Explanation Given: For earth, R = 3960 mi = 20.909 × 10 6 ft g = 32.2 ft 2 G M = g R 2 G M = ( 32.2 ) ( 20.909 × 10 6 ) 2 G M = 14.077 × 10 15 ft 3 / s 2 Speed on a circular orbit of radius r , r A and r B : F = m a n G M m r 2 = m v 2 r v 2 = G M r v = G M r r A = 3960 + 185 = 4145 mi = 21.886 × 10 6 ft ( v A ) c i r c = 14.077 × 10 15 21.886 × 10 6 = 25.362 × 10 3 ft / s r B = 3960 + 22230 = 26190 mi = 138.283 × 10 6 ft ( v B ) c i r c = 14.077 × 10 15 138.283 × 10 6 = 10.089 × 10 3 ft / s Calculate speeds A and B for path AB, Conservation of angular momentum: m r A v A sin ϕ A = m r B v B sin ϕ A v B = r A v A sin 90 ∘ v B sin 90 ∘ = 21.886 × 10 6 v A 138 To determine (b) Increase in velocity required at B to place the satellite in its final orbit.

12th Edition

ISBN: 9781260265521

Author: BEER

Publisher: MCG

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Chapter 13.2, Problem 13.101P

To determine

(a)

Velocity of IUS relative to the shuttle after its engine has been fired at A.

To determine

(b)

Increase in velocity required at B to place the satellite in its final orbit.

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Chapter 13.2, Problem 13.100P

Chapter 13.2, Problem 13.102P

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(a) Velocity of IUS relative to the shuttle after its engine has been fired at A.

Solution Summary: The author compares the velocity of IUS relative to the shuttle after its engine has been fired at A.