Length of Lunar Month

Question

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?

Answer 1

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 $365.256\text{ d}$.

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

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

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.

Answer 2

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.

Answer 3

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).