This answer to Will the Earth ever be tidally locked to the Moon? supports the widely held thinking that during the Sun's red giant phase or later the Earth and the Moon should be tidally locked to each other. The Earth is said to have a slower rotation then.

Do we have any idea how long a sidereal day and the mean solar day on Earth would be in several billion years?

  • 1
    $\begingroup$ This is an interesting question, but can you mention where you read that they should be tidally locked when the Sun is in a red giant phase? Thanks! $\endgroup$
    – uhoh
    Commented Jul 14, 2020 at 6:54
  • $\begingroup$ @uhoh On one occasion it was in an answer to one of the questions in the following link. But it's the general assumption among many scientists that the Earth and the Moon are to be tidally locked once in case they'd survive the red giant phase. astronews.com/frag/index.shtml $\endgroup$
    – Ioannes
    Commented Jul 14, 2020 at 7:13
  • $\begingroup$ In Stack Exchange it is a good idea to support each question's premise with facts or supporting links. In this case there's some information here already! :-) $\endgroup$
    – uhoh
    Commented Jul 14, 2020 at 7:21
  • 1
    $\begingroup$ It would surely be well after the Sun becomes a white dwarf. There's not enough time in six or seven billion years for the Earth-Moon system to become tidally locked. $\endgroup$
    – Brian Tung
    Commented Jul 18, 2020 at 2:12

2 Answers 2


The last part of the answer you linked to actually says (right at the end) that tidal locking will never be achieved, with reasoning similar to what I gave in this answer.

That said, even though the Moon and the Earth will never actually achieve tidal synchronization, we can still do the thought experiment and ask, "If there were enough time for the current Earth-Moon system to achieve tidal synchronization, what would the length of the day be at which the rotation of the Earth and the orbit of the Moon would be synchronized?"

To do this, we can assume that the Moon spirals outward due to an exchange of angular momentum between the Earth's rotation and the Moon's orbit. The Earth's spin slows down as it loses angular momentum, and the Moon moves into a larger (and thus higher angular momentum) orbit as it gains that same angular momentum. The rotation of the Moon would presumably stay locked to the Moon's orbital period, so it would slow down as well.

So, using $L$ to represent angular momentum, the key equation is

$$ L_{\rm now} = L_{\rm then} $$

where "then" is some time in the future when lock is achieved. The total angular momentum in the system is constant.

The angular momentum of any object is $L = I\omega$, where $I$ is the moment of inertia, and $\omega = \frac{2\pi}{P}$ is the orbital frequency, related to the orbital period $P$. For a constant-density sphere of mass $M$ and radius $R$ rotating on its axis, $I = 0.4 M R^2$. The Earth and the Moon are somewhat more centrally condensed, so their moments of (rotational) inertia are a little smaller than the 0.4 for a uniform sphere. The leading coefficient is 0.33 for the Earth, and 0.39 for the Moon.

For the Moon orbiting the Earth, it's a good approximation to just treat it as a point mass (since its size is small compared to its distance from Earth), so it has $I = M_{\rm Moon}R_{\rm Earth-Moon}^2$.

Putting all three of these motions (Earth rotation, Moon orbit, Moon rotation) together, we get

$$L_{\rm now} = 2\pi \left( \frac{0.33 M_\oplus R_\oplus^2}{P_\oplus} + \frac{0.39 M_{\rm Moon} R_{\rm Moon}^2}{P_{\rm Moon}} + \frac{M_{\rm Moon} R_{\rm Earth-Moon}^2}{P_{\rm Moon}} \right) $$

All of the values there represent current, known values, i.e. $P_\oplus = 1$ day, and $P_{\rm Moon} = 1$ month = 27.3 days. Similarly, at some point in the (hypothetical) future, we would have

$$ L_{\rm then} = 2\pi \left( \frac{0.33 M_\oplus R_\oplus^2}{P_{\rm then}} + \frac{0.39 M_{\rm Moon} R_{\rm Moon}^2}{P_{\rm then}} + \frac{M_{\rm Moon} R_{\rm Earth-Moon,\ then}^2}{P_{\rm then}} \right) $$


$$ L_{\rm then} = \frac{2\pi}{P_{\rm then}} \left(0.33 M_\oplus R_\oplus^2 + 0.39 M_{\rm Moon} R_{\rm Moon}^2 + M_{\rm Moon} R_{\rm Earth-Moon,\ then}^2 \right) $$

Notice that there's just a single period here, since everything is now assumed to be synchronized. So we could set this equal to $L_{\rm now}$ and solve for $P_{\rm then}$ – except that we have a second unknown in the equation, $R_{\rm Earth-Moon,\ then}$, the new orbital distance of the Moon from the Earth. Fortunately, we can use Kepler's third law to relate this distance to the orbital period:

$$ P_{\rm then}^2 (M_\oplus + M_{\rm Moon} ) = \frac{4 \pi^2 }{G} R_{\rm Earth-Moon,\ then}^3 $$

To make life a little easier when we substitute into the equation, we could write this as a proportion with the current values, which makes some of the constants cancel:

$$ \frac{P_{\rm then}^2}{P_{\rm Moon}^2} = \frac{R_{\rm Earth-Moon,\ then}^3}{R_{\rm Earth-Moon}^3} $$

which means that

$$ R_{\rm Earth-Moon,\ then}^2 = R_{\rm Earth-Moon}^2 \left(\frac{P_{\rm then}}{P_{\rm Moon}}\right)^{4/3} $$

