Lagrange Points

All the formulas used here are in orbital. We are specifically focusing on $F_g$ for this calculation though.

What is a Lagrange point? §

A Lagrange point is essentially a point between two bodies in space where the gravitational force of both bodies is the same; aka, $F_g=F_g$, which also means $F_c=F_c$. This essentially means that an object experiences zero net force, since the same force is acting on it from both directions.

How can you calculate a Lagrange point? §

To find a Lagrange point, you need to determine the point between body one and body two that makes it so that $F_g=F_g$. We find the first Lagrange point between the Earth and the Moon as an example:

$$\begin{array}{rl}{F}_{g_{Earth}}& =F_{g_{Moon}}\\ & \\ \frac{G{m}_{Earth}{m}_{Object}}{(r_{Earth}-x{)}^2}& =\frac{G{m}_{Moon}{m}_{Object}}{x^2}\end{array}$$

Where $x$ is equivalent to the distance from the moon to the Lagrange point, and also what we are solving for.

For this example we will just solve for a ratio between Earth and the Moon. We’ll start by factoring out what we can:

$$\begin{array}{rl}\frac{m_E{m}_O}{(r_E-x{)}^2}& =\frac{m_M{m}_O}{x^2}\text{ (We can factor }G\text{ out through division.)}\\ \frac{m_E}{(r_E-x{)}^2}& =\frac{m_M}{x^2}\text{ (We can do the same with }{m}_O.)\end{array}$$

Now, to set the equation to be equivalent to a ratio of $\frac{m_E}{m_M}$, something we can more easily calculate:

$$\begin{array}{rl}{m}_E& =(r_E-x{)}^2\frac{m_M}{x^2}\\ \frac{m_E}{m_M}& =\frac{(r_E-x{)}^2}{x^2}\end{array}$$

And to substitute in the following values:

• $m_E$ = $5.98\text{E}24$ kg
• $m_M$ = $7.35\text{E}22$ kg
• $r_E$ = $6.38\text{E}6$ meters

$$\begin{array}{rl}\frac{m_E}{m_M}& =81.361\\ & \\ 81.361& =\frac{(r_E-x{)}^2}{x^2}\\ & \\ \sqrt{81.361}& =\frac{r_E-x}{x}\\ & \\ 9.02x& =r_E-x\\ & \\ 10.02x& =(6.38\text{E}6)\\ & \\ x& =\frac{(6.38\text{E}6)}{10.02}\\ & \\ x& =636726\text{ meters}\end{array}$$