7
$\begingroup$

Problem

The Moon tidally brakes Earth's rotation, and as a result distances itself away from Earth as a result of conservation of total angular momentum*. Certainly, tidal braking slows the rotation of the Earth. However, and specifically, does the distancing of the Moon away from the Earth over time contribute to the increase or decrease of the Earth's rotational period (day length)?

Context

We are physics laypersons trying to understand what to us appears a dilemma, thought probably has a simple answer that escapes us. We understand that the Earth experiences tidal breaking from the Moon, i.e. the Moon slows down Earth's rotation velocity, rendering days longer. Over time, this causes the Moon to have a further orbital distance from Earth*. However, as the Moon distances itself from Earth, tidal breaking should lessen, deceleration decreases, and the Earth's rotation should increase, rendering days shorter. However, we disagree whether this is the case, and whether the contribution of (lessening) tidal braking has any impact over the conservation of total angular momentum, which suggests that a drop in Earth's rotational momentum due to tidal braking necessarily means that Earth must slow down as the Moon recedes, no matter what, which would render days longer as the Moon gradually recedes. Some of us also think that a receding Moon would result in the pulling in of Earth's ovoid tidal shape toward a more spherical shape, which would increase the rotational velocity (just as pulling in the arms of a figure skater would cause them to spin faster), rendering days shorter.

I have found this sketchy-looking paper from what may be a predatory journal which states

The earth rotation period is going to be faster [decreased, i.e. shorter days?] when the earth-moon distance increases.

In parallel, a friend (BSc in astrophysics) has given us the exact opposite answer:

as the Moon gains "orbital" angular momentum (distance from Earth), you need to subtract angular momentum from the "rotation" angular momentum of the Earth, to conserve total angular momentum of the (closed) system

Could someone please explain how to correctly understand this problem, and settle the confusion between all of us curious about the physics? We are trying our best to understand the complete picture. Apologies for the layman's naiveté. Thank you.


*Total angular momentum is always conserved; as Earth's (spin/rotational) angular momentum decreases due to tidal braking, the Moon's (orbital) angular momentum increases to maintain total angular momentum; resultantly, the Moon's orbit grows further out.

$\endgroup$
5
  • 2
    $\begingroup$ I'm not sure how you come to the conclusion that "as the Moon distances itself from Earth, tidal breaking should lessen, deceleration decreases, and the Earth's rotation should increase, rendering days shorter." Deceleration decreasing in rate does not imply that it will reverse sign and turn into acceleration; it just means that the rate at which the rotation slows will itself become slower. $\endgroup$
    – Lawton
    Aug 16 at 12:54
  • $\begingroup$ Hi! And thanks. Is the fallacy I am making then the fact that removing the Moon entirely would not speed up the Earth's rotation at all (to take a hypothetical case)? Perhaps I was reasoning that there is something intrinsically accelerating Earth's rotation and the Moon is counteracting that, by decelerating Earth's rotation via tidal braking. The analogy would be lifting the break of a car going downhill would make it accelerate. But the reality is that the car is not at all going downhill, it's level. Do I explain the fallacy correctly? $\endgroup$
    – S Pr
    Aug 16 at 16:17
  • $\begingroup$ Your explanation of your fallacy sounds correct. The cause of Earth's rotation is angular momentum, not a driving force. Removing the brake will not make it faster, it will stop it from getting slower. Over time, the brake will be weaker. So the first derivative is to make the Earth's period slower. The second derivative is to make the first derivative slower, but you can decelerate and move forward at the same time, or accelerate and move backward at the same time. $\endgroup$
    – Alwin
    Aug 17 at 0:54
  • $\begingroup$ The details of Earth's rotation are complex. From iers.org/IERS/EN/Science/EarthRotation/EarthRotation.html "The variability of the earth-rotation vector relative to the body of the planet or in inertial space is caused by the gravitational torque exerted by the Moon, Sun and planets, displacements of matter in different parts of the planet and other excitation mechanisms". $\endgroup$
    – PM 2Ring
    Aug 17 at 1:02
  • $\begingroup$ (cont) "The observed oscillations can be interpreted in terms of mantle elasticity, earth flattening, structure and properties of the core-mantle boundary, rheology of the core, underground water, oceanic variability, and atmospheric variability on time scales of weather or climate". $\endgroup$
    – PM 2Ring
    Aug 17 at 1:03
6
+100
$\begingroup$

A first look

Let’s call the angular momentum of the Earth’s rotation about its axis $L_E = I_E \omega_E$, where $I_E \lesssim \frac25 M_E R_E^2$ is the moment of inertia for a sphere whose density is greater at the core, and the frequency $\omega_E$ is $2\pi$ radians per day. This is somewhat less than half of the angular momentum in the Earth-Moon system.

