<p>Consider a non-rigid rotating system where the rotational velocity is given by the Kepler speed. In such system, adjacent orbits have a different orbital speed of <span class="math-container">$$ v \approx \sqrt{\frac{GM}{r}}.$$</span>
Thus the orbital velocity drops the further you go outward.</p>
<p>Place yourself onto one particle and you see that the elements outward are slower than you, while the elements further inward are faster. A velocity shear is <span class="math-container">$dv / dr$</span>. Adding &quot;Keplerian&quot; tells you (1) how fast the velocity is changing, and (2) that the shear is in the radial direction. A shear <span class="math-container">$dv / dz$</span> would be in a different direction, so it would not make sense to call it Keplerian because the Keplerian velocity depends only on radial distance.</p>
<p>In a protoplanetary disk, such shear motion counter-acts self-gravity of the small particles in a way it has mass further inside rotate quicker. With sufficiently dense disk, thus solid mass, and sufficient gas in the disc, this may actually lead to hydrodynamic instabilities forming vortices which locally counter this shear - and essentially only then allow planetesimals to form. In order to assess stability in such rotating disc with shear, one uses the <a href="https://en.wikipedia.org/wiki/Toomre%27s_stability_criterion" rel="nofollow noreferrer">Toomre criterion</a> (which is basically the <a href="https://en.wikipedia.org/wiki/Jeans_instability" rel="nofollow noreferrer">Jeans stability criterion</a> expanded to a rotating disc).</p>
<p>Similarily Keplerian shear can be observed in the <a href="https://medium.com/@fabiograetz/today-i-take-you-on-a-trip-to-the-icy-world-of-saturns-rings-and-then-450-lightyears-further-f26819451d0b" rel="nofollow noreferrer">Saturnian rings</a> where in the absence of gas and hydrodynamics it basically stops the growth of the so-called moonlets.</p>

CC BY-SA 4.0
2022-02-02T12:27:44.057Z
Seems to be closely related to the Roche Limit for solid bodies.
Carl Witthoft
CC BY-SA 4.0
2022-02-02T12:45:51.207Z
@CarlWitthoft yes and no. If you have a self-gravitating mass in a Keplerian orbit, that helps to overcome that body being torn apart again; You have the Keplerian shear also when you are outside any Roche limit or the central body. It just is differential Kepler velocity.
planetmaker
CC BY-SA 4.0
2022-02-02T14:12:28.670Z
Quite right. But just to point out that this condition has been superseded by the Toomre criterion when discussing gaseous disks.
ProfRob
CC BY-SA 4.0
2022-02-02T15:00:25.030Z
good point @ProfRob. I amended the explanation
planetmaker
CC BY-SA 4.0
2022-02-02T07:02:44.893Z
2022-02-14T02:44:33.977Z
2022-02-14T02:44:33.977Z
planetmaker
<p>I was reading an article on Gas Giants and how some form close to their host star.</p>
<p>One line in the article says:</p>
<blockquote>
<p>A region of the proto-planetary disk will be susceptible to gravitational
instability if the free-fall time due to self gravity is sufficiently rapid
to overcome Keplerian shear.</p>
</blockquote>
<p>I tried to find a wiki on what Keplerian sheer was, not much luck. Does it go by another name, or can some one explain what it is ? Saw a paper that mentioned it about ergodic theory (no idea what that theory is) it's way above my knowledge to get a general idea of what it is from the paper.</p>

CC BY-SA 4.0
2022-02-02T14:08:36.847Z
"how some form close to their host star". Insert the word "may". There is no strong evidence that this mechanism works, especially not close to a star.
ProfRob
CC BY-SA 4.0
2022-02-03T02:41:36.593Z
Well the phrase was talking about formation from in situ. Rather than migrating in after formation.
WDUK
<exoplanet><planetary-formation>
What is a Keplerian shear
WDUK
# What is a Keplerian shear

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

I was reading an article on Gas Giants and how some form close to their host star.

One line in the article says:

> A region of the proto-planetary disk will be susceptible to gravitational
> instability if the free-fall time due to self gravity is sufficiently rapid
> to overcome Keplerian shear.

I tried to find a wiki on what Keplerian sheer was, not much luck. Does it go by another name, or can some one explain what it is ? Saw a paper that mentioned it about ergodic theory (no idea what that theory is) it's way above my knowledge to get a general idea of what it is from the paper.

**Tags:** exoplanet, planetary-formation
> **Comments:**
> **username_1**: "how some form close to their host star". Insert the word "may". There is no strong evidence that this mechanism works, especially not close to a star.  
> **username_0**: Well the phrase was talking about formation from in situ. Rather than migrating in after formation.  

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## 1 answers

### Answer by username_2 on 2022-02-02. Score: 7 | Accepted answer

Consider a non-rigid rotating system where the rotational velocity is given by the Kepler speed. In such system, adjacent orbits have a different orbital speed of $$ v \approx \sqrt{\frac{GM}{r}}.$$
Thus the orbital velocity drops the further you go outward.

Place yourself onto one particle and you see that the elements outward are slower than you, while the elements further inward are faster. A velocity shear is $dv / dr$. Adding "Keplerian" tells you (1) how fast the velocity is changing, and (2) that the shear is in the radial direction. A shear $dv / dz$ would be in a different direction, so it would not make sense to call it Keplerian because the Keplerian velocity depends only on radial distance.

In a protoplanetary disk, such shear motion counter-acts self-gravity of the small particles in a way it has mass further inside rotate quicker. With sufficiently dense disk, thus solid mass, and sufficient gas in the disc, this may actually lead to hydrodynamic instabilities forming vortices which locally counter this shear - and essentially only then allow planetesimals to form. In order to assess stability in such rotating disc with shear, one uses the [Toomre criterion](https://en.wikipedia.org/wiki/Toomre%27s_stability_criterion) (which is basically the [Jeans stability criterion](https://en.wikipedia.org/wiki/Jeans_instability) expanded to a rotating disc).

Similarily Keplerian shear can be observed in the [Saturnian rings](https://medium.com/@fabiograetz/today-i-take-you-on-a-trip-to-the-icy-world-of-saturns-rings-and-then-450-lightyears-further-f26819451d0b) where in the absence of gas and hydrodynamics it basically stops the growth of the so-called moonlets.

> **Comments:**
> **username_3**: Seems to be closely related to the Roche Limit for solid bodies.  
> **username_2**: @CarlWitthoft yes and no. If you have a self-gravitating mass in a Keplerian orbit, that helps to overcome that body being torn apart again; You have the Keplerian shear also when you are outside any Roche limit or the central body. It just is differential Kepler velocity.  
> **username_1**: Quite right. But just to point out that this condition has been superseded by the Toomre criterion when discussing gaseous disks.  
> **username_2**: good point @ProfRob. I amended the explanation  

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