp. 
Home the protractor or ruler on the diagram accurately and read the 
scale to the tenth degree if feasible. For a small protractor of only 
a few centimeter diameter you'll have to make do with readings of the 
quarter degree at best. 
    The results will be a bit rough but the concepts are fully 
An other trick 
    Kepler wanted other places where Mars is known to occupy at each 
lap of the orbit. He used the two nodal crossings, ascending and 
descending. He verified the period of Mars by noting the interval 
between successive ascending or descending node passes. He then 
captured the ecliptic longitude of Mars at each pass to construct more 
    If you try this trick you hunt for dates when Mars crosses the 
ecliptic. This is easiest done with an ephemeris giving ecliptic 
coordinates. The node pass occurs when the latitude of Mars passes 
zero degrees. 
Working with angles 
    In this simple example of triangulation we came across some 
troubles with angles. That's why I stress that you must sketch out the 
geometry to catch silly, or stupid, blunders. It's in the nature of 
angles that the troubles come from and there are two special ones to e 
on watch for. 
    First is the walk between angle and trig function. Each angle has 
its own unique value for the trig functions. On a calculette you can 
key in the angle and get back the proper one and only function value. 
    It doesn't work so well from function back to angle! Every value 
for a trig function has TWO angles, not just one. In general the 
calculette gives you the smaller or positive one of the two possible 
angles. You must thru supplemental considerations decide which of the 
two is the proper result. That's what the sketch is for.
    The other issue is that the angle so returned may be in any of the 
four quadrants of 0-90 deg, 90-180 deg, 180-270 deg, 270-360 deg. You 
have to inspect the geometry carefully to get the proper quadrant for 
the returned angle. 
    I warn that this chore is an integral part of astronomy, a 
discipline that lives off of angles. 
    This is an exercise for a high school class versed in trigonometry 
or a freshman college class. It shows how it is possible to fathom the 
solar system by exploiting the stability of planet orbits. It also 
shows the cautions in manipulating angular measures and trig 
    The simulated observations from a planetarium program adds some 
realism by taking measurements off of the sky on the computer screen.