Author: jlromano ()
Date: 2003-Oct-11 16:00:36

Here's a puzzle I made up that stumped me for quite a while:

   I had read that Neptune and Pluto (the eighth and ninth planets from our
sun, respectively), are locked in a 3:2 orbit with each other.  This means
that for every three orbits around the sun that Neptune makes (which is the
"faster" orbiting planet), Pluto makes only two.  (I verified this and it's 
accurate to about one percent.)

   So here's the puzzle:  In the amount of time it takes for Neptune to 
orbit the sun three times (which is the same as the amount of time it takes 
Pluto to orbit the sun twice), how many times do Neptune and Pluto meet?  
(By "meet", I mean align themselves with the sun at one end and Neptune 
between the sun and Pluto, assuming that both Neptune and Pluto have 
circular orbits on the same plane and that Pluto is always farther than the 
sun than Neptune is.)

   Once you've figured out the answer to this puzzle, what would the answer 
be for any other orbit ratio, like 4:7?  And what if the planets were 
orbiting the sun in opposite order?  How would that affect the answer?

   -- Jean-Luc   


Author: tomc (Tom Carmichael) Date: 2003-Oct-14 18:21:23 I'm not going to post an answer here to the puzzle yet, but there are other problems in this 'family' that I have encountered before. Usually the questions I have seen are related to clocks...problems such as "What time is it when the hour hand and the minute hand of a clock are exactly on top of each other between 9 and 10 o'clock." The reverse orbit is an interesting twist though, I'll need to look at that closely. :-)
Author: bandsmer (Michael Bandsmer) Date: 2003-Oct-14 21:35:01 Consider the 2 planets as 2 points A and B moving in circles around the (0,0) point of the x,y coordinate system. Let a(t) and b(t) be the angle in radians, relative to the positive x-axis, of points A and B at time t. If planet A takes time Ta for 1 orbit, and planet B takes time Tb for 1 orbit, then we have a(t) = 2*Pi * t/Ta b(t) = 2*Pi * t/Tb (This assumes that the initial conditions are a(0) = b(0) = 0, i.e. that the planets start aligned on the x-axis.) E.g. For A=Neptune and B=Pluto, we have Ta=2*T, Tb=3*T, for some unit of time T. Notice that a(t) - b(t) = 2*Pi* t*(1/Ta-1/Tb) = 2*Pi * t/Tc [1] where Tc = Ta*Tb/(Tb-Ta). [2] |Tc| is the time between alignments. To see this, note that when A and B are aligned, their angles must be the same, i.e. a(t) - b(t) = 2*Pi*k [3] for some integer k. Equating [1] and [3], the alignment times 't' must satisfy a(t)-b(t) = 2*Pi * t/Tc = 2*Pi*k t = k*Tc So the only times that the planets are aligned are when t = a multiple of Tc. For the Neptune/Pluto example, Ta=2*T, Tb=3*T, so from [2] we have the alignment time Tc=6*T. So in the time that it takes Neptune to orbit 3 times (3*Ta = 6*T), we have only 1 alignment. For the 4:7 example, Ta=4*T and Tb=7*T, so time between alignments is Tc=(28/3)*T, so they will align 3 times in time 28*T (= 7 of A's orbits = 4 of B's orbits). There's no reason why the same equation [2] wouldn't work for reverse orbits as well; just negate one of Ta or Tb. E.g. For the -4:7 case (reverse orbits), we have time between alignments Tc=(-4*7/(-4-7))*T = (28/11)*T. So A and B would align 11 times in (7 of A's orbits = 4 of B's orbits). Mike
Author: mbays (Martin Bays) Date: 2003-Oct-15 10:18:55 Woah! I don't see you need to start with cartesian crap. If you've got A rotating positively at a rate of a full rotations in time t and B at b rotations in t, then the relative rate is (a-b) rotations in time t. If you're given a ratio a:b, then you've got this with t the time for a rotations of A = time for b rotations of B. So A rotates positively relative to B (a-b) times in time t, and meet up |a-b| times (or |a-b|+1 if you count start and end) So that's 1 for 3:2, 3 for 7:4, 11 for -7:4 etc. No?
Author: bandsmer (Michael Bandsmer) Date: 2003-Oct-15 16:22:00 Yes, that's certainly an easier way of looking at things.