Problem 32 Easy Difficulty

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 $\approx 1.496 \times 10^{8} \mathrm{km} .$ . 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 $r=m_{2} /\left(m_{1}+m_{2}\right),$ it turns out that the $x$ -coordinate of $L_{1}$ is the unique root of the fifth-degree equation
$$p(x)=x^{5}-(2+r) x^{4}+(1+2 r) x^{3}-(1-r) x^{2}$$
$$+2(1-r) x+r-1=0$$
and the $x$ -coordinate of $L_{2}$ is the root of the equation
$$p(x)-2 r x^{2}=0$$
Using the value $r \approx 3.04042 \times 10^{-6},$ find the locations of the libration points (a) $L_{1}$ and $(\mathrm{b}) L_{2}$ .

Answer

(a) $p(x)=x^{5}-(2+r) x^{4}+(1+2 r) x^{3}-(1-r) x^{2}+2(1-r) x+r-1$
$p^{\prime}(x)=5 x^{4}-4(2+r) x^{3}+3(1+2 r) x^{2}-2(1-r) x+2(1-r)$
So $x_{n+1}=x_{n}-\frac{x^{5}-(2+r) x^{4}+(1+2 r) x^{3}-(1-r) x^{2}+2(1-r) x+r-1}{5 x^{4}-4(2+r) x^{3}+3(1+2 r) x^{2}-2(1-r) x+2(1-r)}$
We substitute $r \approx 3.04042 \times 10^{-6} . \quad L_{1}$ is a little less than $1 \mathrm{Au},$ so let
$x_{1}=0.95$ This will give us $x_{8} \approx x_{9} \approx 0.98999$
Therefore $L_{1}$ is located 0.98999 $\mathrm{AU}$ from the Sun

(b) $p(x)-2 r x^{2}=x^{5}-(2+r) x^{4}+(1+2 r) x^{3}-(1+r) x^{2}+2(1-r)+r-1$
$\left(p(x)-2 r x^{2}\right)^{\prime}=5 x^{4}-4(2+r) x^{3}+3(1+2 r) x^{2}-2(1+r) x+2(1-r)$
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 $(\mathrm{a}),$ but $L_{2}$ is a little more than 1 $\mathrm{AU}$ 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 $\mathrm{AU}$ from the Sun

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Video Transcript

question 42. We are given you have X, which equals X to the fifth minus x to the poor. Times two plus r waas x Q one plus two are minus ex cleared one minus lower plus thanks one minus. R less R minus one, which gives our you prime be five x to the fore. Minus were X Q times two butts are plus three x weird one plus two are minus legs. One Linus are plus two one bar. So again I put, uh, this in why one in white too. And then we're using our equals a 3.4042 times 10 to the negative six. And when we do that, we get it x one starting at approximately yeah, 0.95 when we use oh X and fo X plus one equals x of n minus crying, which I have as mine. It's why one divided by y two in my wife three position. And then when I plug in, I goto ah table and I said it to ask. So what? I type in 0.95 The next number it gives me core Exit two equals 396682 Exit three gives me quite nine 7770 except for 90 war exit five 0.9 83 b ro six 98 Milling seven Exit eight People's Exit nine. Which equals Wait. Knowing a 999 heartbeat, we do the same process. Except for for ex of one. We're gonna start with 1.2 on equals, X one on one. Thieving Vero. Hey.

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