Explanation Given information: The gross weight of the spacecraft is W 0 = 11 , 600 lb . The change in velocity is v 1 − v 0 = 360 ft / s . The fuel consumed rate is W = 1 , 000 lb . Calculation: Calculate the weight of the spacecraft ( W 1 ) as shown below. W 1 = W 0 − W Substitute 11 , 600 lb for W 0 and 1 , 000 lb for W . W 1 = 11 , 600 − 1 , 000 = 10 , 600 lb Sketch the Free Body Diagram of the spacecraft as shown in Figure 1. Refer to Figure 1. Apply the conservation of momentum to the fuel plus the fuel as shown below. m v = ( m − Δ m ) ( v + Δ v ) + ( Δ m ) ( v + Δ v − u ) m v = m v + m Δ v − Δ m v − Δ m Δ v + Δ m v + Δ m Δ v − Δ m u m Δ v − u Δ m = 0 Divide both sides of the Equation into Δ t as shown below. m Δ v Δ t − u Δ m Δ t = 0 m Δ v Δ t = u Δ m Δ t Δ v Δ t = u m Δ m Δ t Substitute − d m d t for Δ m Δ t . d v d t = − u m d m d t Integrate both sides of the Equation and apply the limits as shown below

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ISBN: 9781260804638

Author: BEER

Publisher: MCG

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Chapter 14.3, Problem 14.97P

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The relative velocity of the fuel ejected.

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Chapter 14.3, Problem 14.96P

Chapter 14.3, Problem 14.98P

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Chapter 14 Solutions

VECTOR MECHANICS FOR ENGINEERS W/CON >B

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The relative velocity of the fuel ejected.

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