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The figure shows the sun located at the origin and the earth at the point $ (1, 0) $. (The unit here is the distance between the centers of the earth and the sun, called an astronomical unit: $ 1 AU \approx 1.496 x 10^8 $ km.) There are five locations $ L_1 $, $ L_2 $, $ L_3 $, $ L_4 $, and $ L_5 $ in this plane of rotation of the earth about the sun where a satellite remains motionless with respect to the earth because the forces acting on the satellite (including the gravitational attractions of the earth and the sun) balance each other. These locations are called libration points. (A solar research satellite has been placed at one of these libration points.) If $ m_1 $ is the mass of the sun, $ m_2 $ is the mass of the earth, and $ r = m_2/(m_1 + m_2) $, it turns out that the $ x $-coordinate of $ L_1 $ is the unique root of the fifth-degree equation

$$

\begin{equation}

p(x) = x^5 - (2 + r)x^4 + (1 + 2r)x^3 - (1 - r)x^2 \\

+ 2(1 - r)x + r - 1 = 0

\end{equation}

$$

and the $ x $-coordinate of $ L_2 $ is the root of the equation

$$ p(x) - 2rx^2 = 0 $$

Using the value of $ r = 3.04042\: X\: 10^{-6} $, find the locations of the libration points of (a) $ L_1 $ and (b) $ L_2 $.

So $x_{n+1}=x_{n}-\frac{x_{n}^{5}-(2+r) x_{n}^{4}+(1+2 r) x_{n}^{3}-(1+r) x_{n}^{2}+2(1-r) x_{n}+r-1}{5 x_{n}^{4}-4(2+r) x_{n}^{3}+3(1+2 r) x_{n}^{2}-2(1-r) x_{n}+2(1-r)}$ $r$ is the same as in $(a),$ but $L_{2}$ is a little more than 1 AU from the Sun, so let $x_{1}=1.02 .$ This will give us $x_{5} \approx x_{6} \approx 1.01008$ Therefore $L_{2}$ is located 1.01008 AU from the Sun

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