Explanation Given information: h 1 = 225 m i ϕ B = 60 o h 2 = 40 m i Formula used: G M = g R 2 T = 1 2 m v 2 V = − G M m r L = m v r Calculation: As we have, r A = 3960 m i + 225 m i = 4185 m i r A = 4185 m i × 5280 f t / m i = 22097 × 10 3 f t And, r B = 3960 m i × 40 m i = 4000 m i ⇒ r B = 4000 × 5280 = 21120 × 10 3 f t ⇒ R = 3960 m i = 20909 × 10 3 f t We know that: G M = g R 2 Put the values in the above equation; ⇒ G M = ( 32.2 f t / s 2 ) ( 20909 × 10 3 f t ) 2 ⇒ G M = 14.077 × 10 15 f t 3 / s 2 According to conservation of energy; ⇒ T A = 1 2 m v A 2 ⇒ V A = − G M m r A Put the values in the above equation; ⇒ − 14.077 × 10 15 m 22097 × 10 3 ⇒ − 637.1 × 10 6 m ⇒ T B = 1 2 m v B 2 ⇒ V B = − G M m r B ⇒ − 14 To determine (b) The sped at point B at given condition.

Vector Mechanics for Engineers: Dynamics

11th Edition

ISBN: 9780077687342

Author: Ferdinand P. Beer, E. Russell Johnston Jr., Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Chapter 13.2, Problem 13.110P

To determine

(a)

The speed of the vehicle as it leaves its circular orbit at A at given condition.

To determine

(b)

The sped at point B at given condition.

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Chapter 13.2, Problem 13.109P

Chapter 13.2, Problem 13.111P

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(a) The speed of the vehicle as it leaves its circular orbit at A at given condition.

Solution Summary: The author calculates the speed of the vehicle as it leaves its circular orbit at A at given condition.