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Derive the relationship between a planet's synodic period and its sidereal period

  1. Jun 27, 2012 #1

    s3a

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    1. The problem statement, all variables and given/known data
    The question is: "Derive the relationship between a planet's synodic period and its sidereal period (Eq. 1.1). Consider both inferior and superior planets." Figure 1.7 and Equation 1.1 are attached.


    2. Relevant equations
    Attached.


    3. The attempt at a solution
    The solution states:
    "From Fig. 1.7, Earth makes S/P_⊕ orbits about the Sun during the time required for another planet to make S/P orbits. If that other planet is a superior planet then Earth must make one extra trip around the Sun to overtake it, hence

    S/P_⊕ = S/P + 1.


    Similarly, for an inferior planet, that planet must make the extra trip, or

    S/P = S/P_⊕ + 1.

    Rearrangement gives Eq. (1.1)."

    I am VERY confused. I found definitions for some things that were confusing me.:

    (1) The sidereal period is considered to be an object's true orbital period.
    (2) The synodic period is the time period between two successive astronomical conjunctions of the same celestial objects.
    (3) Planets closer to the Sun than to Earth are inferior planets.
    (4) Planets that are farther away from the Sun than Earth are the superior planets.

    So from that, I feel pretty confident about (3) and (4), think I get (1) but I'm heavily confused about (2). Also, could someone relate (1) and (2) to Figure 1.7 for me please?

    If I left something out and/or more information is needed, just ask me and I will add whatever more is needed.

    Any help would be GREATLY appreciated!
    Thanks in advance!
     

    Attached Files:

  2. jcsd
  3. Jun 27, 2012 #2

    cepheid

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    You can consider the sidereal period to be the true orbital period of the planet. In other words, it's the time it takes to go around the sun once. It should be pretty clear what this looks like on the diagram.

    We say that Mars is at opposition (or that there is an opposition of Mars) when it appears in the exact opposite part of the sky as the sun does (as viewed from Earth). In the diagram, this is true at t = 0, and at t = 2.135. The opposite of an opposition is a conjunction: when a planet appears in the same part of the sky (i.e. in the same direction) as the sun.

    Oppositions are only possible for superior planets, for obvious reasons. Inferior planets instead have two types of conjunctions: superior conjunctions (when the planet is in the same general direction as the sun, but in behind it), and inferior conjunction (when the planet is in the same general direction as the sun, but in front of it). Just look up "conjunction" on Wikipedia and you'll see a helpful diagram illustrating what I said.

    The synodic period is the time from one opposition to the next opposition (or the time from one conjunction to the next conjunction).

    If the Earth didn't move (if it stayed where it was at t = 0), then the synodic and sideral periods would be the same, since, after one Mars orbit, Mars would be back where it started at t = 0, and therefore would be lined up with the stationary Earth again.

    But Earth does move, and your diagram clearly illustrates why this leads to the synodic and sidereal periods of Mars not being the same again. After one Mars orbit, it's back where it started at t = 0, but Earth is not, therefore you don't get an opposition this time.

    Is that enough for you to solve the problem, or is there something else you'd like to know?
     
  4. Jun 28, 2012 #3

    s3a

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    You have helped but there is a bit more I need to know. Also, sorry if my questions are stupid and/or have been addressed in another way, but, could you please answer me these?:

    #1. When taking the synodic/sidereal ratio, does that gives us the period of each planet in sidereal years? (I'm thinking yes based on mathematical logic but I would just like confirmation if I am right, please.)

    #2. Why does the Earth need to take an extra trip around the Sun in order to overtake a superior planet?
     
  5. Jun 28, 2012 #4

    cepheid

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    Yes, you can certainly interpret the ratio of S/P as being the synodic period in units of "number of orbits." But rather than just accepting it and trying to make sense of it (or justify it) in this way, a much better thing to do is to understand why this ratio appears in the expression in the first place. You can derive it mathematically. Just set up a coordinate system in which the x-axis is a horizontal line (a line that passes through the sun and both planets at t = 0. Consider the angular position θ of each planet, as measured clockwise from the positive x-axis. An opposition occurs when θ is the same for both planets (or when θ for one planet is different from θ for the other planet by a multiple of 2π, which means they're not really different at all). So, (assuming circular orbits), if you write down the expression for θ vs. t for both planets, assuming that they are coincident at t = 0, and equate these two expressions, you can find the values of t for which θ is the same for both planets (again, modulo 2π). This will give you the synodic period. It's not a hard derivation at all.

