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Hypothetical Question
If the earth’s rotation slowed down to align to the lunar cycle so that the orbital period began and ended at the same lunar phase, what other things would result?
For example, take the orbital period as exactly 360 earth rotations. The rotation would have decelerated. Would the speed of the lunar cycle also change, and by how much?
How might a change in rotation speed effect the orbital period?
How would all that be estimated?
Thank you!
This is not a hypothetical question since the Earth's rotation rate is slowing due to tidal interactions with the Moon. Currently, the Earth rotates once every 86164 seconds (sidereal time) for about 366 days in a year. If there were 360 rotations a year, then a day would last $86164*366/360\approx 87600$ seconds which is 1436 seconds slower than now. The Earth's rotation rate is slowing by about 2 seconds every hundred thousand years according to Deines and Williams. This means there will be 360 days in a year in about 71.8 million years.
The Earth's current rotational angular momentum is about $7.1 \times 10^{33} \text{kg} \text{m}^2 \text{s}^{-1}$. The momentum 71.8m years from now will be about 360/366 of this, or $6.98 \times 10^{33} \text{kg} \text{m}^2 \text{s}^{-1}$. This is a loss of $1.2 \times 10^{32} \text{kg} \text{m}^2 \text{s}^{-1}$. From conservation of angular momentum, the momentum lost by the slowing of Earth is gained in the Moon's orbit around the Earth.
The angular momentum of the Moon in orbit is defined as
$L=I\omega = ma^22\pi/p$
where $I=ma^2$ is the moment of inertia, and $\omega = 2\pi/p$ is the angular orbital speed and $m = 7.3459 \times 10^{22} \text{kg}$, $a=3.85 \times 10^8 \text{m}$, and $p = 2.36 \times 10^6 \text{s}$ are the mass, semi-major axis, and period of the Moon respectively.
Plugging these into the above formula for angular momentum gives us the current angular momentum of the Moon as: $L=2.9 \times 10^{34} \text{kg} \text{m}^2 \text{s}^{-1}$.
If we add Earth's angular momentum loss to the Moon's current angular momentum, in 71.8m years, the Moon's angular momentum will be $L=2.912 \times 10^{34} \text{kg} \text{m}^2 \text{s}^{-1}$. We know from Kepler's 3rd law that $p^2 = ka^3$ for some constant $k=9.7598202 \times 10^{-14}$. Which means $p=\sqrt{k}a^{3/2}$.
Substituting $p$ back into our equation for angular momentum and simplifying gives us:
$L= m\sqrt{a}2\pi/\sqrt{k}$
Solving for $a$:
$a=(\frac{L\sqrt{k}}{2m\pi})^2\approx 3.885 \times 10^8 m$
solving for a new $p$ gives us $p=2.39 \times 10^6 \text{s}$
So, when the Earth's rotation rate slows to 360 rotations per orbital revolution, the Moon's average distance will increase by almost a percentage from 385,000 km to 388,500 km (a total change of 3,500km). The Moon's orbital period will increase by more than a percentage by about 30,000 seconds or a bit over 8 hours.
In principle, nothing. There is no direct link between the length of a day, the length of a month, and the length of a year. If you have a wizard on hand to do the magic, you can change the rotation speed without changing the orbit of the moon, or the orbit of the Earth.
If you don't have a wizard, then the outrageous amounts of energy required to change any of these three values may have a secondary effect on the others. How much of an effect might depend on how you slow down the rotation period of the Earth.
