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The formula below is used to calculate the synodic period between two bodies orbiting a same third one:

$$\dfrac{1}{T_\mathrm{syn}} = \dfrac{1}{T_{1}} - \dfrac{1}{T_{2}}$$

This formula can be used, for instance, to deduce the period of a planet in the Solar System, given that $T_\mathrm{syn}$ can be measured (period between two consecutive observations of a planet in the same relative position in the sky) and taking $T_1$ as 365.26 days.

I have read that this formula is derived from:

$$w_{\mathrm{syn}} = w_{1} - w_{2} = \dfrac{2\pi}{T_\mathrm{syn}} = \dfrac{2\pi}{T_{1}} - \dfrac{2\pi}{T_{2}} \therefore \dfrac{1}{T_\mathrm{syn}} = \dfrac{1}{T_{1}} - \dfrac{1}{T_{2}}$$

My question is: which approximations are required for this result to hold true? Must the orbits be circular? Must these two bodies be in the same plane around the third? Also, considering the case of elliptical orbits but still assuming that the mass of the third body $M ≫ m_{1}$ and $M ≫ m_{2}$, would the synodic period of body 2 as observed from body 1 be indeed constant?

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    $\begingroup$ The calculation is made assuming circular orbits where the angle from the central body increases linearly with time. I'm pretty sure the orbiting bodies must be coplanar for this to work, but not 100% sure. $\endgroup$ – barrycarter Sep 17 '16 at 15:36
  • $\begingroup$ @barrycarter Hi Barry, thanks for the comment. In that case, would you know if is there a formula which takes into consideration elliptical orbits? $\endgroup$ – gilbertohasnofb Sep 17 '16 at 18:31
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    $\begingroup$ That gets complicated because the time between syncs can vary for each orbit depending on position. Ping me on google talk and I'll show you some formulas for ellipses (it's not easy: equal area over equal time gets messy) $\endgroup$ – barrycarter Sep 17 '16 at 21:04
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The formula :-

\begin{equation}\tag{A} \frac{1}{T_{syn}} = \frac{1}{T_{1}} - \frac{1}{T_{2}} \end{equation}

for the synodic period between two planets orbiting the Sun is derived in a similar way to the derivation of the Moon's synodic period :-

\begin{equation}\tag{B} \frac{1}{T_{syn}} = \frac{1}{T_{sid}} - \frac{1}{T_{e}} \end{equation}

where $T_{sid}$ is the sidereal perod of the Moon and $T_{e}$ is the sidereal period of the Earth.

Formula (B) is derived by making the following simplifying assumptions about the motions of the Sun, Earth and Moon :-

  1. Earth has a circular orbit around the Sun of constant speed and a fixed period $T_{e}$

  2. Moon has a circular orbit around the Earth of constant speed and a fixed period $T_{sid}$. (Thus the moon's path wrt the fixed Sun frame is an epicycle, ie a small circle whose center moves along the circumference of a larger circle, in this case the Earth's orbit).

  3. The above two orbits are confined to a common plane (the Ecliptic Plane)

  4. From the direction we are viewing (ie `North') both Earth and Moon orbit in the anti-clockwise direction

and (A) is derived with similar assumptions about the planetary orbits.

(See this answer for the derivation of formula (B) for the Moon - for the planets the derivation of (A) would be virtually the same).

In reality the orbits are not perfectly circular and they are not in the exact same plane. The circular orbit approximation means the speed of the planet is constant, but in the case of an elliptical orbit the speed varies.

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