The time it takes the Moon to return to a given position as seen against the background of fixed stars, $27.3$ days, is called sidereal month. The time interval between identical phases of the Moon is called a lunar month. A lunar month is longer than a sidereal month. Why and by how much?

I do understand the concept that in $27.3$ days, the moon may have orbited $360^\circ$ but since the Earth moved through $(27.3/365) \times 360^\circ = 27^\circ$. So, the Moon needs to move $27^\circ$ to catch up. That should take $(27^\circ / 360^\circ) \times 27.3$ days $= 2.05$ days , but in that time Earth would have been moved on yet much farther.

Is there any method to calculate the exact time taken i.e. (the taken by the Moon to be in the same phase and the same background positioning among the fixed set of stars) rather than doing this cumbersome process over and over again?

Are there any new methods to solve this question?

  • $\begingroup$ Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. $\endgroup$
    – Community Bot
    Sep 12, 2022 at 16:24
  • $\begingroup$ You should include units completely, and correctly. The 1st equation is days/days times degrees and comes out in degree*days. $\endgroup$
    – JEB
    Sep 12, 2022 at 17:23
  • $\begingroup$ Ah sorry about that. Let me correct it. $\endgroup$
    – AshCAD
    Sep 12, 2022 at 17:58

3 Answers 3


The sidereal month (one lunar cycle relative to the background stars) is $27.321661\text{ d}$.
The synodic month (one lunar cycle from full moon to full moon) is $29.530589\text{ d}$.
They are connected by the sidereal year (one solar cyle relative to the backgound stars), which is $365.256\text{ d}$.

$$\frac{1}{27.321661\text{ d}} =\frac{1}{29.530589\text{ d}} +\frac{1}{365.256\text{ d}}$$

To understand where this relation comes from, consider moon and sun moving across the background of the fixed stars.

enter image description here

During a time-interval $\Delta t$ (let us say a few days) the moon advances by the angular distance $\Delta\phi_\text{sidereal,moon}$, while the sun advances by $\Delta\phi_\text{sidereal,sun}$ relative to the stars. And the angular distance between sun and moon increases by $\Delta\phi_\text{synodic,moon}$. From the drawing you see $$\Delta\phi_\text{sidereal,moon} = \Delta\phi_\text{synodic,moon} + \Delta\phi_\text{sidereal,sun}$$

Dividing by the time interval $\Delta t$ you get the angular speeds $$\frac{\Delta\phi_\text{sidereal,moon}}{\Delta t} = \frac{\Delta\phi_\text{synodic,moon}}{\Delta t} + \frac{\Delta\phi_\text{sidereal,sun}}{\Delta t}$$

Inserting the measured angular speeds for the $\frac{\Delta\phi}{\Delta t}$ this becomes $$\frac{360°}{27.321661\text{ d}} = \frac{360°}{29.530589\text{ d}} + \frac{360°}{365.256\text{ d}}$$

which (after dividing by $360°$) is just the formula given at the beginning.

  • $\begingroup$ Does that mean on sidereal year, the moon's in identical phase with fixed background stars? $\endgroup$
    – AshCAD
    Sep 12, 2022 at 17:29
  • $\begingroup$ @AshCAD I'm not sure I understand what you mean. After one sidereal year the moon has done a non-integer number of sidereal months and a non-integer number of synodic months. The moon phases are (by definition) relative to the sun. They have nothing to do with the background stars. $\endgroup$ Sep 12, 2022 at 17:40
  • $\begingroup$ Oh! I checked the wiki! It says sidereal year is onger than a sloar year by 20 min and 24.5s. But in that case a sidereal year should end up repeating itself every x years. Then what is the frequency of a sidereal year? $\endgroup$
    – AshCAD
    Sep 12, 2022 at 17:41
  • $\begingroup$ @AshCAD As I said: A sidereal year is one solar cyle relative to the backgound stars. $\endgroup$ Sep 12, 2022 at 17:43
  • $\begingroup$ I didn't understand the term 'relative' over here. What do you exactly mean by 'relative to back ground stars' when you said it completes a number of non integer synodic and sidereal months. $\endgroup$
    – AshCAD
    Sep 12, 2022 at 17:46

You don't need a new method - simple algebra works just fine.

In $x$ days the line between the Earth and the Sun has moved through an angle of approximately $\frac x {365.25} \times 360$ degrees. The angle between the Moon and the Earth has moved through an angle of $\frac{(x - 27.3)} {27.3} \times 360$ degrees. For the Moon to return to the same position relative to the line between the Earth and the Sun, these two angles must be equal, So $x$ must satisfy the equation

$\displaystyle \frac x {365.25} = \frac {(x - 27.3)} {27.3} \\ \Rightarrow 27.3 x = 365.25(x-27.3) \\ \displaystyle \Rightarrow x = \frac {365.25 \times 27.3} {365.25-27.3} \approx 29.5 \text{ days}$

This is known as the Moon's synodic period.

  • $\begingroup$ Uh there's something I don't seem to understand here. Does Moon's phase depend on the line between the Sun and the Earth? And if it is then does that mean Moon's Synodic period is the position in which it almost remains in the same phase and with the same background position? $\endgroup$
    – AshCAD
    Sep 12, 2022 at 16:55
  • 1
    $\begingroup$ @AshCAD Yes, the Moon's phase (as seen from the Earth) depends on the direction of the line from the Earth to the Sun. The Moon will be full when the line from the Moon to the Sun is in the same direction as the line from the Earth to the Sun and it will be new when the the line from the Moon to the Sun is in the opposite direction to the line from the Earth to the Sun. $\endgroup$
    – gandalf61
    Sep 12, 2022 at 17:08
  • $\begingroup$ That's interesting! I never knew we could represent them on lines and can calculate the estimated phases using the lines itself. $\endgroup$
    – AshCAD
    Sep 12, 2022 at 17:33

Just work in the angular frequency domain:

$$ \omega_{\rm sideral} = \frac{360^{\circ}}{27.321661\,\rm days} = 13.176358494456103\,\rm deg/day$$

$$ \omega_{\rm year} = \frac{360^{\circ}}{(365 + \frac 1 4 - \frac 1 {100} + \frac 1 {400})\,\rm days}=0.9856198070156418\,\rm deg/day$$

and set:

$$ \omega_{\rm lunar} = \omega_{\rm sideral} - \omega_{\rm year} = 12.190738687440462\,\rm deg/day$$

Which gives:

$$ 1{\,\rm Lunar Month} = \frac{360^{\circ}}{12.190738687440462\,\rm deg/day} = 29.530614118641626\,{\rm days}$$

which is good to the 1st 5 digits (sorry I didn't clean up the insignificant digits).

  • $\begingroup$ But this doesn't solve the Question does it? The thing is if we take some fixed stars and then we say that the full moon is at 50m from star A and 80m from the star B from the naked eye then how many days will it take for the same full moon to be at 50m from star A and 80m from star B? $\endgroup$
    – AshCAD
    Sep 12, 2022 at 18:24
  • $\begingroup$ @AshCAD 19 years, which is 1 lunar cycle. $\endgroup$
    – JEB
    Sep 12, 2022 at 20:00
  • $\begingroup$ I see...that clears the problem. (⁠・⁠∀⁠・⁠) $\endgroup$
    – AshCAD
    Sep 13, 2022 at 2:26

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