<p>I can't comment here because I don't have enough reputation, but this is very related to your problem.</p>

<p><a href="https://en.wikipedia.org/wiki/Tidal_locking#Timescale" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Tidal_locking#Timescale</a></p>

<p>Tidal locking will always occur albeit potentially not in a Hubble time. In order for there to be no tidal locking, you would need:</p>

<p>$$\frac{\mathrm{d}F}{\mathrm{d}r} = 0$$</p>

<p>Which any extended object will not have. Problems like this always involve timescales and not definite yes/no answers. I hope this helped.</p>

CC BY-SA 3.0
2016-10-21T13:54:32.347Z
Going by that logic, Is Earth tidally locked to Sun because it's not zero.
fahadash
CC BY-SA 3.0
2016-10-22T01:15:47.520Z
No just because dF/dr > 0 doesn't mean that it currently is tidally locked, it just means that EVENTUALLY, possibly longer than the age of the universe, it will become tidally locked. Read the timescale section on that Wikipedia article.
Doug
CC BY-SA 3.0
2016-10-21T00:28:14.897Z
2017-01-04T22:15:24.907Z
2017-01-04T22:15:24.907Z
Doug
<p>I am a beginner, please pardon my non-use of proper astronomical terms. I would appreciate proper edits to my questions.</p>

<p>I have learned that when the $\frac{\mathrm{d}F}{\mathrm{d}r}$ is a very big negative number, that would mean that the far side will have much less gravitational tug than the near side of a satellite. The differential gravity or $\frac{\mathrm{d}F}{\mathrm{d}r}$ will be as follows</p>

<p>$$\frac{\mathrm{d}F}{\mathrm{d}r} = \frac{-2 GMm}{r^3}$$</p>

<p><a href="https://www.wolframalpha.com/input/?i=-2%20*%20(Gravitational%20Constant)%20*%20(mass%20of%20Earth)%20*%20(Mass%20of%20Moon)%20%2F%20(Semimajor%20axis%20of%20moon)%5E3" rel="nofollow">If we compute $\frac{\mathrm{d}F}{\mathrm{d}r}$ of Earth-Moon system</a> we get $-1.308 \times 10^{-5} \ \mathrm{M_o^2G/AU^3} $</p>

<p>I was surprised to know that Ganymede, being almost 3 times as far away from Jupiter as our Moon is from Earth, still is tidally locked.</p>

<p><a href="https://www.wolframalpha.com/input/?i=-2*(Gravitational%20Constant)%20*%20(mass%20of%20Jupiter)%20*%20(Mass%20of%20Ganymede)%20%2F(Ganymede%20orbital%20distance)%5E3" rel="nofollow">If we compute $\frac{\mathrm{d}F}{\mathrm{d}r}$ of Jupiter-Ganymede system</a> we get $-1.565 \times 10^{-6} \ \mathrm{M_o^2G/AU^3}$</p>

<p>I then wondered what is the $\frac{\mathrm{d}F}{\mathrm{d}r}$ of Sun-Earth that our Earth is not tidally locked to it and <a href="https://www.wolframalpha.com/input/?i=-2*(Gravitational%20Constant)%20*%20(mass%20of%20Sun)%20*%20(Mass%20of%20Earth)%20%2F(Earth%20orbital%20distance)%5E3" rel="nofollow">I found $-2.422 \times 10^{-8} \ \mathrm{M_o^2G/AU^3}$</a> </p>

<p>Which is a very small negative number so the difference in tug is not significant.</p>

<p>So what is the minimum $\frac{\mathrm{d}F}{\mathrm{d}r}$ (or I guess "maximum") value or a boundary value at which the object can be tidally locked and beyond that, the tidal locking won't occur?</p>

CC BY-SA 3.0
2016-06-15T03:14:27.657Z
Try Sun-Venus.   The sun appears to have slowed down Venus rotation significantly, though it's hard to get a reading on how fast Venus rotated in the past.  The malleability of the surface and/or liquid or heavy atmosphere on the surface is also a factor, it's not just the tidal force but how much the planet bends/flows in response to the force.    I don't think there's a clear cut answer as to where tidal locking stops either, because, as tidal force decreases, the time it takes for tidal locking to happen increases, so there's no neat and tidy cut-off, just a steady increase in time it takes
userLTK
CC BY-SA 3.0
2016-06-15T10:47:08.600Z
@userLTK Thank you. I like your idea. Edited my question.
fahadash
CC BY-SA 3.0
2016-10-17T16:05:53.203Z
I think it will largely depend on the material of the body rotating. In principle, a body composed of a fluid, whether gas or liquid, will respond to tidal forces, no matter how small -- it's just a question of whether the process has time to happen before something else interferes. If the body deforms elastically, that is without generating heat, it will never lock.
Sean Lake
<tidal-forces>
What is the cut-off point of $\frac{\mathrm{d}F}{\mathrm{d}r}$ beyond which no tidal locking will occur?
fahadash
# What is the cut-off point of $\frac{\mathrm{d}F}{\mathrm{d}r}$ beyond which no tidal locking will occur?

