Robert Clark wrote:
Is there a formula for computing the period of Jupiter as seen from Mars? That is, if one were an observer on Mars what would be the formula for determining how long it would take for Jupiter to return to the same position in the sky?
Both Mars and Jupiter have rather large eccentricities, as planets go, taking Mars as fixed would Jupiter describe an ellipse in its motion? I'm thinking Jupiter would describe some retrograde motion as viewed from Mars so that it would not describe a simple ellipse. But perhaps its overall shape is elliptical.
Someone else has already answered this question by drawing up the equation for synodic period. However, I'm not sure that's what you mean by "return to the same position in the sky." The synodic period is, roughly, the time between opposition dates; however, the position of the superior planet (that is, the outside planet—Jupiter in this case) at opposition can be anywhere along the ecliptic.
The page referenced in the other response can be a little cryptic for someone not used to reading almost only algebra, so here's a synopsis: Jupiter's orbital period is about 12 years; that of Mars is about 2 years. (Those aren't precise, by any means, but will suffice to show the method, and you can substitute more precise values later.)
Suppose Jupiter were at opposition now, as seen from Mars; that is, the sun, Mars, and Jupiter are all in a straight line. If Jupiter stood stock still, Mars would return to its current position in the orbit in another 2 years, and Jupiter would once again be at opposition.
But Jupiter does not stand still; in those 2 years, it has made its way around (2 years)/(12 years) or 1/6 of its orbit. It takes Mars 1/6 times 2 years, or 1/3 year, to make its way around 1/6 of its orbit. But in that 1/3 year, Jupiter has swung around another (1/3 year)/(12 years) or 1/36 of its orbit. It takes Mars 1/36 of 2 years, or 1/18 year, to make its way around that much of its orbit. And so on, and so on…
In this way, Mars and Jupiter play a little Achilles-and-the-tortoise catch-up game. And just as Achilles eventually does catch up with the tortoise, Mars catches up with Jupiter too, and the time it takes is, all told,
2 + 1/3 + 1/18 + 1/108 + … = 2 + 2(1/6) + 2(1/6)^2 + 2(1/6)^3 + …
From the algebraic expression for infinite geometric series
a + ar + ar^2 + ar^3 + … = a/(1-r)
that series is equal to
2/(1-(1/6)) = 2/(5/6) = 12/5
or 2-2/5 years. After that much time, the sun, Mars, and Jupiter are again all in a straight line.
Now, since Mars takes just 2 years to go around one orbit, it goes more than one orbit in those 2-2/5 years. In fact, it goes through (2-2/5 years)/(2 years), or 1-1/5 orbits. So the new straight line is not pointing in the same direction as the old straight line; it's offset by 1/5 of an orbit, or 72 degrees. At the third opposition, it will be offset by 2/5 of an orbit, or 144 degrees.
Finally, after 5 synodic periods with Jupiter, Mars has gone around 5 times 1-1/5 or 6 complete orbits. So after 6 orbits, which combined take 6 times 2 years or 12 years total, Mars and Jupiter are back in the same place as their original opposition, more or less. Because Mars and Jupiter do not change their orbits noticeably, we can ignore their respective eccentricities to a first-order approximation.
However, to go back to your original question: when does Jupiter return to the same place in the sky? If you mean, how long before Jupiter returns to the same RA/Dec area of the sky, then the synodic period is most definitely not the right answer, since what you want is the straight line to point in the same direction each time.
Because of retrograde motion, this can happen more than once per Martian orbit, depending on which position you're returning to. Consider the position below, with only the sun (S), Mars (M), and Jupiter (J) marked down. By convention, planets revolve counter-clockwise around the sun; this is the view as seen from high above the earth's north pole.
S M ----- J -->
Here, Jupiter is at opposition as seen from Mars. In a little less than one Earth year (a little less than half a Martian year, in other words), their positions will be as follows:
M ------------- J --> S
Then, a little more than one Earth year later, the configuration is now like this:
M ------ J --> S
In each case, Jupiter is in the same position along the ecliptic.
However, this situation can also go for several Martian years before repeating. Consider the positions below:
M ------- J --> S
Now, Mars is as far "up" (on this screen) as it can go; however, Jupiter is still moving "up." Jupiter will not be in this position with respect to Mars—that is, directly to its "right"—until some Martian years later, when the position is inverted:
S M ------- J -->
and the time in between is around 3/4 of a Jovian orbit, around 9 Earth years (about 4 Martian years).
The rules for determining the recurrence of this happening, as you can probably guess, are not simple, and it makes no sense to ignore the orbital eccentricities, since you are probably consigned to working out their precise positions and computing the position of Jupiter on the Martian sky. Sorry!
Copyright (c) 2000 Brian Tung