It is my understanding that the tidal forces of the Moon acting on Earth cause it to slow down its rotation and, because angular momentum is conserved, the Moon's orbit subsequently expands. This continues until the Earth's rotation is synchronous with the Moon's orbit leaving both bodies tidally locked to each other.

However once that happens, how will the tidal forces exerted by the Sun on Earth (which are weaker than the Moon's) affect the Earth's rotation? Would they cause the Earth to start rotating again and thus cause the Moon's orbit to shrink?

P.S. I do realize the Earth is not destined to be tidally locked to the Moon until after the Sun evolves into a red giant, so for the sake of simplicity assume the Earth-Moon system survives unscathed without getting disturbed.

  • $\begingroup$ Regarding the PS, you are asking us to scientifically explain something that cannot scientifically happen. That doesn't make sense. Voting to close. $\endgroup$ May 20, 2020 at 13:04
  • $\begingroup$ I’m voting to close this question because the question asks us to use science to explain something that science says will not happen. $\endgroup$ May 20, 2020 at 13:05
  • 1
    $\begingroup$ Although hypothetical scenarios are typically off-topic for this site, this seems like an answer that could get interesting answers. The evolution of the Sun into a red giant has no impact on the tidal locking of the Earth and the Moon. So tidal locking can be considered independently from the Sun's evolution. $\endgroup$
    – usernumber
    May 20, 2020 at 13:34
  • $\begingroup$ All problems in physics make some simplifying assumptions. This one doesn't seem exuberant, so I vote to leave open. $\endgroup$
    – usernumber
    May 20, 2020 at 13:36
  • 2
    $\begingroup$ @usernumber - One key reason that the Earth will never become tidally locked to the Moon is that the Earth's oceans are responsible for almost all of the Earth's rotational deceleration. Thank's to the ever increasing so-called solar constant, the Earth will lose its oceans in about a billion years, well before the Sun turns into a red giant, and well before the Earth could ever become tidally locked to the Moon. $\endgroup$ May 20, 2020 at 13:48

1 Answer 1


user177107 should be commended for asking this clever question.

In short, the described configuration is possible, at least in principle.

First of all, let us recall that the solar tides in a planet whose spin is synchronised with the star are always working to ensure descent, see eqn (150) in this work. On the other hand, the solar tides in a nonsynchronised planet are working to repel the planet from the star.

A large moon orbiting a not too large planet may synchronise this planet's rotation. Planet's rotation synchronised with the moon will not be synchronised with the star. As a result of this, the solar tides in the planet may be contributing either to the tidal ascent or to the tidal descent to the star, dependent on whether the planet's (to be exact, the planet-moon barycentre's) mean motion $n$ about the star is smaller or larger than the planet's spin rate. As demonstrated here the spin rate of a planet synchronised by a moon will, typically, be several times higher than the mean motion $n$ of the planet about the star. So the solar tides in the planet will be repulsive -- and this is one possible mechanism helping a close-in planet to avoid engulfment.

Two technical warnings to those who decide to integrate this problem.

(1) If a prograde moon starts above the synchronous radius (like Deimos or the Moon), it will usually be ascending. I am saying `usually', because different is a situation where the eccentricity is noticeable. (And it is likely that Phobos was at some point in this exceptional situation -- which helped it to fall down through the synchronous orbit.) Assuming that nothing prevents the orbit from circularisation, and that the moon is ascending, we have to take into account that at some point the moon may be lost to the star. Naively, this should happen when its semimajor axis crosses the Hill radius $r_H$. In reality, the orbit becomes unstable already at $0.49 r_H$, for a prograde-orbiting moon, and $0.93r_H$, for a retrograde-orbiting one. Keep this detail in mind if you try to integrate.

(2) If a prograde moon starts below the synchronous radius, it will be falling down and, at the same time, accelerating the planet's rotation. It may result in synchronisation of the planet with the moon -- provided this end state is attained before the descending moon crosses the Roche radius.

  • $\begingroup$ Ok, but as David Hammen mentioned in the question comments, the tidal motion on Earth will be extremely reduced in a billion years, when the oceans are gone. (Also, tectonic plate motion will virtually cease). And in a few more billion years, the mass loss of the Sun will surely have an effect on planetary orbits, especially in the inner Solar system. Perhaps the Earth & Moon can survive the early stages of the Sun's red giant phase, I doubt they'll survive the later stages (but of course that's just a guess). $\endgroup$
    – PM 2Ring
    Jul 2, 2022 at 16:07
  • 1
    $\begingroup$ Surely the effects also depend on the rotation rate of the Sun and whether that is faster or slower than the Earth/Moon orbital period. The Sun's rotation rate is basically unaffected by its tidal interaction with the Earth. $\endgroup$
    – ProfRob
    Jul 21, 2023 at 20:06
  • $\begingroup$ @ProfRob The rotation rate of the Sun will determine in which direction the solar tides will be pushing the Earth. Given that the Sun's rotation is faster than the Earth's mean motion about it, the Earth-generated tides in the Sunn will be working to push the Earth away (a feeble effect actually, given the large distance). I do not see, however, how the rotation rate of Sun can influence the working of the solar tides in the Earth. That can depend only on the Earth's rotation rate. Please correct me if I am missing something. $\endgroup$ Jul 21, 2023 at 21:22

You must log in to answer this question.

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