Similarly the moon has angular momentum $L_M=I_M \omega_M$ from spinning around its axis once per month. That’s negligible compared to $L_E$, because $I_M \approx \frac{I_E}{80 \cdot 4^2}$ is small and $\omega_M \approx \omega_E/30 $ is slow.

Somewhat more than half of the angular momentum in the Earth-Moon system is from the orbital term. We have $I_{EM} / I_E \approx \frac{60^2}{80} \approx 60$, because the Moon is about 1/80th of Earth’s mass but it is 60 Earth radii away.* But the month $\omega_{ME} = \omega_M$ is about thirty times slower than the day $\omega_E$, so the angular momentum product comes out comparable.

If we approximate that the Moon orbits in Earth’s equatorial plane (it doesn’t), so that all the angular momenta are parallel, we have the conserved angular momentum

$$ L_\text{total} = I_E \omega_E + I_{EM} \omega_M = \text{constant} $$

and the two terms contribute similarly.

Tidal locking happens as rotational energy is extracted to do dissipative work. The rotational energy is

\begin{align} T &= \frac12 I_E \omega_E^2 + \frac12 I_{EM} \omega_M^2 \\ &= \frac{L_E^2}{2 I_E} + \frac{L_{EM}^2}{2 I_{EM}} \end{align}

Since the angular momenta $L_E$ and $L_{EM}$ are within a factor two or so, but the Earth has a smaller moment $I_E$ and larger frequency $\omega_E$, then nearly all of the kinetic energy is from the spinning-Earth term. Furthermore Earth’s moment of inertia is fixed, while the Earth-Moon distance is free to vary. Therefore any increase in Earth’s rotational speed would increase the total rotational energy of the system:

\begin{align} L_E &\to L_E + \delta L \\ L_{EM} &\to L_{EM} - \delta L \\ T &\to T + \delta L\cdot \left( \frac{L_E}{I_E} - \frac{L_{EM}}{I_{EM}} \right) + \frac{\mathcal O (\delta L^2)}{I} \\ &\approx T + \delta L \cdot ( \omega_E - \omega_M ) \end{align}

Tidal locking therefore happens as the total kinetic energy $T$ is decreased by the transfer of angular momentum in the direction $L_E \to L_{EM}$. This will continue until the frequencies $\omega_E, \omega_M$ become the same. If $I$ is fixed, the way to release kinetic energy is to transfer angular momentum from the higher-frequency subsystem to the lower-frequency one.

More careful: use total energy

What about the Moon, whose orbital moment $I_{EM}$ is not fixed? Could the tidal locking process perhaps speed up its orbit? To answer this, we have to remember that the Moon is held in its orbit by gravity, with centripetal acceleration

\begin{align} \frac{\text{constant}}{R_{EM}^2} &= R_{EM} \omega_{EM}^2 \\ \text{constant} &= R_{EM}^3\omega_{EM}^2 \end{align}

If we increase the orbital angular momentum via some interaction, the new orbit will still follow this constraint:

\begin{align} L & \propto R^2\omega & L &\to (1 + \epsilon) L \\ & & \frac{L^2}{R^3\omega^2} & \to \frac{1+2\epsilon} {\text{constant}} \frac{L^2}{R^3\omega^2} \\ R &\propto \frac{L^2}{R^3\omega^2} & R &\to (1 + 2\epsilon) R \\ \omega &\propto \frac{L}{R^2} & \omega &\to \frac{1+\epsilon}{1-4\epsilon}\omega \approx (1 - 3\epsilon)\omega \end{align}

So for a gravitational orbit, increasing its angular momentum will necessarily make it bigger and slower. If you allow the size of the orbit to change you have to include changes in the gravitational potential energy $U_{EM} = -\frac{G M_E M_M}{ R_{EM}}$ in your energy calculations. After a nontrivial amount of algebra, we end up with the same result for the total energy $E = T + U$ that we got above for the fixed-distance case,

$$ \frac{\mathrm dE}{\mathrm dL_E} = \omega_E - \omega_M, $$

or in words, the total energy of the system is reduced when angular momentum flows from the faster rotation to the slower one.

Note that this is a "mechanism-free" argument. Another answer here, and the Wikipedia page about Phobos (on which more below), go into detail about the locations of the "tidal bulges." But the "tidal bulge" model is mostly bogus, especially for Earth's oceans. This model says that if the mechanical energy of the Earth-Moon system is reduced, without changing $L$, it must also transfer angular momentum out of Earth's spin and into the Moon's orbit. You can push the Moon away by improving the efficiency of your local tidal-powered electrical generator.

