<blockquote>
<p>Why is the L3 Lagrangian point not perfectly stable?</p>
</blockquote>
<p>In the <a href="https://en.wikipedia.org/wiki/Three-body_problem#Restricted_three-body_problem" rel="noreferrer">circular restricted three-body problem</a> (CR3BP or CRTBP) an object at any of the first Lagrange points L1, L2, L3 is unstable <em>mathematically</em>. Yes, a ball at the exact top of a hill will sit there, but any tiny offset in position or tiny nonzero velocity will lead to it accelerating down the hill.</p>
<p>As <a href="https://astronomy.stackexchange.com/a/46804/7982">@JamesK points out</a> these three points are unstable against displacements along the Sun-Earth line, <em>but stable against displacements</em> in the perpendicular direction.</p>
<p>That's why the diagram shows blue arrows along the Earth-Sun line and red arrows perpendicularly for these three points.</p>
<p>It is important to point out that this is the zero-velocity (pseudo)potential surface. The diagram for stability against small velocities will look different!</p>
<blockquote>
<p>And why is the Earth-Sun L3 point a bit less than one A.U.?</p>
</blockquote>
<p>The Lagrange points are mathematical points in the CR3BP where forces exactly cancel. They can be calculated in either an inertial frame (where the Earth and Sun rotate around their common center of mass) or in a rotating frame where centripetal and gravitational forces exactly cancel.</p>
<p>The equations for this are a little complicated when you try to solve for the exact positions. They are given for example in Wikipedia's <a href="https://en.wikipedia.org/wiki/Lagrange_point#Mathematical_details" rel="noreferrer">Lagrange point; mathematical details</a>. It is important to remember that the two massive bodies <strong>orbit around a common center of mass</strong> rather than the smaller one orbiting exactly around an immobile larger one.</p>
<p>The big <span class="math-container">$R$</span> is the distance between the two massive bodies, and the little <span class="math-container">$r$</span> is the distance from a given Lagrange point to the smaller one which is in this case, from Earth.</p>
<p>The locations are where the forces cancel, and for L3 that cancellation happens at...</p>
<p><strong>uhoh!</strong> The equations in that Wikipedia link for L3 <strong>seem wrong!</strong></p>
<p>Looking elsewhere in the internet I've found conflicting solutions for the location of L3. The solution requires finding roots of a fifth-order polynomial which is hard on paper but may be easy with a computer (Wolfram alpha, Maple, Mathematica...) or just by numerical searching.</p>
<p>But considering the disagreements I've found I'm not sure right now what the correct polynomial even is, so I'll have to derive it.</p>
<p><strong>stay tuned for updates!</strong></p>

CC BY-SA 4.0
2021-09-21T06:00:14.443Z
**uhoh!** indeed. I did however recover the factor of $\frac7{12}\frac{M_2}{M_1}$ mentioned a bit later in the Wikipedia article in the case $M_2 \lll M_1$ by a couple of different approaches.
David Hammen
CC BY-SA 4.0
2021-09-21T09:26:00.207Z
@DavidHammen if you'd like, please feel free to add an answer with that, or edit in here, or anything else.
uhoh
CC BY-SA 4.0
2021-09-21T02:37:17.567Z
2021-09-21T02:37:17.567Z
uhoh
<p>L1, L2 and L3 are saddle points in the effective potential of the gravitational field in a rotating frame of reference.</p>
<p>That is if you combine gravity (of Earth and Sun) with the centrifugal force on an object that is moving around a point at one orbit per year you find that there are three saddle points, and these are L1, L2 and L3</p>
<p>If an object is in orbit around the Sun and close to L3, if it is behind L3 it will tend to orbit faster and so catch up to L3.  If it ahead it will orbit slower.  If it is outside L3 it will move further out and if it is inside it will move in towards the sun.</p>
<p>This is typical of a saddle point: in two directions it is attractive and in two directions it is repulsive.  If you could place an object exactly on L3 it would remain there, but that is impossible (remember that there are 7 other planets to tug on it)</p>
<p>So L3 is unstable in the way that a ball on a col, or pass, between mountains is unstable. If you can place it exactly on the col it would stay there, but if it is even slightly off it will roll downhill. If you put the ball on the mountain, it will first roll towards the col, but (unless perfectly positioned) it will slip off sideways towards the valleys.</p>
<p>It's a bit less than one AU from the Sun because you need to consider also the gravity from the Earth.  If there was no Earth, then a body would be stable at one AU from the sun and orbiting once per year.  But the gravity of the Earth shifts that point at which the combined gravity of the sun and Earth balance the centrifugal force.</p>
<p>This is due to Newtonian gravity, there is no need for relativity (though it would also work if you used GR)</p>

CC BY-SA 4.0
2021-09-21T06:05:28.570Z
Re "No need for GR": I removed the general-relativity tag from the question because there is indeed no need for GR here.
David Hammen
CC BY-SA 4.0
2021-09-21T06:07:01.920Z
Good explanation about the col
Reversed Engineer
CC BY-SA 4.0
2021-09-21T06:10:57.773Z
Yeah, better than the one with a pencil that I wrote first.
James K
CC BY-SA 4.0
2021-09-21T06:17:51.663Z
@JamesK I agree that this is a much better explanation. An inverted pendulum (e.g., a pencil balanced on its tip) is unstable in both dimensions, while a saddle point is stable in one dimension, unstable in the other.
David Hammen
CC BY-SA 4.0
2021-09-21T10:00:03.607Z
Here's a classic saddle surface: a [hyperbolic paraboloid](https://sagecell.sagemath.org/?z=eJwtjEEOwiAURPecYnYFpWirG03YegLXTb6Uqg1IQ6uB24umm8mbvMwkiQyND0VeJZkrwajURjUsvu6FapIgZoaCjvytJ6SyOIOnrsUWuWsFduCNOm6oMJtcWA7939Y_K1F-1ggTmeeS9V6dJExwIWpuhhU9TbMa7SIkvJ0f-hrfVmKI5K2-kJut-AKfLTH2&lang=sage)
PM 2Ring
CC BY-SA 4.0
2021-09-21T19:01:05.027Z
But this leaves the follow-up question open: why are the Langrangian points after L3 stable? Aren't they exposed to the same tugging forces of other bodies outside the reference frame?
D. Kovács
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2021-09-21T19:32:50.083Z
@D.Kovács That is a good question,  considering only gravity and centifugal forces the pseudo potential is unstable at L4 and L5, but you also need to consider coriolis forces which are dependent on the object's velocity in the rotating frame, so doesn't form a pseudo potential.  However when this force is considered, stable trojan and horseshoe orbits can exist.
James K
CC BY-SA 4.0
2021-09-20T18:59:19.203Z
2021-09-22T15:21:16.750Z
2021-09-22T15:21:16.750Z
James K
<p>I suppose the answer might involve general relativity, but still....</p>
<p>The L4 and L5 points are considered, theoretically, long-term stable, but not L3, on the exact opposite side of the Sun... And it is not, apparently, located exactly one A.U. from the Sun, as one would assume....</p>

CC BY-SA 4.0
2021-09-21T06:20:02.560Z
There is no need to invoke general relativity. The five Lagrange points are a consequence of good old Newtonian mechanics.
David Hammen
<earth><lagrange-point><time-domain-astronomy>
Why is the L3 Lagrangian point not perfectly stable? And why is the Earth-Sun L3 point a bit less than one A.U.?
Kurt Hikes
# Why is the L3 Lagrangian point not perfectly stable? And why is the Earth-Sun L3 point a bit less than one A.U.?

(Asked by: username_0 on 2021-09-20)

I suppose the answer might involve general relativity, but still....

The L4 and L5 points are considered, theoretically, long-term stable, but not L3, on the exact opposite side of the Sun... And it is not, apparently, located exactly one A.U. from the Sun, as one would assume....

**Tags:** earth, lagrange-point, time-domain-astronomy
> **Comments:**
> **username_1**: There is no need to invoke general relativity. The five Lagrange points are a consequence of good old Newtonian mechanics.  

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## 2 answers

### Answer by username_3 on 2021-09-20. Score: 31 | Accepted answer

L1, L2 and L3 are saddle points in the effective potential of the gravitational field in a rotating frame of reference.

That is if you combine gravity (of Earth and Sun) with the centrifugal force on an object that is moving around a point at one orbit per year you find that there are three saddle points, and these are L1, L2 and L3

If an object is in orbit around the Sun and close to L3, if it is behind L3 it will tend to orbit faster and so catch up to L3. If it ahead it will orbit slower. If it is outside L3 it will move further out and if it is inside it will move in towards the sun.

This is typical of a saddle point: in two directions it is attractive and in two directions it is repulsive. If you could place an object exactly on L3 it would remain there, but that is impossible (remember that there are 7 other planets to tug on it)

So L3 is unstable in the way that a ball on a col, or pass, between mountains is unstable. If you can place it exactly on the col it would stay there, but if it is even slightly off it will roll downhill. If you put the ball on the mountain, it will first roll towards the col, but (unless perfectly positioned) it will slip off sideways towards the valleys.

It's a bit less than one AU from the Sun because you need to consider also the gravity from the Earth. If there was no Earth, then a body would be stable at one AU from the sun and orbiting once per year. But the gravity of the Earth shifts that point at which the combined gravity of the sun and Earth balance the centrifugal force.

This is due to Newtonian gravity, there is no need for relativity (though it would also work if you used GR)

> **Comments:**
> **username_1**: Re "No need for GR": I removed the general-relativity tag from the question because there is indeed no need for GR here.  
> **username_4**: Good explanation about the col  
> **username_3**: Yeah, better than the one with a pencil that I wrote first.  
> **username_1**: @JamesK I agree that this is a much better explanation. An inverted pendulum (e.g., a pencil balanced on its tip) is unstable in both dimensions, while a saddle point is stable in one dimension, unstable in the other.  
> **username_5**: Here's a classic saddle surface: a [hyperbolic paraboloid](https://sagecell.sagemath.org/?z=eJwtjEEOwiAURPecYnYFpWirG03YegLXTb6Uqg1IQ6uB24umm8mbvMwkiQyND0VeJZkrwajURjUsvu6FapIgZoaCjvytJ6SyOIOnrsUWuWsFduCNOm6oMJtcWA7939Y_K1F-1ggTmeeS9V6dJExwIWpuhhU9TbMa7SIkvJ0f-hrfVmKI5K2-kJut-AKfLTH2&lang=sage)  
> **username_6**: But this leaves the follow-up question open: why are the Langrangian points after L3 stable? Aren't they exposed to the same tugging forces of other bodies outside the reference frame?  
> **username_3**: @D.Kovács That is a good question,  considering only gravity and centifugal forces the pseudo potential is unstable at L4 and L5, but you also need to consider coriolis forces which are dependent on the object's velocity in the rotating frame, so doesn't form a pseudo potential.  However when this force is considered, stable trojan and horseshoe orbits can exist.  

---

### Answer by username_2 on 2021-09-21. Score: 6

> Why is the L3 Lagrangian point not perfectly stable?

In the [circular restricted three-body problem](https://en.wikipedia.org/wiki/Three-body_problem#Restricted_three-body_problem) (CR3BP or CRTBP) an object at any of the first Lagrange points L1, L2, L3 is unstable *mathematically*. Yes, a ball at the exact top of a hill will sit there, but any tiny offset in position or tiny nonzero velocity will lead to it accelerating down the hill.

As [@JamesK points out](https://astronomy.stackexchange.com/a/46804/7982) these three points are unstable against displacements along the Sun-Earth line, *but stable against displacements* in the perpendicular direction.

That's why the diagram shows blue arrows along the Earth-Sun line and red arrows perpendicularly for these three points.

It is important to point out that this is the zero-velocity (pseudo)potential surface. The diagram for stability against small velocities will look different!

> And why is the Earth-Sun L3 point a bit less than one A.U.?

The Lagrange points are mathematical points in the CR3BP where forces exactly cancel. They can be calculated in either an inertial frame (where the Earth and Sun rotate around their common center of mass) or in a rotating frame where centripetal and gravitational forces exactly cancel.

The equations for this are a little complicated when you try to solve for the exact positions. They are given for example in Wikipedia's [Lagrange point; mathematical details](https://en.wikipedia.org/wiki/Lagrange_point#Mathematical_details). It is important to remember that the two massive bodies **orbit around a common center of mass** rather than the smaller one orbiting exactly around an immobile larger one.

The big $R$ is the distance between the two massive bodies, and the little $r$ is the distance from a given Lagrange point to the smaller one which is in this case, from Earth.

The locations are where the forces cancel, and for L3 that cancellation happens at...

**uhoh!** The equations in that Wikipedia link for L3 **seem wrong!**

Looking elsewhere in the internet I've found conflicting solutions for the location of L3. The solution requires finding roots of a fifth-order polynomial which is hard on paper but may be easy with a computer (Wolfram alpha, Maple, Mathematica...) or just by numerical searching.

But considering the disagreements I've found I'm not sure right now what the correct polynomial even is, so I'll have to derive it.

**stay tuned for updates!**

> **Comments:**
> **username_1**: **uhoh!** indeed. I did however recover the factor of $\frac7{12}\frac{M_2}{M_1}$ mentioned a bit later in the Wikipedia article in the case $M_2 \lll M_1$ by a couple of different approaches.  
> **username_2**: @DavidHammen if you'd like, please feel free to add an answer with that, or edit in here, or anything else.  

---