(Asked by: username_0 on 2016-06-14)

I am a beginner, please pardon my non-use of proper astronomical terms. I would appreciate proper edits to my questions.

I have learned that when the $\frac{\mathrm{d}F}{\mathrm{d}r}$ is a very big negative number, that would mean that the far side will have much less gravitational tug than the near side of a satellite. The differential gravity or $\frac{\mathrm{d}F}{\mathrm{d}r}$ will be as follows

$$\frac{\mathrm{d}F}{\mathrm{d}r} = \frac{-2 GMm}{r^3}$$

[If we compute $\frac{\mathrm{d}F}{\mathrm{d}r}$ of Earth-Moon system](https://www.wolframalpha.com/input/?i=-2%20*%20(Gravitational%20Constant)%20*%20(mass%20of%20Earth)%20*%20(Mass%20of%20Moon)%20%2F%20(Semimajor%20axis%20of%20moon)%5E3) we get $-1.308 \times 10^{-5} \ \mathrm{M\_o^2G/AU^3} $

I was surprised to know that Ganymede, being almost 3 times as far away from Jupiter as our Moon is from Earth, still is tidally locked.

[If we compute $\frac{\mathrm{d}F}{\mathrm{d}r}$ of Jupiter-Ganymede system](https://www.wolframalpha.com/input/?i=-2*(Gravitational%20Constant)%20*%20(mass%20of%20Jupiter)%20*%20(Mass%20of%20Ganymede)%20%2F(Ganymede%20orbital%20distance)%5E3) we get $-1.565 \times 10^{-6} \ \mathrm{M\_o^2G/AU^3}$

I then wondered what is the $\frac{\mathrm{d}F}{\mathrm{d}r}$ of Sun-Earth that our Earth is not tidally locked to it and [I found $-2.422 \times 10^{-8} \ \mathrm{M\_o^2G/AU^3}$](https://www.wolframalpha.com/input/?i=-2*(Gravitational%20Constant)%20*%20(mass%20of%20Sun)%20*%20(Mass%20of%20Earth)%20%2F(Earth%20orbital%20distance)%5E3)

Which is a very small negative number so the difference in tug is not significant.

So what is the minimum $\frac{\mathrm{d}F}{\mathrm{d}r}$ (or I guess "maximum") value or a boundary value at which the object can be tidally locked and beyond that, the tidal locking won't occur?

**Tags:** tidal-forces

> **Comments:**
> **username_1**: Try Sun-Venus.   The sun appears to have slowed down Venus rotation significantly, though it's hard to get a reading on how fast Venus rotated in the past.  The malleability of the surface and/or liquid or heavy atmosphere on the surface is also a factor, it's not just the tidal force but how much the planet bends/flows in response to the force.    I don't think there's a clear cut answer as to where tidal locking stops either, because, as tidal force decreases, the time it takes for tidal locking to happen increases, so there's no neat and tidy cut-off, just a steady increase in time it takes  
> **username_0**: @username_1 Thank you. I like your idea. Edited my question.  
> **username_2**: I think it will largely depend on the material of the body rotating. In principle, a body composed of a fluid, whether gas or liquid, will respond to tidal forces, no matter how small -- it's just a question of whether the process has time to happen before something else interferes. If the body deforms elastically, that is without generating heat, it will never lock.  

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## 1 answers

### Answer by username_3 on 2016-10-21. Score: 5

I can't comment here because I don't have enough reputation, but this is very related to your problem.

<https://en.wikipedia.org/wiki/Tidal_locking#Timescale>

Tidal locking will always occur albeit potentially not in a Hubble time. In order for there to be no tidal locking, you would need:

$$\frac{\mathrm{d}F}{\mathrm{d}r} = 0$$

Which any extended object will not have. Problems like this always involve timescales and not definite yes/no answers. I hope this helped.

> **Comments:**
> **username_0**: Going by that logic, Is Earth tidally locked to Sun because it's not zero.  
> **username_3**: No just because dF/dr > 0 doesn't mean that it currently is tidally locked, it just means that EVENTUALLY, possibly longer than the age of the universe, it will become tidally locked. Read the timescale section on that Wikipedia article.  

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