<p><strong>Tidal locking</strong> is a <strong>primary</strong> effect due to a direct torque on a body from its tidal bulge lagging its rotation.  This is pictured nicely on <a href="https://en.wikipedia.org/wiki/Tidal_acceleration" rel="noreferrer">wikipedia</a> with an example from the Earth/Moon system.</p>
<p><a href="https://i.sstatic.net/JFVsm.png" rel="noreferrer"><img src="https://i.sstatic.net/JFVsm.png" alt="enter image description here" /></a></p>
<p><strong>Tidal Orbital Circularization</strong> is a <strong>secondary</strong> effect.  To quote Daddy Kropotkin's excellent answer to</p>
<p><a href="https://astronomy.stackexchange.com/questions/28962/is-the-moons-orbit-circularizing-why-does-tidal-heating-circularize-orbits">Is the moon&#39;s orbit circularizing? Why does tidal heating circularize orbits?</a></p>
<blockquote>
<p>tidal torque drives dissipation, and this dissipation brings the
binary to a minimum kinetic energy state, i.e. circular orbit,
synchronized spins with the orbit, aligned spins with the orbit.</p>
</blockquote>
<p>There is really no reason to expect the timescales for these two effects to be similar since there are different mechanisms at play.</p>
<p>Another way to think about tidal circularization is to imagine a tidally locked planet in an eccentric orbit.  From Kepler's 2nd law, we know the planet moves fastest at periapsis and slowest at apoapsis.  So the planet's tidal bulge won't always quite face its star.  Instead, the tidal bulge will lead and lag its orientation to the star.  This offset will reduce orbital energy at periapsis and increase orbital energy at apoapsis, circularizing the orbit. Tidal heating will also continue to occur, since the tidal bulge will continue to move until the orbit comes to a minimum energy state (a circular orbit).</p>
<p>Is there <em>a simple way to quantify the timescale of circularization</em>?</p>
<p><strong>No.</strong> There is a way, but it is not simple. From <a href="https://arxiv.org/abs/0903.0763" rel="noreferrer">Rodriguez and Ferraz-Mello 2009</a>, the time scale for orbital circularization of a short period planet in a 2-body system is</p>
<p><span class="math-container">$$\tau_e=\frac{3n^{-1}a^5}{18\hat{s}+7\hat{p}}$$</span></p>
<p>Here, <span class="math-container">$n$</span> is the mean orbital motion, <span class="math-container">$a$</span> is the orbital semi-major axis, and
<span class="math-container">$$\hat{s} = \frac{9k_{d*}m_p}{4Q_*m_*}R_*^5$$</span>
<span class="math-container">$$\hat{p} = \frac{9k_{dp}m_*}{2Q_pm_p}R_p^5$$</span></p>
<p>where <span class="math-container">$Q$</span> are quality factors defined by <span class="math-container">$Q_*=|\epsilon'_{0*}|^{-1}$</span> and <span class="math-container">$Q_p=|\epsilon'_{2p}|^{-1}$</span>.  Here, <span class="math-container">$\epsilon'_{0*}$</span> and <span class="math-container">$\epsilon'_{2p}$</span> are lag angles associated to the tidal waves whose frequencies are <span class="math-container">$2\Omega_*-2n$</span> for the star and <span class="math-container">$2\Omega_p-n$</span> for the planet.  The angular velocity of the rotation of the tidally deformed body is <span class="math-container">$\Omega$</span>.  The masses, radii, and dynamical Love numbers for the star and planet are respectively <span class="math-container">$m_*$</span>,<span class="math-container">$R_*$</span>,<span class="math-container">$k_{d*}$</span> and <span class="math-container">$m_p$</span>,<span class="math-container">$R_p$</span>,<span class="math-container">$k_{dp}$</span>.</p>
<p>The above equations also assume no obliquities.  That is, the axis of rotation of each body is perpendicular to the orbital plane.  Without this assumption, things really start to get complicated.</p>
<p>The <a href="https://en.wikipedia.org/wiki/Tidal_locking#Timescale" rel="noreferrer">locking time equation</a> for a planet around a star is much simpler:
<span class="math-container">$$\tau_{l}=\frac{2\Omega_p a^6m_pQ}{15Gm_*^2k_{dp}R_p^3}$$</span></p>
<p>Here, <span class="math-container">$G$</span> is the gravitational constant, and <span class="math-container">$Q$</span> is the dissipation function of the planet.</p>
<p><strong>Notes:</strong></p>
<ol>
<li><p>The Earth-Moon system is not an ideal system to discuss tidal orbital circularization since the gravitational effects of the Sun are acting against such circularization.</p>
</li>
<li><p>My tidal locking time equation looks different from Wikipedia because I wanted to be consistent with notation internal to this answer.</p>
</li>
<li><p>In general, tidal lock will occur faster than orbital circularization because tidal lock is dependent on the amount of lag in the tidal bulge, whereas circularization is dependent on the change in the lag in the tidal bulge as a function of orbital speed change.  However, there may be cases in which tidal circularization occurs sooner than tidal lock.  Take Venus, for example, which has the least eccentric orbit of the planets, but may be prevented from tidal lock with the Sun due to torque from its atmospheric tides.</p>
</li>
</ol>

