14.6Ex1-19

# 14.6Ex1-19 - 572 C H A P T E R 14 C A L C U L U S O F VE C...

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572 CHAPTER 14 CALCULUS OF VECTOR-VALUED FUNCTIONS (ET CHAPTER 13) SOLUTION Kepler’s Third Law states that the period T of the orbit is given by: T 2 = Ã 4 π 2 GM ! a 3 or T = 2 a 3 / 2 If a is increased four-fold the period becomes: 2 ( 4 a ) 3 / 2 = 8 · 2 a 3 / 2 That is, the period is increased eight-fold. Exercises 1. Kepler’s Third Law states that T 2 / a 3 has the same value for each planetary orbit. Do the data in the following table support this conclusion? Estimate the length of Jupiter’s period, assuming that a = 77 . 8 × 10 10 m. Planet Mercury Venus Earth Mars a (10 10 m) 5.79 10.8 15.0 22.8 T (years) 0.241 0.615 1.00 1.88 Using the given data we obtain the following values of T 2 / a 3 ,whe re a , as always, is measured not in meters but in 10 10 m: Planet Mercury Venus Earth Mars T 2 / a 3 2 . 99 · 10 4 3 · 10 4 2 . 96 · 10 4 2 . 98 · 10 4 The data on the planets supports Kepler’s prediction. We estimate Jupiter’s period (using the given a )a s T a 3 · 3 · 10 4 11 . 9 years. 2. A satellite has initial position r = h 1 , 000 , 2 , 000 , 0 i and initial velocity r 0 = h 1 , 2 , 2 i (units of kilometers and seconds). Find the equation of the plane containing the satellite’s orbit. Hint: This plane is orthogonal to J . The vectors r ( t ) and r 0 ( t ) lie in the plane containing the satellite’s orbit, in particular the initial position r = h 1000 , 2000 , 0 i and the initial velocity r 0 = h 1 , 2 , 2 i . Therefore, the cross product J = r × r 0 is perpendicular to the plane. We compute J : J = r × r 0 = ¯ ¯ ¯ ¯ ¯ ¯ ij k 1 , 000 2 , 000 0 12 2 ¯ ¯ ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ 2 , 000 0 22 ¯ ¯ ¯ ¯ i ¯ ¯ ¯ ¯ 1 , 000 0 ¯ ¯ ¯ ¯ j + ¯ ¯ ¯ ¯ 1 , 000 2 , 000 ¯ ¯ ¯ ¯ k = 4 , 000 i 2 , 000 j = h 4000 , 2000 , 0 i We now use the vector form of the equation of the plane with n = J = h 4000 , 2000 , 0 i and h x 0 , y 0 , z 0 i = r = h 1000 , 2000 , 0 i , to obtain the following equation: h 4000 , 2000 , 0 i · h x , y , z i = h 4000 , 2000 , 0 i · h 1000 , 2000 , 0 i 2000 h 2 , 1 , 0 i · h x , y , z i = 2000 h 2 , 1 , 0 i · h 1000 , 2000 , 0 i 2 x y = 2 , 000 2 , 000 + 0 = 0 2 x y = 0 The plane containing the satellite’s orbit is, thus: P = { ( x , y , z ) : 2 x y = 0 } 3. The earth’s orbit is nearly circular with radius R = 93 × 10 6 miles (the eccentricity is e = 0 . 017). Find the rate at which the earth’s radial vector sweeps out area in units of ft 2 / s. What is the magnitude of the vector J = r × r 0 for the earth (in units of squared feet per second)?

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SECTION 14.6 Planetary Motion According to Kepler and Newton (ET Section 13.6) 573 SOLUTION The rate at which the earth’s radial vector sweeps out area is dA dt = 1 2 k J k; J = r ( t ) × r 0 ( t ) (1) Since J is a constant vector, its length is constant. Moreover, if we assume that the orbit is circular then r ( t ) lies on a circle, and therefore r ( t ) and r 0 ( t ) are orthogonal. Using properties of the cross product we get: k J k=k r ( t ) × r 0 ( t ) r ( t ) kk r 0 ( t ) k= R k r 0 ( t ) const We conclude that the speed v =k r 0 ( t ) k is constant. We Fnd the speed using the following equality: 2 π R = v T v = 2 R T .
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## This note was uploaded on 02/13/2010 for the course MATH MATH 32A taught by Professor Park during the Fall '09 term at UCLA.

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14.6Ex1-19 - 572 C H A P T E R 14 C A L C U L U S O F VE C...

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