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I was tasked in my programming lesson to find a value of L1 correct to 5 significant figures, but in my discussion section I began to wonder why there was no second solution to the equation I was trying to find the root of (L2). The equation was $$\frac{GM}{x^2}-\frac{Gm}{(R-x)^2}-w^2x=0.$$ Where $M$ was the mass of the Earth, $m$ was the mass of the moon, $x$ was the distance from the earth to the satellite, $R$ was the distance to the moon and $w$ is the angular velocity of the moon.

My question is, along the line connecting the earth and the moon there should be 2 points where the net force is pointing towards the Earth with a magnitude that gives the net accleration to be the same as the moon's. These points are what I understand to be L1 and L2. From my code only L1 was found. I plot the function on desmos and saw this:

enter image description here

According to desmos, there was no L2 point.

Is this because of my assumption that the orbits are circular? How does one find L2?

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  • $\begingroup$ $x$ should be the distance of the point you are considering from the earth and $R$ the earth-to- moon distance! $\endgroup$ – AoZora Apr 14 at 20:12
  • $\begingroup$ I have made the edit thank you. $\endgroup$ – Vishal Jain Apr 15 at 9:27
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HINT: At L1, the gravitational force from the Earth and the Moon point in opposite directions. However, at L2, the gravitational force from the Earth and the Moon point in the same direction.

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  • $\begingroup$ But the second Lagrange point is in the region where x > R, so the signs should adjust automatically. $\endgroup$ – Vishal Jain Apr 15 at 9:26
  • $\begingroup$ @VishalJain: The sign of $- \frac{Gm}{(R-x)^2} < 0$ is always negative (i.e., away from the Earth), regardless of whether $x$ is less than or greater than $R$. If $x > R$, then "away from the Earth" also means "away from the Moon". $\endgroup$ – Michael Seifert Apr 15 at 17:54
  • $\begingroup$ I completely forgot about the squaring there... Yes that works thanks. $\endgroup$ – Vishal Jain Apr 16 at 11:04
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In the restricted three-body problem, the five Lagrangian points are determined based on the effective potential $$U(x,y,\mu)=\frac{(x^2 + y^2)}{2} + \frac {\mu }{\sqrt {(x - 1 + \mu)^2 + y^2}} + \frac {(1 - \mu)}{\sqrt {(x + \mu)^2 + y^2}}$$ $\mu $ is the relative mass of the moon, $1-\mu $ is the relative mass of the earth. The moon is at $(x,y)=(1-\mu ,0)$. Earth is at $(x,y)=(-\mu ,0)$. Triangular and collinear libration points are determined from the system of equations $\nabla U=0$. For the Earth-Moon system, these points and contour plot of the effective potential are shown in Figure 1. Fig1 Some authors define effective potential as $$ U_{eff}(x,y,\mu )= \frac {\mu r_1^2}{2} + \frac {(1 - \mu)r_2^2}{2}+ \frac{\mu}{r_1}+ \frac{(1 - \mu)}{r_2}$$ $$r_1= \sqrt {(x + 1 - \mu)^2 + y^2}, r_2= \sqrt {(x - \mu )^2 + y^2}$$ In this case the moon is located at $(x,y)=(-1+\mu ,0)$. Earth is at $(x,y)=(\mu ,0)$. Triangular and collinear libration points are determined from the system of equations $\nabla U_{eff}=0$. For the Earth-Moon system, these points and contour plot of the effective potential are shown in Figure 2. fig2

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  • $\begingroup$ Ah ok i see why this works, but in the case of finding L2, why doesnt the approach I used work? It seems to work fine for L1? $\endgroup$ – Vishal Jain Apr 15 at 9:28
  • $\begingroup$ @VishalJain Check your equation, see en.wikipedia.org/wiki/Lagrangian_point $\endgroup$ – Alex Trounev Apr 15 at 11:33

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