Explanation Given information: ε = 0.999887 Minimum distance between the comet and the sun = 0.230 R E , Mean distance from the sun to earth, R E = 150 × 10 6 k m Motion of the comet, 1 r = G M h 2 ( 1 + ε cos θ ) Angular momentum of a unit mass, h = r . v Periodic time τ = 2 π a b h a = 1 2 ( r 0 + r 1 ) b = r 0 r 1 Calculation: g = 9.81 m s − 2 R E a r t h = 6.37 × 10 6 m Consider the motion of the comet, 1 r = G M h 2 ( 1 + ε cos θ ) At the closet point to the sun, θ = 0 r 0 = 0.230 R E = 0.230 × 150 × 10 9 m r 0 = 34.5 × 10 9 m 1 34.5 × 10 9 m = G M h 2 ( 1 + 0.999887 cos 0 0 ) G M h 2 = 1.449 × 10 − 11 m − 1 At the farthest point to the sun, θ = 180 0 1 r 1 = G M h 2 ( 1 + 0.999887 cos 180 0 ) 1 r 1 = 1.449 × 10 − 11 × ( 1 + 0.999887 cos 180 0 ) r 1 = 6.1 × 10 14 m Periodic time of the comet can be calculated using the following equation. τ = 2 π a b h a = 1 2 ( r 0 + r 1 ) a = 1 2 ( 34.5 × 10 9 + 6

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 12.3, Problem 12.109P

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The periodic time of the comet.

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Chapter 12.3, Problem 12.108P

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Vector Mechanics for Engineers: Dynamics

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The periodic time of the comet.

Solution Summary: The author explains the periodic time of the comet and the angular momentum of a unit mass.

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