## FANDOM

155 Pages

Lagrange points are equilibrium points in the presence of the gravitational field of two rotating point masses. We denote one point mass by $M_1$ and the other by $M_2$. We denote the separation between the two point masses by $a$. The centre of mass lies between the two point masses, such that the distance between the first point mass to the centre of mass is $r_1 = a \frac{M_2}{M_1 + M_2}$, and to the second is $r_2 = a \frac{M_1}{M_1 + M_2}$. We choose an coordinate system whose origin coincides with the centre of mass, and whose x axis aligns with the line connecting the two point masses. The position of the first mass on this axis is $\left( -r_1, 0 \right)$, and that of the second point mass is $\left( r_2, 0 \right)$. In a co - rotating frame one must also take into account the centrifugal force. The rotation frequency is $\omega = \sqrt{\frac{G\left(M_1+M_2\right)}{a^3}}$. Hence the potential (in polar coordinates) is

$\varphi = -\frac{G M_1}{\sqrt{r_1^2+r^2+2 r r_1 \cos \theta}} -\frac{G M_2}{\sqrt{r_2^2+r^2 - 2 r r_2 \cos \theta}} - \frac{1}{2} \omega^2 r^2$

This expression can be simplified by introducing dimensionless parameters $\mu = \frac{M_2}{M_1}$, $\xi = \frac{r}{a}$ and $\hat{\varphi} = \frac{\varphi}{G M_1 / a}$.

$\hat{\varphi} = -\frac{1}{\sqrt{\left(\frac{\mu}{1+\mu}\right)^2+ \xi^2 + 2 \xi \frac{\mu}{1+\mu} \cos \theta}} - \frac{\mu}{\sqrt{\left(\frac{1}{1+\mu}\right)^2+ \xi^2 - 2 \xi \frac{1}{1+\mu} \cos \theta}} - \frac{1}{2} \left(1+\mu \right) \xi^2$

It turns out that this expression has 5 extrema: 3 on the x axis (L1, L2 and L3) and 2 at an oblique angle (L4 and L5). Here we will present a somewhat simplified description. A more detailed analysis can be found elsewhere.

## Aligned Points Edit

In this section we will obtain an equation for the location of the Lagrange points that lie on the line connecting the two masses. On that line the potential is

$\hat{\varphi} = -\frac{1}{\left| \xi - \frac{\mu}{1+\mu} \right|} - \frac{1}{\left|\xi + \frac{1}{1+\mu} \right|} - \frac{1}{2} \left( 1+ \mu \right) \xi^2$

Finding the extrema in the most general case involves solving a quintic equation. Instead, we will examine limiting cases. In the first case we examine an extremely large mass ratio $\mu \rightarrow \infty$. In this case the centre of mass coincides with the position of the heavier mass. If we disregard the effect of the lighter mass, then the positions of the Lagrange points are at $\xi = 1$. This means that one point is opposite to the lighter mass, and the other two are very close to it. The other extreme limit is when both masses are equal $\mu = 1$. From symmetry considerations, onte Lagrange point coincides with the centre of mass at $\xi = 0$. The other two points (which are at equal distances from the centre of mass) can be found by explicit calculation. The dimensionless potential is

$\hat{\varphi} = -\frac{1}{\xi - \frac{1}{2}} - \frac{1}{\xi+\frac{1}{2}} - \xi^2$

Deriving with respect to $\xi$

$\frac{\partial \hat{\varphi}}{\partial \xi} = \frac{1}{\left(\xi-\frac{1}{2}\right)^2} + \frac{1}{\left(\xi + \frac{1}{2}\right)^2} - 2 \xi = 0$

The only real solution for this equation is $\xi \approx 1.2$. All in all the outer Lagrange points vary in the range 1 to 1.2, and the inner point in the range 0 to 1.

## Oblique Points Edit

To find the other two point which are not on the same axis, we need to also find the extrema in the angle

$\frac{\partial \hat{\varphi}}{\partial \theta} = \frac{-\frac{\mu}{\mu+1} \xi \sin \theta }{\left[ \left( \frac{\mu}{1+\mu} \right)^2 + \xi^2 -2 \xi \frac{\mu}{\mu+1} \cos \theta \right]^{3/2}} + \frac{\frac{\mu}{\mu+1} \xi \sin \theta }{\left[ \left( \frac{1}{1+\mu} \right)^2 + \xi^2 + 2 \xi \frac{1}{\mu+1} \cos \theta \right]^{3/2}} = 0$

After some algebra

$\cos \theta = \frac{1}{2 \xi} \frac{\mu - 1}{\mu + 1}$

Substituting back into the condition for equilibrium in the radial direction

$\frac{\partial \hat{\varphi}}{\partial \xi} = \frac{\xi-\frac{\mu}{1+\mu} \cos \theta}{\left[ \left(\frac{\mu}{1+\mu}\right)^2 + \xi^2 -2 \frac{\mu}{1+\mu} \cos \theta \right]^{3/2}} + \frac{\mu \left(\xi-\frac{\mu}{1+\mu} \cos \theta \right)}{\left[ \left(\frac{1}{1+\mu}\right)^2 + \xi^2 + 2 \frac{1}{1+\mu} \cos \theta \right]^{3/2}} - 2 \left(1+\mu \right) \xi =$

$= \frac{\xi-\frac{\mu}{1+\mu} \frac{1}{2 \xi} \frac{\mu - 1}{\mu + 1}}{\left[ \left(\frac{\mu}{1+\mu}\right)^2 + \xi^2 -2 \frac{\mu}{1+\mu} \frac{1}{2 \xi} \frac{\mu - 1}{\mu + 1} \right]^{3/2}} + \frac{\mu \left(\xi-\frac{\mu}{1+\mu} \frac{1}{2 \xi} \frac{\mu - 1}{\mu + 1} \right)}{\left[ \left(\frac{1}{1+\mu}\right)^2 + \xi^2 + 2 \frac{1}{1+\mu} \frac{1}{2 \xi} \frac{\mu - 1}{\mu + 1} \right]^{3/2}} - 2 \left(1+\mu \right) \xi =$

$= \frac{\left(1+\mu \right) \xi}{ \left[ \frac{\mu}{\left(1+\mu \right)^2} + \xi^2 \right]^{3/2}} - \left( 1+ \mu \right) \xi = 0$

Solving for $\xi$

$\xi = 1 - \frac{\mu}{\left(1 + \mu \right)^2}$