Phys - Lecture on lagrange-points

Lagrange Points

Preliminaries

Any conservative force can be related to potential energy by the following (in 1-D):

\begin{equation} F(x) = - \frac{dU}{dx} \end{equation}

or in 3-D:

\begin{equation} \mathbf{F} = - \nabla U \end{equation}

For example, mass on a spring: \begin{equation} F_x= - k x \end{equation} \begin{equation} U_\textrm{spring} = \frac{1}{2}k x^2 \end{equation}

Or, for gravity near the surface of the Earth: \begin{equation} F_y = - m g \end{equation} \begin{equation} U_\textrm{grav} = m g y \end{equation}

Gravitational Potential

Potential Energy and Force due to a single gravitational source.

And, if we consider any distance away from the Earth (or central body), we would have: \begin{equation} F(r) = - G \frac{M_E m}{r^2} \end{equation} \begin{equation} U(r) = - G \frac{M_E m}{r} \end{equation}

Adding another source makes a more complicated landscape

Now, if we have 2 source bodies, the functions have more features. Now, we ask what happens to a 3rd body in this landscape.

Three Bodies

A different type of problem

Restricted 3 body

$m_1$ much greater than $m_2$
Circular Orbits
$m_3$ is negligible in mass

Coordinates for a restricted 3 body system

Forces in Rotating Reference Frame

\begin{equation} \mathbf{F}_\textrm{rot} = \mathbf{F}_\textrm{in}\; \underbrace{- m \boldsymbol{\omega} \times \left( \boldsymbol{\omega } \times \mathbf{r} \right)_\textrm{rot} }_\textrm{centrifugal}\; \underbrace{- 2 m \left( \boldsymbol{\omega} \times \mathbf{v}_\textrm{rot} \right)_\textrm{rot}}_\textrm{Coriolis} \; \underbrace{- m \left( \dot{\boldsymbol{\omega}} \times \mathbf{r} \right)_\textrm{rot}}_\textrm{Euler} \end{equation}

For the case of gravity: \begin{equation} U_G(r) = - \frac{G m_1 m_2}{r} \end{equation} and so our effective potential is: \begin{equation} U_\textrm{eff} = \frac{l^2}{2 \mu r^2} -\frac{G m_1 m_2}{r} \end{equation}

The effective potential

Lagrange Points

The basic FBD and L1 point

All 5 Lagrange points

Find L1 →

The sum of forces acting on the earth is equal to its mass time the acceleration, which for an object moving in a circular orbit is centripetal and equal to $v^2/R$: $$\Sigma F_\textrm{on Earth} = \frac{G M m_E}{R^2} = \frac{m_E v^2}{R}$$ Thus we can say: $$\frac{GM}{R} = v^2$$ but, based on the relation between speed and period, $T$: $$v = \frac{2 \pi R}{T}$$ or $$v^2 = \frac{4 \pi^2 R^2}{T^2}$$ So we can rewrite as: $$\frac{GM}{R} = \frac{4 \pi^2 R^2}{T^2}$$ Rearranging yields Kepler's Third Law $$ \begin{equation} \frac{GM}{R^3} = \frac{4 \pi^2}{T^2} \end{equation} $$

Now consider the sum of forces on the satellite: $$\Sigma F_\textrm{on Sat} = \frac{G M m_\textrm{sat}}{(R-r)^2}-\frac{G m_E m_\textrm{sat}}{r^2} = \frac{m_\textrm{sat}v_\textrm{sat}^2}{(R-r)}$$ This we can simplify to, using Kepler's Third law for the satellite: $$\frac{GM}{R-r}-\frac{Gm_E(r-R)}{r^2} = v_\textrm{sat}^2 = \frac{4 \pi^2 (R-r)^2}{T_\textrm{sat}^2}$$ or $$ \begin{equation} \frac{GM}{(R-r)^3}-\frac{Gm_E}{r^2(R-r)} = \frac{4\pi^2}{T_\textrm{sat}^2} \end{equation} $$

Lastly, since we want their periods to be the same: $T_\textrm{sat} = T$ $$\begin{equation} \frac{GM}{(R-r)^3}-\frac{Gm_E}{r^2(R-r)} = \frac{GM}{R^3} \end{equation}$$

If we put in the masses of the two major bodies, the Earth and Sun for example, we can calculate the value of $r$ in terms of $R$. For our earth-sun system, the L1 point would be located at $$r = 0.009969 \; \textrm{AU}$$ or about 1/100 of the earth-sun distance. Changing signs in the sums of forces could allow to calculate the L2 and L3 positions as well.