Stability

Let's emphasize that this analysis also applies to other planets and their orbiting moons with a change of notation. We'll talk about "a planet" with mass $M$ and radius $R$, with "a moon" having mass $m<M$ and radius $r$, in an equatorial, circular orbit with radius $a > R+r$. The angular momenta for the planet $L_p$ and the orbit $L_o$ are related to the total energy $E = T+U$ by

\begin{align} L &= L_p + L_o = I_p\omega_p + I_o \omega_o \\ E &= \frac{L_p^2}{2I_p} - \frac{L_o^2}{2I_o} \end{align}

This formulation makes it clear that the angular momentum $L_m = I_m \omega_m$ is negligible unless the moon in spinning very fast, or is very large and very close. The moment of inertia for the orbit is $I_o = \mu a^2$, where $\mu = (\frac1m + \frac1M)^{-1} \lesssim m < M$ is the "reduced mass." This orbital moment is always§ larger than the moon's rotational moment $I_m \lesssim mr^2 < m(r+R)^2 < ma^2$.

Shifting angular momentum to the planet while holding $L$ constant changes the energy by

$$ \frac{\mathrm dE}{\mathrm dL_p} = \omega_p - \omega_o $$

That is, energy is released when angular momentum flows from the faster-rotating part of the system to the slower-rotating part of the system. For the Earth-Moon system this has the effect of making both the spin and the orbital frequencies slower: the extra angular momentum causes the Moon's orbit to move out to a larger radius.

But for a system with $\omega_p < \omega_m$, energy will be released when angular momentum flows to the planet. The extra spin energy for the planet, as well as whatever energy was released, has to come from the gravitational energy $U=-\frac{Gmm}{a}$ of the moon, which would move into a lower, faster orbit. It's fair to ask, then: is a tidally-locked orbit a stable configuration? Or, having achieved a tidal lock, will the system continue to release energy by having the moon move closer and faster, until it crashes into the planet's surface?

The condition for stability is that the second derivative of the energy

$$ \frac{\mathrm d^2E}{\mathrm dL_p^2} = \frac{1}{I_p} - \frac{1}{I_o} $$

be positive. We've already seen that, in the Earth-Moon system, $\frac{I_{EM}}{I_E} \approx 10^2$, and getting larger as the Moon moves away. But for a different system with a smaller, nearer moon, it's possible for the orbital moment $I_o \approx ma^2$ to be smaller than the planet's moment $I_p \lesssim MR^2$.

Note that since we always have $I_m<I_o$, it's always stable for the moon to become tidally locked to the planet.

Interesting cases

Let’s look at the consequences of these relationship in a few interesting scenarios.

  1. A planet spinning faster than the orbit of its large moon will evolve towards a stable tidal lock. This is the Earth-Moon system. We can figure out the eventual separation $a$ and frequency $\omega$ by fixing the constraints \begin{align} (I_p + \mu a^2) \omega &= L \\ a^3 \omega^2 &= G(M+m) \end{align} Using this value of $I_E \approx \frac13 M_E R_E^2$ for Earth's moment of inertia, I get an equilibrium distance of about 40% larger than the current Earth-Moon distance, with a period of about 47 days.

    I have read that the Earth-Moon tidal lock may not finish happening before the Sun turns into a red giant and swallows them both, but I haven't tried to compute that.

    On the graph below, this is the middle region.

  2. An planet spinning very much faster than is moon is only stable in isolation. If the equilibrium distance is too far, the moon will eventually wander out of the planet's "Hill sphere" and be lost. I think this is related to the fact that the rapidly-spinning planet can give the total system energy $E>0$, which is gravitationally unbound. The Earth-Moon system currently has $E>0$, but would have $E<0$ at the equilibrium distance I described. The radius of Earth's Hill sphere is about three or four current Earth-Moon distances.

    Mars's outer moon, Deimos, could migrate out of Mars's Hill sphere without materially changing the length of the Martian day.

  3. A planet rotating more slowly than the orbit of a large moon can evolve towards a stable, tidally-locked equilibrium. However, the equilibrium is achieved by speeding up both the planet's rotation and the moon's orbit. On the graph below, this is the region on the right.

  4. A planet rotating more slowly than the orbit of a small moon is in for an adventure.

    If the orbital moment of inertia $I_o = \mu a^2$ is less than the planet's moment of inertia $I_p$, the tidally-locked equilibrium $\omega_p = \omega_o$ is unstable. Any system where the moon is too close (small $a$) will release energy by transferring angular momentum to the planet, moving the moon closer. If the moon is too small, this inward motion is away from equilibrium, instead of towards, and will end with the moon crashing into the planet's surface.

    This is why atmospheric drag makes low-Earth satellites fall out of orbit: they trade angular momentum with Earth by interacting dissipatively with the atmosphere, rather than tidally. In the longer term, this is the fate of Mars's moon Phobos, which rises in Mars's western sky about three times a day.

