Expert: Philip Stahl - 12/12/2005

Question
How do they find the positions of planets and stars (mathematically)...i.e without a star map. and what is the angle of the sun in relation to the earth at sunrise and sunset, and what would negative angles mean?

Answer
Hello.

Determining the position of any planet in the orbital plane is not trivial at all, nor can it be answered in a simple response format. The reason is that whole courses in what is called celestial mechanics are devoted to it.

Students cannot take this course without first having studied (or taken courses in) calculus, advanced calculus and differential equations, as well as advanced computer programming. Hence, if you haven't taken or been exposed to such - it is impossible to convey what is necessary to compute orbit positions!

Basically, position determination entails identifying - calculating the seven elements that make up a given planetary orbit, including: the eccentricity of orbit (how much it deviates from circularity); the semi-major axes (or mean distance from the Sun); the longitude of the ascending node, the inclination (i) of the orbit, the true anomaly, and the mean anomaly, and the time of perihelion passage (T) or alternately the 'mean motion' n = M/ (t - T).

The true anomaly is perhaps most difficult to obtain since it requires one take a Fourier expansion (again, this is taught in advanced calculus!) of the difference (w - M), e.g. the mean anomaly (M) from the true anomaly.

Showing the final result (to four terms) one has:

w = M + (2e - e^3/4)sin M  + 5/4(e^2) sin 2M + 13/12 (e^3)sin 3M +   ......

Where the ellipses (. . ) at the end indicate more terms (for higher accuracy), and e in the expansion is the eccentricity.


If you would like to find out more details of planetary position computation and exactly how it's done - there is an excellent and detailed site (with tutorials) at:


http://www.stjarnhimlen.se/comp/tutorial.html

This will make more or less sense to you, again, depending on the sophistication of your mathematical background. Before tackling this you may therefore want to get hold of one or more texts at your local library, including: 'Introduction to Celestial Mechanics' by Forrest Ray Moulton', and 'Principle of Celestial Mechanics' by Philip Fitzpatrick. The latter book is especially good as it has all the mathematical tools you need to use these position-finding methods.

Another good book that may be easier to obtain today is: 'Modern Astrodynamics' by Victor R. Bond and Marc C. Allman (Princeton University Press). The text has an excellent chapter on the two body problem (2), as well as Kepler's laws (3) and also how they are put to use ('Methods of computation', Ch. 4).

Regarding your other questions, bear in mind that the measurement of angles etc. depends on the coordinate system being used, or the frame of reference. In an Earth-based (geocentric-horizon) system, with the observer positioned at the center of a flat horizontal plane, then the angle of the Sun in relation to the Earth would be the 'hour angle'. At sunrise it would be known as 'the hour angle at sunrise'(H_sr), and at sunset, the 'hour angle at sunset(H_st)'

The value for the hour angle of the Sun at sunset is in general:

cos (H_st) =  - tan (d) tan (L)

showing that the Sun's hour angle depends on the Sun's declination (d) and the observer's latitude (L). If the observer's latitude is NORTH (above the equator) and the sun's declination is positive, one has in general that:

6 h  <  H_st  <  12 h

e.g. the hour angle is between 6h (90 degrees or 6h x 15 deg/h) and 12 h (180 degrees or 12 h x 15 deg/h).

If the latitude is north and the sun's declination is *negative* (e.g. negative angle wrt celestial equator) :

0h <  H_st < 6 h

Thus, note that the angle being negative must always be taken in context, and one must be aware of exactly *what coordinate system* one is using and wherein the negative angle occurs!

The Sun's hour angle at sunrise is given by:

H_sr =  360 - h

where:   cos (h) =  -tan(L) tan (d)  


In the context of the above observer-centered system, a negative angle of say (h = - 30 degrees) would mean the Sun is 30 degrees *below* the observer's eastern horizon, or 30 deg/15 deg/h = 2 h or about 2 hr. until sunrise time at the observer's latitude.

As you can see from all the above, mathematics - including advanced calculus and trig, plays a very important role in much of astronomy. This is why - to get even a basic astronomy degree- one requires very nearly a minor in mathematics!