# How to find Lagrange point $L_2$ of the Earth-Moon system?

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: According to desmos, there was no L2 point.

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

• $x$ should be the distance of the point you are considering from the earth and $R$ the earth-to- moon distance! – AoZora Apr 14 at 20:12
• I have made the edit thank you. – Vishal Jain Apr 15 at 9:27

## 2 Answers

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.

• But the second Lagrange point is in the region where x > R, so the signs should adjust automatically. – Vishal Jain Apr 15 at 9:26
• @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". – Michael Seifert Apr 15 at 17:54
• I completely forgot about the squaring there... Yes that works thanks. – Vishal Jain Apr 16 at 11:04

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. 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. • 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? – Vishal Jain Apr 15 at 9:28
• @VishalJain Check your equation, see en.wikipedia.org/wiki/Lagrangian_point – Alex Trounev Apr 15 at 11:33