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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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)?
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

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.

Page1 / 8

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online