Explanation Given information: Radius of the earth = 6370 k m Altitude to the point C = 1500 k m A and B on the earth’s surface. R ϕ of the trajectory = 6000 k m Angular momentum of a unit mass, h = r . v Motion of satellites and spacecraft 1 r = G M h 2 ( 1 + ε cos θ ) Calculation: Initial radius of the trajectory at point C , r 0 = 6.37 × 10 6 + 1.5 × 10 6 r 0 = 7.87 × 10 6 m g = 9.81 m s − 2 R = 6.37 × 10 6 m Motion of the projectile, 1 r = G M h 2 ( 1 + ε cos θ ) C is the apogee. Thus at point C , θ = π r a d Applying the equation at point C , 1 7.87 × 10 6 = G M h 2 ( 1 + ε cos π ) ......... ( 1 ) Given that, R ϕ = 6000 k m Therefore, ϕ = 6 × 10 6 6.37 × 10 6 = 0.9418 r a d ϕ 2 = 0.471 r a d At point B , θ = ( π − ϕ 2 ) r a d θ = ( π − 0.471 ) r a d θ = 2 To determine (b) The eccentricity ε of the trajectory.

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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Question

Chapter 12.3, Problem 12.115P

To determine

(a)

The velocity of the projectile at C.

To determine

(b)

The eccentricity ε of the trajectory.

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Chapter 12.3, Problem 12.114P

Chapter 12.3, Problem 12.116P

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Vector Mechanics for Engineers: Dynamics

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