    Look at the diagram. A t = 0, the two planets are lined up, like runners on a track at the start of a race. Earth is in the inside lane, and Mars is in the outside lane. Earth's track is shorter. But not only that, but Earth also travels faster in its orbit than Mars does in its orbit. This is always true: orbital speed decreases as orbital radius increases. So the outer planets move more slowly than the inner ones (see Kepler's Third Law or Newton's Law of Gravitation). This means that Earth is "ahead" of Mars for the entire first lap. So Earth is not going to be next to Mars at any time during its first lap, which explains why, by the time they do end up side by side again, Earth has gone through at least one more orbit than Mars has. Referring to the diagram, we see that by the time Earth finishes one lap (at t = 1.0), Mars has only gone a little over halfway through its first orbit (lap).

    Earth doesn't pass (or lap) Mars during its second orbit either. As you can see from looking at the t = 1.5 and t = 2.0 positions of the two planets, Earth's angular position is less than Mars (i.e. it's chasing after it) throughout its second orbit. It's not until a little bit into Earth's third orbit (and a little bit into Mars' second orbit) that the two meet up again (Earth "laps" Mars).

    So, in summary, the reason why Earth will have gone for one more orbit than Mars has by the time they meet up again is because Earth's orbit is closer in, and therefore Earth moves faster than Mars does.

    The opposite is true when looking at the synodic period of an inferior planet, whose "lane" is inside of Earth's, meaning that it moves faster than Earth does.
     
  6. Jun 29, 2012 #5

    s3a

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    Thanks again :) but I have more questions (sorry):

    Firstly, the equation being derived for the initial question is solely for our solar system, right?

    Secondly, how do I know that the +1 period comparison is universal to all planets in our solar system (rather than just Mars) without just being told that it is (since the equation being derived is attempting to be universal)? Kepler's third law is an acceptable way for me to "just accept" that the planet further from the Sun is the one that moves more slowly but how do I know whether or not the outer planet is only slightly slower such that let's say "a million" (fake number just to emphasize my point) rotations need to occur for the inner planet to overtake the outer planet such that I replace the +1 by +10^6?

    If that didn't make sense, tell me to rephrase it.
     
  7. Jun 29, 2012 #6

    cepheid

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    Not really. It applies to a system with two planets with periods P1 and P2, for which you want to find the synodic period. It's pretty general.

    No particular assumptions are being made here. The method that I asked you to try in order to derive the synodic period in terms of the two sidereal periods is a general approach. If you try it (which I urge you to do), you will find that the first time that the planets are lined up (after t = 0) always occurs when the faster, inner planet, has completed one more full revolution than the slower, outer planet. The reason for this is that the two planets are lined up when their accumulated angular displacements are the same (modulo 2pi). Since the inner planet moves faster than outer planet, and since they both start at angular position 0, the faster planet's angular displacement must always be larger than the slower planets (at any given time). When the difference in displacement is an integer multiple of 2pi, they will be lined up. The earliest that this can happen is when that integer equals 1 (i.e. the angular displacement of the faster planet is 2pi radians or 1 full revolution greater than the angular displacement of the slower planet). The next time it can happen is for 2*2pi radians and then 3*2pi radians etc. etc.

    Still don't believe me? Consider the case for P1 = 1 yr and P2 = 100 yr (two periods that are very different from each other). The fast planet will go around once and then catch up to the slow planet when both of them are slightly ahead where they started. To put that more quantitatively, the synodic period for this system will be S = 1.01010101... years, so the first opposition (after the one at t = 0) will occur when the fast planet has completed 1.0101010... orbits, and the slow planet has completed 0.0101010101... orbits. The fast planet has completed exactly one orbit more than the slow planet at this point.

    Consider the case for P1 = 1.00 yr, and P2 = 1.01 yr. (Two periods that are very close to each other). The synodic period for this system will be S = 101 yr. So, the faster planet will go around 101 times in the same amount of time that it takes the slow planet to go around 100 times, and in this case they happen to meet up back where they started. Once again, the fast planet has completed exactly one more orbit than the slow planet has.

    Try it with any numbers you like. What if P1 = 1.23875 yr, and P2 = 5.68459 yr. The synodic period for this system will be 1.5839 yr. After one synodic period, the fast planet will have completed 1.5839/1.23875 = 1.279 orbits. The slow planet will have completed 1.5839/5.68459 = 0.279 orbits. Once again, the faster planet has completed exactly one more orbit than the slower planet has.
     
    Last edited: Jun 29, 2012
  8. Jun 30, 2012 #7

    s3a

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    I fully get it now and I love the way you explain yourself! Thank you very much!
     
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