Explanation Given: h 1 = 200 mi h 2 = 500 mi Elliptical orbit between A and B Conservation of angular momentum: r A m v A = r B m v B v A = r B r A v B = 7.170 6.690 v B r A = 6370 km + 320 km = 6690 km r A = 6.690 × 10 6 m v A = 1.0718 v B … (1) r B = 6370 km + 800 km = 7170 km r B = 7.170 × 10 6 m R = ( 6370 km ) = 6.37 × 10 6 m Conservation of energy: G M = g R 2 = ( 9.81 m / s 2 ) ( 6.37 × 10 6 m ) 2 G M = 398.060 × 10 12 m 3 / s 2 Point A: T A = 1 2 m v A 2 V A = − G M m r A = − ( 398.060 × 10 12 ) m ( 6.690 × 10 6 ) = 59.501 × 10 6 m Point B: T B = 1 2 m v B 2 V B = − G M m r B = − ( 398.060 × 10 12 ) m ( 7.170 × 10 6 ) = 55.5 × 10 6 m T A + V A = T B + V B 1 2 m v A 2 − 59 To determine (b) Total energy per unit mass required to execute the transfer.

12th Edition

ISBN: 9781259977305

Author: BEER, Ferdinand P. (ferdinand Pierre), Johnston, E. Russell (elwood Russell), Cornwell, Phillip J., SELF, Brian P.

Publisher: Mcgraw-hill Education,

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Chapter 13.2, Problem 13.105P

To determine

(a)

Increases in speed required at A and at B.

To determine

(b)

Total energy per unit mass required to execute the transfer.

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Chapter 13.2, Problem 13.104P

Chapter 13.2, Problem 13.106P

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(a) Increases in speed required at A and at B.

Solution Summary: The author explains the increase in speed required at A and at B.