To determine The change of the g spacecraft approaching earth at 4,500 ft/s at 25,500 mi from the surface. Explanation The acceleration due to the gravity g on a spacecraft is inversely proportional to its distance from the center of Earth. i.e. g ∝ 1 r + h g = k r + h Where k is the proportionality constant. Substitute the radius of the earth r is 3,960 mi = 20.9 × 10 6 ft and the gravity of g is g = 32.2 ft/s 2 at h = 0 in g = k r + h . 32.2 = k 20.9 × 10 6 + 0 k = 32.2 × 20.9 × 10 6 = 673 × 10 6 ft 2 /s 2 Differentiate the equation g = k r + h with respect to t : d d t ( g ) = d d t ( k r + h ) d g d t = k d d t ( r + h ) − 1 d g d t = k × ( − 1 ) ( r + h ) − 2 d h d t = − k ( r + h ) 2 d h d t Substitute the value of constant k = 673 × 10 6 ft 2 /s 2 , the radius of the earth r is 3,960 mi = 20

Basic Technical Mathematics with Calculus (11th Edition)

11th Edition

ISBN: 9780134437736

Author: Allyn J. Washington, Richard Evans

Publisher: PEARSON

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Chapter 24.4, Problem 30E

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The change of the g spacecraft approaching earth at 4,500 ft/s at 25,500 mi from the surface.

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Chapter 24.4, Problem 29E

Chapter 24.4, Problem 31E

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Chapter 24 Solutions

Basic Technical Mathematics with Calculus (11th Edition)

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