# At what rate does my planet become tidally locked?

I am designing an alien planet for a speculative evolution project. It is slightly less massive than Earth and is orbiting around a red dwarf, within the Goldilocks zone. It takes 220 Earth days to orbit this star. The planet rotates as well as orbits counterclockwise. It seems likely that it would become tidally locked, but at what rate would this happen? I imagine that its rotation speed would decrease exponentially since it is easier to stop something moving slower. I also imagine that at some point it would come to a complete halt and face its star.

At this point, I should state that I am by no means a mathematician and that the following might be totally bonkers. However, I made a simple graph to visualize my thinking. The y-axis shows the speed of rotation, which is measured in Earth days it takes for one full rotation, meaning that when it reaches 220 Earth days for one rotation, it will rotate once each year and thereby be tidally locked. The x-axis is time, measured in millions of years from a point in its history when the surface became habitable. At this start point, it is one Earth day per rotation. The equation for the curve is the following: y + (y * 0.08) for every step, so y increases for each iteration. X is a fixed amount of time for each step. Probably not the way to do things, might be bonkers, but it made a nice graph though.

At 285 million years, the curve goes to 220 Earth days. The planet should then be tidally locked. The 0.08 number is just something I used to get the millions of years that I wanted; I might be way off on timescales. So, I understand that this model cannot be used 100%. But could this curve give a general idea of at what rate my planet might become tidally locked?

• Head over to our worldbuilding resources page, scroll down to "Celestial Mechanics" and click on "Shagomir's Planet Calculator." This opens a Google Docs spreadsheet which includes the ability to calculate years to a planet becoming tidally locked. Cheers!
– JBH
Commented May 28 at 20:22
• Late tidal-locking planets on Worldbuilding Pasta. Commented May 28 at 20:55
• @AlexP Oooh, that's a good page! I added it to the resource list.
– JBH
Commented May 28 at 21:25
• agreed! Very convenient, straightforward with math. Certainly could be applied to this question. Commented May 29 at 12:43

## 1 Answer

There is an equation to find the time taken for a satellite to become tidally locked. I believe it can be rearranged to make a function to show how day length will change over time.

$$t = \frac{wa^6IQ}{3Gmp^2kR^5}.$$

• $$t$$ is the time until tidal locking (in years, I think)

• $$w$$ is initial spin rate of the satellite around its axis, expressed in radians per second (a full rotation takes $$2\pi$$ radians),

• $$a$$ is the semi-major axis,

• $$I$$ is the satellite’s moment of inertia, equal to approx. $$0.4mR^2$$, where $$R$$ is the radius of the satellite and $$m$$ is the mass of the object being orbited,

• $$Q$$ is the dissipation function of the satellite (not usually known, apparently using $$Q=100$$ is common for estimation),

• $$G$$ is the gravitational constant,

and $$k$$ is the ‘tidal Love number’ of the satellite (can be estimated with $$k=\frac{1.5}{1+\frac{19µ}{2pgR}}$$ with $$p$$ being the satellite’s density, $$g$$ being the surface gravity of the satellite, $$R$$ being the radius of the satellite and $$µ$$ being rigidity of the satellite (estimates around $$3\cdot10^{10}$$ for rocky objects, $$4\cdot10^9$$ for icy objects. I imagine your planet would be rocky, but I’m not sure how close it would be to that specific value, and can’t find earth’s rigidity value for reference.)

Rearranging the equation to make w the subject, and changing out the parts which need to be broken down, we would get:

$$w = \frac{40a^6mR^2}{3Gm^2kR^5t}.$$

(I have not broken down $$k$$, so as to keep the equation at least slightly readable.)

From what I can find, earth has a spin rate of about $$7.272\cdot10^{-5}$$ radians per second, so you should switch out $$w$$ for $$\frac{7.272\cdot10^{-5}}{w}$$ to get results in earth days.

So, the bits of information you need are:

• semimajor axis
• mass of body being orbited
• radius of satellite
• density of satellite
• surface gravity of satellite.

And the resulting equation should turn out to be something like:

$$\frac{7.272\cdot10^{-5}}{w}=\frac{a}{bt}$$, where $$L$$ = day length (in earth days) $$a$$ = constant 1, $$b$$ = constant 2, and $$t$$ is the time until the planet becomes tidally locked.

You can then work backwards to find the point at which $$\frac{7.272\cdot10^{-5}}{w}$$ is whatever you want it to start at (such as one earth day, when the function = 1).

If you want to substitute in a day length value to find out how far from locking the planet would be when it would have that day length, you'll have to rearrange the function into:

$$w=\frac{7.272\cdot10^{-5}}{w}/\frac{a}{bt}$$

and then substitute your day length in as w. Then you'll get an answer as a constant (such as 1.66 billion years or something like that).

Hope that helps! I would have put the equation together for you, but I don't have the info about the parameters. Will happily discuss things in comments on this answer if you have any questions.

• Pretty sure the earth doesn't rotate 7.3 rad/s, since that would imply a rotational period of about 0.86 seconds as opposed to the 86,400 seconds in a day. I think you dropped a $\times10^{-5}$ in your answer. Commented May 29 at 12:38
• @controlgroup meaning 86,164.1... seconds I suppose. Commented May 31 at 15:40
• Of course. Silly rounding errors always get me. Commented May 31 at 17:27