Find Synodic Period of Jupiter | Astronomy Help for Introductory Course

[cassiopeiae]

I am taking an introductory astronomy course and am having problems finding the calculations needed for this problem. It is not a homework problem, just an extra question at the end of a chapter. I have asked my professor about this, and he is not sure either...any help is greatly appreciated. I know the answer is approx 11.9 years, I have no reference in either textbooks I have consulted. It is a "trial by error" problem, but I don't have a clue where to start.

Question:
The synodic period of Jupiter is 399 days.* What is the interval of time between successive times that Jupiter appears in opposition at the same position among the stars?* (this is a trial by error problem)

 

[HallsofIvy]

"Synodic Period" (I had to look it up. Here's the website I went to:
http://www.sweb.cz/vladimir_ladma/english/cycles/reson/synodp.htm)
of Jupitor is the time until Jupitor has exactly the same position relative to the Earth as it has now (imagine that Jupitor is exactly over your head at midnight. It will be exactly over your head at midnight again in 399 days). Of course the "synodic period" of the fixed stars is 365.24 days (approximately). Jupitor will be in the same postion relative to the stars again when those- 365.25 days and 399 days- match up again?

Since your book says "this is a trial and error problem", you might try calculating multiples of both 399 and 365.25 days and see when they come out the same.

 

[cassiopeiae]

Ok, that's where I got confused. [the "synodic period" of the fixed stars is 365.24 days (approximately)] I knew that, but for some reason, that part of the question was eluding me...as I smack myself on the forehead...duh.

Thanks for your help!

 

[turin]

Why is this trial and error? Shouldn't you be able to calculate it analytically given these two synodic periods? In other words, take the Earth as a fixed point (like the 12 o'clock point) and then take Jupiter and the stars to be like two different hands of a clock. The question is then basically asking how long will it take both hands to point to "12" again.

Probably you would want to round off the Earth's year to 365 days.

 

[cassiopeiae]

Actually, I did find a formula...

P = sidereal motion of a planet
P_Earth = sidereal motion of Earth = 1 year
S = synodic period

1/P = 1/P_Earth - 1/S for superior planets
1/P = 1/P_Earth + 1/S for inferior planets

That would have helped me in the beginning