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If given the sidereal day, is it possible to derive the synodic day of Earth? If a derivation is possible, could anyone illustrate it or point me in the right direction?

Wikipedia's Synodic day begins:

A synodic day is the period it takes for a planet to rotate once in relation to the star it is orbiting (its primary body).

Question: How can one calculate that?

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    $\begingroup$ What have you tried? $\endgroup$ – fasterthanlight Feb 27 at 15:29
  • $\begingroup$ What do you mean by "given the sidereal day"? This is a constant value (or nearly constant except in the very long term), as is the synodic day. It is constant at 24 hours. Do you mean given the Sidereal day length and orbital period of any planet? $\endgroup$ – James K Feb 27 at 16:05
  • $\begingroup$ Sidereal and synodic days do not depend on one another. The sidereal day depends on the rotation speed alone. The synodic day depends on both, the orbital speed and the rotational speed, thus are independent, given an arbitrary choice of planetary parameters $\endgroup$ – planetmaker Feb 27 at 23:59
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    $\begingroup$ Welcome to Stack Exchange @Shaun and congratulations on your first question! I've edited it a bit to make it fit the site better; the cryptic and unhelpful "What have you tried" comment should have explained that questions should contain some evidence of prior research. It's likely that you did a quick search first and found the Wikipedia article and noticed it doesn't tell you how to calculate this, so I've added that to your otherwise excellent question. $\endgroup$ – uhoh Feb 28 at 3:14
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    $\begingroup$ Thank you so much @uhoh. Got my intro to astro exam tomorrow so was busy preparing for that. I appreciate the answers and sorry for the naivety as it is my first question in this forum. $\endgroup$ – Shaun Ethan C. Phangcho Mar 2 at 14:11
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Surprisingly your linked Wikipedia article does not link to an explanation of how to calculate it, and neither does this.

The equation for a synodic period is based on subtracting two frequencies, the same way a radio down-converts to an intermediate frequency by mixing two frequencies nonlinearly and using the difference in frequencies as the new signal:

$$f_{\text{diff}} = f_> - f_<$$

where the > and < indicate the larger and smaller frequencies.

In orbital mechanics we usually don't talk about orbital frequency, but instead orbital period where $T = 1/f$, so:

$$\frac{1}{T_{syn}} \ = \ \frac{1}{T_<} - \frac{1}{T_>} \ = \ \frac{T_> - T_<}{T_> \ T_<}.$$

Use that if the two rotations are in the same direction which is the case for Earths rotation around its axis and around the Sun.

could anyone illustrate it or point me in the right direction?

Put the sidereal day for $T<$ and a year for $T>$ and you'll have your answer.

But what if they are rotating in opposite directions?

In that case you can define one as having negative frequency or a negative period, and replace the minus sign with a plus sign.

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  • $\begingroup$ @ConnorGarcia yes of course! Please feel free to just go ahead and edit typos like that yourself in the future, thank you! $\endgroup$ – uhoh Feb 28 at 6:15
  • $\begingroup$ +1 Very nice answer. Maybe add a tl;dr as preface : you need to know the length of the sidereal day and the length of the year? $\endgroup$ – planetmaker Feb 28 at 8:06
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    $\begingroup$ @planetmaker it's only eight sentences total. It's definitely not "tl" so if someone "dr" it, it's their loss. $\endgroup$ – uhoh Feb 28 at 8:26

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