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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 C H A P T E R 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 = 4 π 2 GM a 3 or T = 2 π GM a 3 / 2 If a is increased four-fold the period becomes: 2 π GM ( 4 a ) 3 / 2 = 8 · 2 π GM 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 SOLUTION Using the given data we obtain the following values of T 2 / a 3 , where 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 ) as T a 3 · 3 · 10 4 11 . 9 years. 2. A satellite has initial position r = 1 , 000 , 2 , 000 , 0 and initial velocity r = 1 , 2 , 2 (units of kilometers and seconds). Find the equation of the plane containing the satellite’s orbit. Hint: This plane is orthogonal to J . SOLUTION The vectors r ( t ) and r ( t ) lie in the plane containing the satellite’s orbit, in particular the initial position r = 1000 , 2000 , 0 and the initial velocity r = 1 , 2 , 2 . Therefore, the cross product J = r × r is perpendicular to the plane. We compute J : J = r × r = i j k 1 , 000 2 , 000 0 1 2 2 = 2 , 000 0 2 2 i 1 , 000 0 1 2 j + 1 , 000 2 , 000 1 2 k = 4 , 000 i 2 , 000 j = 4000 , 2000 , 0 We now use the vector form of the equation of the plane with n = J = 4000 , 2000 , 0 and x 0 , y 0 , z 0 = r = 1000 , 2000 , 0 , to obtain the following equation: 4000 , 2000 , 0 · x , y , z = 4000 , 2000 , 0 · 1000 , 2000 , 0 2000 2 , 1 , 0 · x , y , z = 2000 2 , 1 , 0 · 1000 , 2000 , 0 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 for the earth (in units of squared feet per second)?

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S E C T I O N 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 d A dt = 1 2 J ; J = r ( t ) × r ( 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 ( t ) are orthogonal. Using properties of the cross product we get: J = r ( t ) × r ( t ) = r ( t ) r ( t ) = R r ( t ) = const We conclude that the speed v = r ( t ) is constant. We find the speed using the following equality: 2 π R = v T v = 2 π R T . Therefore, J = R · 2 π R T = 2 π R 2 T . Substituting in (1) we get: d A dt = 1 2 · 2 π R 2 T = π R 2 T .
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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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