Explanation Given: G M J = 319 G M E Conservation of energy, Point A: T A = 1 2 m ( v A − Δ v A ) 2 V A = − G M J m r A G M J = 319 G M E = 319 g R E 2 R E = 6.37 × 10 6 m G M J = 319 G M E = 319 ( 9.81 m / s 2 ) ( 6.37 × 10 6 m ) 2 G M J = 126.98 × 10 15 m 3 / s 2 r A = 350 × 10 6 m V A = − ( 126.98 × 10 15 m 3 / s 2 ) m ( 350 × 10 6 m ) V A = − ( 362.8 × 10 6 ) m Point B: T B = 1 2 m v B 2 V B = − G M J m r B V B = − ( 126.98 × 10 15 m 3 / s 2 ) m ( 100 × 10 6 m ) V B = − ( 1269.8 × 10 6 ) m T A + V A = T B + V B : 1 2 m ( v A − Δ v A ) 2 − 362

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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Conservation of Energy

In physics, the rule of the conservation of energy states that energy doesn’t disappear from the atmosphere; instead, it changes its form from one type to another. Energy is one of the most important physical quantities in nature, and it cannot be create…

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Chapter 13.2, Problem 13.103P

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Decrease in speed Δv at point A which will place the spacecraft into the required orbit.

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Chapter 13.2, Problem 13.102P

Chapter 13.2, Problem 13.104P

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Decrease in speed Δ v at point A which will place the spacecraft into the required orbit.

Solution Summary: The author explains that the Delta v at point A will place the spacecraft into the required orbit.

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Decrease in speed Δv at point A which will place the spacecraft into the required orbit.

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Chapter 13 Solutions