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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?

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    $\begingroup$ 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. $\endgroup$ – AtmosphericPrisonEscape Mar 25 '19 at 23:49
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    $\begingroup$ 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?" $\endgroup$ – Ingolifs Mar 26 '19 at 2:50
  • $\begingroup$ @Ingolifs Thank you and updated. $\endgroup$ – Muze the good Troll. Mar 26 '19 at 2:53
  • $\begingroup$ 0.166 g in moon and 9.8 g in earth so M1/M2 , 0.0169 g in L4 and L5 $\endgroup$ – Adrian R Mar 27 '19 at 4:37
  • $\begingroup$ @AdrianR if you put that as an answer I'll accept it. $\endgroup$ – Muze the good Troll. Mar 27 '19 at 4:40
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Because The triangular points (L4 and L5) are stable equilibria, provided that the ratio of M1/M2, moon 0.166 g and earth 9.8 g so 0.0169 g in L4 and L5

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    $\begingroup$ 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. $\endgroup$ – uhoh Mar 27 '19 at 5:41
  • $\begingroup$ 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 $\endgroup$ – Adrian R Mar 27 '19 at 8:15
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    $\begingroup$ 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. $\endgroup$ – badjohn Mar 28 '19 at 8:58

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