<p>Really, it's just because the <a href="https://en.wikipedia.org/wiki/Tidal_locking#Timescale" rel="nofollow noreferrer">tidal locking timescale</a> is so long for Earth:
<span class="math-container">$$t\propto\frac{a^6m_{s}}{m_{p}^2R_s^3}$$</span>
where <span class="math-container">$a$</span> is semi-major axis, <span class="math-container">$m_s$</span> is the mass of the secondary object, <span class="math-container">$m_p$</span> is the mass of the primary, and <span class="math-container">$R_s$</span> is the radius of the secondary. If we compare the Sun-Earth system to the Earth-Moon system, we see
<span class="math-container">$$\frac{a_1}{a_2}\approx380,\quad \frac{m_{s_1}}{m_{s_2}}\approx80,\quad \frac{m_{p_1}}{m_{p_2}}\approx333000,\quad \frac{R_{s_1}}{R_{s_2}}\approx3.67$$</span>
where <span class="math-container">$_1$</span> denotes the Sun-Earth system and <span class="math-container">$_2$</span> 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.</p>

CC BY-SA 4.0
2020-12-12T22:34:07.493Z
A reference for this equation would be great!
Daddy Kropotkin
CC BY-SA 4.0
2020-12-14T15:33:27.140Z
@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.
HDE 226868
CC BY-SA 4.0
2019-05-21T15:07:29.367Z
2020-12-12T22:03:41.500Z
2020-12-12T22:03:41.500Z
HDE 226868
<p>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?</p>

CC BY-SA 4.0
2019-05-21T19:50:53.677Z
If it's slowed by 5-6 hours per day since the beginning, theres many billons of years till it stops.
bandybabboon
CC BY-SA 4.0
2019-05-22T22:21:02.610Z
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.
David Hammen
<planet><orbital-mechanics><tidal-locking>
Why isn't Earth tidally locked to the sun?
feetwet
# Why isn't Earth tidally locked to the sun?

(Asked by: username_0 on 2019-05-21)

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?

**Tags:** planet, orbital-mechanics, tidal-locking

> **Comments:**
> **username_1**: If it's slowed by 5-6 hours per day since the beginning, theres many billons of years till it stops.  
> **username_2**: 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.  

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## 1 answers

### Answer by username_3 on 2019-05-21. Score: 8 | Accepted answer

Really, it's just because the [tidal locking timescale](https://en.wikipedia.org/wiki/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.

> **Comments:**
> **username_4**: A reference for this equation would be great!  
> **username_3**: @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.  

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