Given that the moon has long been tidally locked with the Earth, why isn't Earth (or any of our other solar system's planets) tidally locked to the sun?

  • $\begingroup$ If it's slowed by 5-6 hours per day since the beginning, theres many billons of years till it stops. $\endgroup$ Commented May 21, 2019 at 19:50
  • $\begingroup$ Mercury essentially is tidally locked to the Sun. While Mercury is not synchronously locked (in a 1:1 spin orbit resonance) to the Sun, it is nonetheless "locked", in a 3:2 spin orbit resonance. The high eccentricity of Mercury's orbit makes that 3:2 spin orbit resonance much more likely than a 1:1 resonance. $\endgroup$ Commented May 22, 2019 at 22:21

1 Answer 1


Really, it's just because the tidal locking timescale is so long for Earth: $$t\propto\frac{a^6m_{s}}{m_{p}^2R_s^3}$$ where $a$ is semi-major axis, $m_s$ is the mass of the secondary object, $m_p$ is the mass of the primary, and $R_s$ is the radius of the secondary. If we compare the Sun-Earth system to the Earth-Moon system, we see $$\frac{a_1}{a_2}\approx380,\quad \frac{m_{s_1}}{m_{s_2}}\approx80,\quad \frac{m_{p_1}}{m_{p_2}}\approx333000,\quad \frac{R_{s_1}}{R_{s_2}}\approx3.67$$ where $_1$ denotes the Sun-Earth system and $_2$ denotes the Earth-Moon system. To make up for these differences (and to have Earth be tidally locked by now), assuming similar Love numbers and dissipation functions for Earth and the Moon, we would need Earth's initial spin to be substantially smaller than the Moon's initial spin by many, many orders of magnitude, and this just wasn't the case.

  • 1
    $\begingroup$ A reference for this equation would be great! $\endgroup$ Commented Dec 12, 2020 at 22:34
  • $\begingroup$ @N.Steinle It's the same one as on Wikipedia; I just made the assumption that the inertia of the satellite is $I\sim m_sR_s^2$ and simplified. $\endgroup$
    – HDE 226868
    Commented Dec 14, 2020 at 15:33

You must log in to answer this question.

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