   Chapter 3.8, Problem 40E

Chapter
Section
Textbook Problem

# 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 A U ≈ 1.496 × 10 8 km.) There are five locations L 1 , L 2 , L 3 , 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 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 ≈ 3.04042 × 10 − 6 , find the locations of the libration points ( a ) L 1 and ( b ) L 2 To determine

(a)

To find:

Location of liberation point L1

Explanation

1) Concept:

Working rule of Newton’s method is,

Start with initial approximation x1  which is obtained from graph. Evaluate fx1and f'(x1) and apply the Newton’s formula. After the first iteration we get x2. Repeat the process with x1  replaced by x2. Calculate the approximations x1, x2, x3,  until the successive approximations xn, xn+1  agree to the four decimal places.

2) Formula:

Power rule of differentiation:

ddxxn=nxn-1

3) Given:

x-coordinate  of L1 is root of

px= x5-2+rx4+1+2rx3-1-rx2+21-rx+r-1

r=3.04042×10-6

4) Calculation:

Substitute the value of r in px,

px= x5-2+[3.04042×10-6]x4+1+2[3.04042×10-6]x3-1-3.04042×10-6x2+21-3.04042×10-6x+3.04042×10-6-1

px= x5-2.00000304x4+1.000006081x3-0.999996959x2+1.999993919x-0.999996959

Differentiating px with respect to x,

By use of power rule,

p'(x)=5x4-8.00001216x3+3.000018243x2-1.999993918x+1.999993919

Now draw the graph to determine the initial value,

From the graph, initial value is 0.99. So let x1=0.99

So Newton’s formula for nth approximation becomes

xn+1=xn-fxnf'xn

For n=1

To find value of x2,

x2=x1-f(x1 )f'x1

x2=x1-fx1 f'x1=x1-f0

To determine

(b)

To find:

Location of liberation point L2

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