CC BY-SA 4.0
2022-02-23T19:34:35.377Z
Thank you! I see what you say about the timescales being different, but can you please expand on why the secondary effect (circularization) will always be longer timescale? Also, why circular orbit is minimum energy compared to eccentric orbit?
ValientProcess
CC BY-SA 4.0
2022-02-23T20:48:30.980Z
It is better to wait awhile to accept an answer to encourage other answers.  I will add some edits to the post to address your additional questions.
Connor Garcia
CC BY-SA 4.0
2022-02-23T18:43:19.513Z
2022-02-23T21:47:01.457Z
2022-02-23T21:47:01.457Z
Connor Garcia
<p>I'm trying to understand the basic physics of orbital evolution. I know that in a two-body system (a planet orbiting a sun for example), eccentric orbits become circular, and the spin of the planet becomes tidally locked (like the Earth-moon system).
However, I don't understand why the timescales of these two processes are different (as stated for example in <a href="https://doi.org/10.1093/mnras/stt2292" rel="nofollow noreferrer">Storch &amp; Lai (2013)</a>), and if there's a simple way to quantify the timescale of circularization (kind of back of the envelope calculation, like the one for tidal locking found <a href="https://doi.org/10.1006/icar.1996.0117" rel="nofollow noreferrer">here</a>)?</p>

<orbital-mechanics><tidal-forces><tidal-locking>
Why circularization of an orbit has longer time scale than tidal locking?
ValientProcess
# Why circularization of an orbit has longer time scale than tidal locking?

(Asked by: username_0 on 2022-02-23)

I'm trying to understand the basic physics of orbital evolution. I know that in a two-body system (a planet orbiting a sun for example), eccentric orbits become circular, and the spin of the planet becomes tidally locked (like the Earth-moon system).
However, I don't understand why the timescales of these two processes are different (as stated for example in [Storch & Lai (2013)](https://doi.org/10.1093/mnras/stt2292)), and if there's a simple way to quantify the timescale of circularization (kind of back of the envelope calculation, like the one for tidal locking found [here](https://doi.org/10.1006/icar.1996.0117))?

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

---

## 1 answers

### Answer by username_1 on 2022-02-23. Score: 7 | Accepted answer

**Tidal locking** is a **primary** effect due to a direct torque on a body from its tidal bulge lagging its rotation. This is pictured nicely on [wikipedia](https://en.wikipedia.org/wiki/Tidal_acceleration) with an example from the Earth/Moon system.

