Explanation Given information: The mass of the planet is M . The minimum distance from the satellite to the center of the plant is r 0 . The maximum distance from the satellite to the center of the plant is r 1 . The angular momentum per unit mass of the satellite is h . The gravitation constant is G . Calculation: The expression for the initial velocity ( v 0 ) of the satellite as follows: v 0 = h r 0 The expression for the final velocity ( v 1 ) of the satellite as follows: v 1 = h r 1 The expression for the kinetic energy in A ( T A ) as follows: T A = 1 2 m v 0 2 The expression for the potential energy in A ( V A ) as follows: V A = − G M m r 0 Here, m is the mass of the satellite. The expression for the kinetic energy in B ( T B ) as follows: T B = 1 2 m v 1 2 The expression for the potential energy in B ( V B ) as follows: V B = − G M m r 1 The expression for the conservation of energy in the elliptical orbit as follows: T A + V A = T B + V B Substitute 1 2 m v 0 2 for T A , − G M m r 0 for V A , 1 2 m v 1 2 for T B and − G M m r 1 for V B . 1 2 m v 0 2 − G M m r 0 = 1 2 m v 1 2 − G M m r 1 v 0 2 2 − G M r 0 = 1 2 v 1 2 − G M r 1 v 0 2 − 2 G M r 0 = v 1 2 − 2 G M r 1 v 0 2 − v 1 2 = − 2 G M r 1 + 2 G M r 0 v 0 2 − v 1 2 = 2 G M ( 1 r 0 − 1 r 1 ) v 0 2 − v 1 2 = 2 G M ( r 1 − r 0 r 1 r 0 ) Substitute h r 0 for v 0 and h r 1 for v 1

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Author: BEER

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Chapter 13, Problem 13.192RP

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Derive the relation 1r0+1r1=2GMh2 using the principles of conservation of energy and conservation of angular momentum.

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Chapter 13, Problem 13.191RP

Chapter 13, Problem 13.193RP

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Derive the relation 1 r 0 + 1 r 1 = 2 G M h 2 using the principles of conservation of energy and conservation of angular momentum.

Solution Summary: The author explains the derivation relation 1r_0+. The angular momentum per unit mass of the satellite is h.

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Derive the relation 1r0+1r1=2GMh2 using the principles of conservation of energy and conservation of angular momentum.

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Chapter 13 Solutions