Here's a plot showing the total energy $E$ and frequencies for the Earth $\omega_E$ and Moon $\omega_M$, in orbits where the Moon is tidally locked to the Earth and the total angular momentum $L$ for the system is equal to the $L$ that we have today:

plot, see text

The horizontal axis is the Earth-Moon distance, in units of the current Earth-Moon distance $\renewcommand{\anow}{a_\text{now}} \anow$. Currently the Moon orbits at $a = \anow$, and the Earth spins about thirty times faster than the Moon rotates. You can see the stable equilibrium at $a \approx 1.4\anow$. The unstable equilibrium for our large Moon is extremely close to Earth, at only about $0.04\anow \approx 2.2 R_E$. That's actually quite close to where the Moon formed — perhaps the model making that prediction is that the Moon formed from an accretion disk straddling the unstable equilibrium.

If you change this model so that Earth spins more rapidly while the Moon orbits at its current distance, the $\omega_E$ curve shifts up and the two equilibria move further apart. If you give the Earth a very fast retrograde rotation, it's possible to move the $\omega_E$ curve down enough that there is no stable equilibrium.

tl;dr summary

Interactions between co-orbiting objects which reduce their mechanical energy do so by moving angular momentum from the higher-frequency rotation modes to the lower-frequency rotation modes. The Earth and Moon are both slowing towards an equilibrium in which they would be mutually tidally locked; that equilibrium will be stable because the Moon is large and far from the Earth. However, moons that are very small, or very near their planets, or in retrograde orbits, may be unstable against tidal interactions; those moons will eventually fall out of the sky.

Notes

* I’m being intentionally sloppy about the $\frac25$, which is wrong anyway because of Earth’s dense core. I’m also ignoring, for now, Earth’s orbital motion about the Earth-Moon barycenter.

Earth’s moment of inertia changes a little bit as the shape of the Earth changes, e.g. during quakes, landslides, volcanic eruptions, and the falling of leaves in the autumn. In the longer term Earth's moment of inertia is changed by convection currents in the mantle.

§ Unless (possibly) you get into a weird edge case where the two bodies are very close and the lower-mass object has a very low density. We'd have to revisit a lot of our assumptions for such a system.

$\endgroup$
4
  • $\begingroup$ This is a very thorough answer, thanks. Would a fair tl;dr be that Earth's rotation quickens if Moon's rotation is quicker than Earth's, and vice versa? When considering rotation only, you say that when the Moon's rotation is slower than Earth's (as is the case), they eventually slow each other to tidally lock completely. Do I understand correctly? $\endgroup$
    – S Pr
    Aug 18 at 8:12
  • 2
    $\begingroup$ Careful to distinguish between rotation (spin about axis) and orbital motion. You’re correct that a fast planet is slowed by a slow moon, and that a slow planet is sped up by a fast moon. Whether they reach equilibrium is nontrivial (though equilibrium is the fate of our Earth and our Moon). I have a draft of a clarified answer, which I’ll post later. $\endgroup$
    – rob
    Aug 18 at 22:40
  • $\begingroup$ I'm still not sure if I've answered the question in your comment because of the "and vice versa" part. In the language of the edited answer: Consider a system with low-mass moon, $I_o = \mu a^2 < I_p$. It's clear to me that if such a moon is too close / too fast, it'll speed up the planet's rotation until it falls onto the surface. But I'm confused about what happens to a low-mass moon on the "high and slow" side of the unstable equilibrium. $\endgroup$
    – rob
    Aug 21 at 20:00
  • $\begingroup$ If the planet's rotation is faster than the moon's orbit, the constant-$L$ system has a lower/faster orbit which is an unstable equilibrium and a higher/slower orbit which is a stable equilibrium. If the planet's rotation is slower than the moon's orbit, the existence of the two equilibria is a numerical question. If the planet's angular momentum is too low (including, possibly, retrograde rotation) then the orbit may be unstabilizable. Answer updated. $\endgroup$
    – rob
    Aug 22 at 14:24
2
$\begingroup$

Others have answered about the mechanisms, so let me just address where your conceptual confusion comes from:

However, as the Moon distances itself from Earth, tidal breaking should lessen, deceleration decreases, and the Earth's rotation should increase, rendering days shorter.

That last phrase isn’t correct. Yes, the deceleration decreases toward zero. Earth slows down slower and slower, with the day finally becoming constant-length.

But the deceleration never changes sign; it never becomes acceleration. There’s no mechanism to speed the earth back up, so days don’t get shorter.

It’s a bit like a block slowing down on a table: the deceleration stops it, but it never has a way to speed it back up.

$\endgroup$
0
$\begingroup$

The gradient in the gravity from the moon pulls the oceans away from the earth on the near side, and the earth away from the oceans on the far side. Then the rotation of the earth drags the high tides to the east (ahead of the moon). The greater pull of the moon on the near side tide produces a torque which is slowing the rotation of the earth. The greater pull of the near side tide on the moon has a tangential component which tends to accelerate its orbital speed, but the extra energy instead leads to an increase in the orbital radius.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.