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

Question

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?

Answer 1

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.

Answer 2

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.