[![enter image description here](https://i.sstatic.net/JFVsm.png)](https://i.sstatic.net/JFVsm.png)

**Tidal Orbital Circularization** is a **secondary** effect. To quote Daddy Kropotkin's excellent answer to

[Is the moon's orbit circularizing? Why does tidal heating circularize orbits?](https://astronomy.stackexchange.com/questions/28962/is-the-moons-orbit-circularizing-why-does-tidal-heating-circularize-orbits)

> tidal torque drives dissipation, and this dissipation brings the
> binary to a minimum kinetic energy state, i.e. circular orbit,
> synchronized spins with the orbit, aligned spins with the orbit.

There is really no reason to expect the timescales for these two effects to be similar since there are different mechanisms at play.

Another way to think about tidal circularization is to imagine a tidally locked planet in an eccentric orbit. From Kepler's 2nd law, we know the planet moves fastest at periapsis and slowest at apoapsis. So the planet's tidal bulge won't always quite face its star. Instead, the tidal bulge will lead and lag its orientation to the star. This offset will reduce orbital energy at periapsis and increase orbital energy at apoapsis, circularizing the orbit. Tidal heating will also continue to occur, since the tidal bulge will continue to move until the orbit comes to a minimum energy state (a circular orbit).

Is there *a simple way to quantify the timescale of circularization*?

**No.** There is a way, but it is not simple. From [Rodriguez and Ferraz-Mello 2009](https://arxiv.org/abs/0903.0763), the time scale for orbital circularization of a short period planet in a 2-body system is

$$\tau\_e=\frac{3n^{-1}a^5}{18\hat{s}+7\hat{p}}$$

Here, $n$ is the mean orbital motion, $a$ is the orbital semi-major axis, and
$$\hat{s} = \frac{9k\_{d\*}m\_p}{4Q\_\*m\_\*}R\_\*^5$$
$$\hat{p} = \frac{9k\_{dp}m\_\*}{2Q\_pm\_p}R\_p^5$$

where $Q$ are quality factors defined by $Q\_\*=|\epsilon'\_{0\*}|^{-1}$ and $Q\_p=|\epsilon'\_{2p}|^{-1}$. Here, $\epsilon'\_{0\*}$ and $\epsilon'\_{2p}$ are lag angles associated to the tidal waves whose frequencies are $2\Omega\_\*-2n$ for the star and $2\Omega\_p-n$ for the planet. The angular velocity of the rotation of the tidally deformed body is $\Omega$. The masses, radii, and dynamical Love numbers for the star and planet are respectively $m\_\*$,$R\_\*$,$k\_{d\*}$ and $m\_p$,$R\_p$,$k\_{dp}$.

The above equations also assume no obliquities. That is, the axis of rotation of each body is perpendicular to the orbital plane. Without this assumption, things really start to get complicated.

The [locking time equation](https://en.wikipedia.org/wiki/Tidal_locking#Timescale) for a planet around a star is much simpler:
$$\tau\_{l}=\frac{2\Omega\_p a^6m\_pQ}{15Gm\_\*^2k\_{dp}R\_p^3}$$

Here, $G$ is the gravitational constant, and $Q$ is the dissipation function of the planet.

**Notes:**

1. The Earth-Moon system is not an ideal system to discuss tidal orbital circularization since the gravitational effects of the Sun are acting against such circularization.
2. My tidal locking time equation looks different from Wikipedia because I wanted to be consistent with notation internal to this answer.
3. In general, tidal lock will occur faster than orbital circularization because tidal lock is dependent on the amount of lag in the tidal bulge, whereas circularization is dependent on the change in the lag in the tidal bulge as a function of orbital speed change. However, there may be cases in which tidal circularization occurs sooner than tidal lock. Take Venus, for example, which has the least eccentric orbit of the planets, but may be prevented from tidal lock with the Sun due to torque from its atmospheric tides.

> **Comments:**
> **username_0**: Thank you! I see what you say about the timescales being different, but can you please expand on why the secondary effect (circularization) will always be longer timescale? Also, why circular orbit is minimum energy compared to eccentric orbit?  
> **username_1**: It is better to wait awhile to accept an answer to encourage other answers.  I will add some edits to the post to address your additional questions.  

---

