It has been observed since antiquity that, at certain times, some of the planets appear to reverse their direction in the night sky; in other words, their motion appears to be retrograde. This is difficult to explain within the Ptolemian model of the solar system which places the Earth at the centre of the orbits of the other planets. However it is easily explained using the model of Copernicus which places the Sun at the centre of the orbits of all the planets, including the Earth. The retrograde motion results from the relative motion of Earth with respect to the other planets. Maple's computing abilities can be used to predict not only retrograde motion, but, other behaviour as well.
The orbits of most of the planets are quite well approximated by circles
centred at the sun and, moreover, all of these circles can be assumed to
lie in the same plane; let us choose this plane to be the
Letting one AU (Astronomical Unit) denote the radius of the orbit of the
kms), it is possible to set up functions which yield the x and y
coordinates of the Earth as a function of time, t, which will be
measured in earth years:
To keep things simple, assume that the x-axis has been chosen so that at time t=0 both the earth and planet Mars lie on the positive x-axis. Using available data on Mars , set up functions XMars and YMars which yield the x and y coordinates of Mars as a function of time.
Now define a function which yields the angle between the line connecting Mars and the Earth and the horizon as seen at midnight. (Keep in mind that at midnight an astronomer observing the planets would be directly opposite the sun. Hence, the horizon, as seen by the astronomer would be perpendicular to the imaginary line connecting the sun and the earth.) Assuming that all measuremanets are made at a midnight and, disregarding questions of latitude, this angle will correspond to the angle at which one would have to look up to see Mars in the night sky. Define a function which describes this angle. A good idea is to think of the positions of the planets as well as the horizon as vectors since the angle between vectors is readily obtained using dot products and the arccos function. When using the arccos function you must remember to be careful when dealing with angles between 180 and 360 degrees since this is outside the range of the arccos function. You should also keep in mind that the dot product of the horizon vectore and the vector from earth to a planet will not tell you when the angle between horizon and planet has esceeded 180 degrees. For this you might want to consider the dot prouct of the planet vector and the line connecting the sun and earth. What value, positive or negative, will this dot product have if the angle between planet and is greater than 180 degrees?
Explain how this function can be used to locate Mars in the night sky. What additional information about Mars is required to actually be able to make use of this function?
To determine when retrograde motion occurs, we must focus on the motion, rather than the position, of Mars. Hence, the next step is to calculate the derivative of this function. However, the expression Maple yields will be quite complicated. A reasonable course of action at this point is to set the expression obtained by differentiating equal to 0 and then solve to find the points at which motion is reversed. Try this. Depending on which version of Maple you are using you will either fail or come up with a very complicated expression. There are now two ways to appraoch the problem One is to give Maple a hand by suggesting some trigonometric substitutions in order to simplify the derivative and then solve. The other is to try plotting the derivative to determine where it is negative.
Try plotting the position of some of the other planets using data available on the internet. Plotting this information should allow you to answer questions such as the following: How often will Uranus and Venus be within 10° of each other in the night sky?
You should be aware that the assumption of circular orbits is not really justified for Mercury and Pluto. The other assumption made was that the orbitsof all the planets ar in the same plane as the orbit of the earth. This is a reasonably good approximation for most of the planets. For example, checking the data on Saturn reveals that its orbit inclination is only 2.49°; in other words, the angle between the plane of the orbit of Saturn and that of Earth is 2.49 °. Modify the preceding work to take into account the inclination of the orbit of Neptune to define a function which determines its position in the sky more accurately. Compare this function with the one you obtained without taking into account the angle of inclination of Neptune's orbit.