Substituting that into our expression for $I_{\rm then}$, we finally end up with

$$L_{\rm then} = \frac{2\pi}{P_{\rm then}} \left(0.33 M_\oplus R_\oplus^2 + 0.39 M_{\rm Moon} R_{\rm Moon}^2 + M_{\rm Moon} R_{\rm Earth-Moon}^2 \left(\frac{P_{\rm then}}{P_{\rm Moon}}\right)^{4/3} \right) $$

So in principle, we're done - we set this equal to $L_{\rm now}$ and solve for $P_{\rm then}$. It's not a simple equation to solve analytically, but not hard to solve numerically.

Symbolically there's a lot going on, but we know most of these values, so if we plug in numbers for everything we know and simplify, this becomes

$$ P_{\rm then} = 0.16809413\ {\mathrm d} + 0.27626727\ {\mathrm d}^{-1/3} P_{\rm then}^{4/3} $$

where the "d" represents units of days. Solving this equation gives a period of 46.9 days, so that is how long the day, the month (i.e. the orbital period of the Moon), and the rotation period of the Moon would be if all were to become tidally locked to each other.

If you want to see the calculation done in Python, I've posted the code in a gist here. It's a nice example of the usefulness of Python's quantities and astropy's constants.

  • $\begingroup$ Very nice! I didn't believe that the Earth would nearly stop and the Moon would move out so little, so I went and calculated the rotational angular momentum of the Earth and orbital angular momentum of the Moon and was surprised to find out that the latter is five times bigger than the former. Wow that factor of 50 in $r$ really helps! $\endgroup$
    – uhoh
    Commented Jul 18, 2020 at 3:48
  • $\begingroup$ Thank you very much! One question: what is "Python"? $\endgroup$
    – Ioannes
    Commented Jul 18, 2020 at 6:10
  • 1
    $\begingroup$ @uhoh Yes, the long lever arm of the Moon’s orbit makes a big difference. In the solar system overall, most of the mass is in the Sun, but most of the angular momentum is in the orbits of the planets, especially the giant planets. $\endgroup$ Commented Jul 18, 2020 at 12:54
  • $\begingroup$ yep yep, for example despite Jupiter being so much larger, Neptune does quite a job pushing the Sun around the solar system's barycenter. $\endgroup$
    – uhoh
    Commented Jul 18, 2020 at 12:56
  • 1
    $\begingroup$ @Greenhorn Python is a computer programming language. It is used a lot in astronomy research these days (though of course for a lot of other things, too), and astropy is a set of specialized astronomy routines written in Python. $\endgroup$ Commented Jul 18, 2020 at 13:00

Most of the answer is wonderfully written above by Eric Jensen. I shall only add one final touch. As demonstrated by Eric, synchronisation is attained when the Moon's orbital period reaches $P_{then} = 46.9$ days. Implying that $R$ stands for the semimajor axis, we take, for an estimate, $R_{Moon} = 3.84\times 10^8$ m. Also, $P_{Moon} = 27.3$ days. Insertion of these numbers into the Kepler law renders $$ R_{then} = R_{Moon}\,\left(\frac{P_{then}}{P_{Moon}} \right)^{2/3} = 5.72\times 10^8\;\mbox{m}\;\,. $$ This value must be residing within the Earth's Hill sphere, i.e. within the domain where the Moon is attracted by the Earth stronger than by the Sun, so it does not fly away.

Naively, the radius of the Hill sphere of the Earth is $$ r_H = a_E \,(1-e_E) \,\left(\frac{M_E}{3M_\ast} \right)^{1/3} = 1.47\times 10^9\;\mbox{m}\;\,, $$ where $a_E$ and $e_E$ are the Earth's semimajor axis and eccentricity, while $M_E$ and $M_\ast$ are the Earth's and Sun's masses.

However, numerical study by Astakhov et al. (2003) and Domingos et al. (2006) demonstrated that the orbits too closely approaching $r_H$ are unstable in the long term, while those confined to a smaller region (which we may call ''the reduced Hill sphere'') stay stable. The radius of the reduced Hill sphere (call it ''the reduced Hill radius''), for a circular (e = 0) lunar orbit, is $$ r^{\,\prime}_H = B\, a_E\, (1 - b\, e_E)\,\left(\frac{M_E}{3M_\ast} \right)^{1/3} \;\,. $$ For prograde moons, $B=0.49$ and $b = 1.03$, wherefrom
$$ R^{\,\prime}_H = 7.09\times 10^8\;\mbox{m}\;\,. $$ We see that the distance $R_{then}=5.72\times 10^8$ m will barely fit into the reduced Hill sphere. Anyhow, by sheer luck, the synchronous configuration will be stable.

  • 1
    $\begingroup$ You don't have to neglect eccentricity to easily write down the angular momentum of the orbit. $\endgroup$
    – ProfRob
    Commented Oct 23, 2022 at 7:43
  • $\begingroup$ @ProfRob Yes. You are right. I shall remove those words. Thank you. $\endgroup$ Commented Oct 23, 2022 at 8:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .