<p>Because The triangular points (L4 and L5) are stable <a href="https://en.wikipedia.org/wiki/Lagrangian_point" rel="nofollow noreferrer">equilibria</a>, provided that the ratio of M1/M2, moon 0.166 g and earth 9.8 g  so 0.0169 g in L4 and L5</p>

CC BY-SA 4.0
2019-03-27T05:41:05.533Z
I don't believe this can be true. A restoring force of 0.0169 Newtons per kg extending over astronomical scales (millions of km, billions of meters) is unimaginable. I think the answer to "how deep is the potential energy well of L4/L5?" has to be 1) more complicated than a simple ratio and 2) orders of magnitude smaller and 3) have units of energy, or if expressed as $C_3$ units of reduced energy (m^2/s^2) not acceleration.
uhoh
CC BY-SA 4.0
2019-03-27T08:15:53.777Z
maybe, "strong" is more like to force, and you need replace in F= M*g , by any body mass near in a point L4  or L5
Adrian R
CC BY-SA 4.0
2019-03-28T08:58:33.117Z
There seem to be many problems here.  The Moon's surface gravity is approx $1/6$ of Earth's so it might be written as $0.166g$ but in that case the Earth's gravity is just $g$ and not $9.8g$.  Also, if you divide two accelerations you would get a dimensionless ratio not another acceleration.  Try repeating with different units, e.g. $ft/s^2$ and see what you get.
badjohn
CC BY-SA 4.0
2019-03-27T04:46:06.633Z
2019-03-27T04:46:06.633Z
Adrian R
<p>How strongly are objects bound to the Lagrange points of Earth that they inhabit in them in the same way Jupiter's Trojan meteors are entrapped in its Lagrange points?</p>

<p>In the co-moving frame with the two main objects, how deep is the potential energy well of L4/L5?</p>

CC BY-SA 4.0
2019-03-25T23:49:04.857Z
Before you even ask "how is their strength measured?", maybe you should define what that strength should even be? All Lagrange points have "zero" force strength, so you cannot mean that.
AtmosphericPrisonEscape
CC BY-SA 4.0
2019-03-26T02:50:24.373Z
Yes this is a muze question, but it's not a bad one and is perfectly answerable. I'd rephrase it as something like "In the co-moving frame with the two main objects, how deep is the potential energy well of L4/L5?" or "How strongly are objects bound to the Lagrange points that they inhabit?"
Ingolifs
CC BY-SA 4.0
2019-03-26T02:53:51.510Z
@Ingolifs Thank you and updated.
Muze
CC BY-SA 4.0
2019-03-27T04:37:03.233Z
0.166 g in moon and 9.8 g in earth so  M1/M2 ,  0.0169 g in L4 and L5
Adrian R
CC BY-SA 4.0
2019-03-27T04:40:51.683Z
@AdrianR if you put that as an answer I'll accept it.
Muze
CC BY-SA 4.0
2019-03-27T05:41:46.387Z
@Muze you might want to wait a while before accepting, see my comment below.
uhoh
CC BY-SA 4.0
2019-03-28T02:17:50.737Z
Related: [How deep is the force well of L4 and L5 Lagrangian Points of Earth-Sun set?](https://space.stackexchange.com/questions/233/how-deep-is-the-force-well-of-l4-and-l5-lagrangian-points-of-earth-sun-set), [Why are L4L4 and L5L5 lagrangian points stable?](https://physics.stackexchange.com/questions/36092/why-are-l-4-and-l-5-lagrangian-points-stable).
Keith McClary
<magnitude><lagrange-point>
How strong are the Lagrange Points of Earth?
Muze
# How strong are the Lagrange Points of Earth?

(Asked by: username_0 on 2019-03-25)

How strongly are objects bound to the Lagrange points of Earth that they inhabit in them in the same way Jupiter's Trojan meteors are entrapped in its Lagrange points?

In the co-moving frame with the two main objects, how deep is the potential energy well of L4/L5?

**Tags:** magnitude, lagrange-point
> **Comments:**
> **username_1**: Before you even ask "how is their strength measured?", maybe you should define what that strength should even be? All Lagrange points have "zero" force strength, so you cannot mean that.  
> **username_2**: Yes this is a muze question, but it's not a bad one and is perfectly answerable. I'd rephrase it as something like "In the co-moving frame with the two main objects, how deep is the potential energy well of L4/L5?" or "How strongly are objects bound to the Lagrange points that they inhabit?"  
> **username_0**: @Ingolifs Thank you and updated.  
> **username_3**: 0.166 g in moon and 9.8 g in earth so  M1/M2 ,  0.0169 g in L4 and L5  
> **username_0**: @AdrianR if you put that as an answer I'll accept it.  
> **username_4**: @Muze you might want to wait a while before accepting, see my comment below.  
> **username_5**: Related: [How deep is the force well of L4 and L5 Lagrangian Points of Earth-Sun set?](https://space.stackexchange.com/questions/233/how-deep-is-the-force-well-of-l4-and-l5-lagrangian-points-of-earth-sun-set), [Why are L4L4 and L5L5 lagrangian points stable?](https://physics.stackexchange.com/questions/36092/why-are-l-4-and-l-5-lagrangian-points-stable).  

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## 1 answers

### Answer by username_3 on 2019-03-27. Score: 1 | Accepted answer

Because The triangular points (L4 and L5) are stable [equilibria](https://en.wikipedia.org/wiki/Lagrangian_point), provided that the ratio of M1/M2, moon 0.166 g and earth 9.8 g so 0.0169 g in L4 and L5

> **Comments:**
> **username_4**: I don't believe this can be true. A restoring force of 0.0169 Newtons per kg extending over astronomical scales (millions of km, billions of meters) is unimaginable. I think the answer to "how deep is the potential energy well of L4/L5?" has to be 1) more complicated than a simple ratio and 2) orders of magnitude smaller and 3) have units of energy, or if expressed as $C_3$ units of reduced energy (m^2/s^2) not acceleration.  
> **username_3**: maybe, "strong" is more like to force, and you need replace in F= M*g , by any body mass near in a point L4  or L5  
> **username_6**: There seem to be many problems here.  The Moon's surface gravity is approx $1/6$ of Earth's so it might be written as $0.166g$ but in that case the Earth's gravity is just $g$ and not $9.8g$.  Also, if you divide two accelerations you would get a dimensionless ratio not another acceleration.  Try repeating with different units, e.g. $ft/s^2$ and see what you get.  

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