Systems with Varying Mass: A Rocket — Question 10

Section 9-11 · Physics pages 1370–1371

Question 10 A dry-mass constraint limits the deliverable velocity change A rocket has 100 kg of dry structure, 300 kg of propellant and payload mass L ≥0. In force-free straight-line operation its constant relative exhaust speed is u = 2000 m/s. It must supply at least 1500 m/s of velocity gain. 100 300 kg fuel L mi = 400 + L final: structure + payload The dry structure and payload remain after all propellant is expelled. (a) Find the maximum payload mass. (b) Check whether a 200 kg payload meets the requirement, and explain why the ideal logarithm cannot diverge for this rocket.
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Question 10 - Solution A dry-mass constraint limits the deliverable velocity change 100 300 kg fuel L mi = 400 + L final: structure + payload (a) Derivation and calculation The full-burn capability must satisfy 2000 ln 400 + L 100 + L ≥1500. With R = e0.75 > 1, 400 + L ≥R(100 + L) ⇒0 ≤L ≤400 −100R R −1 = 168.58 kg . (b) Result and interpretation For L = 200 kg, ∆v = 2000 ln(600/300) = 1386.29 m/s < 1500 m/s . The final mass obeys mf = 100 + L ≥100 kg. In fact ∆v ≤2000 ln 4 = 2772.59 m/s, so the logarithm cannot diverge within the stated physical model. Physical check Adding payload decreases the mass ratio and therefore decreases the achievable velocity gain.

Question 10

A dry-mass constraint limits the deliverable velocity change

A rocket has 100kg100\,\mathrm{kg} of dry structure, 300kg300\,\mathrm{kg} of propellant and payload mass L≥0L\ge0. In force-free straight-line operation its constant relative exhaust speed is u=2000m/su=2000\,\mathrm{m/s}. It must supply at least 1500m/s1500\,\mathrm{m/s} of velocity gain.

Diagram shown in the original SVG worksheet.

  1. Find the maximum payload mass.

  2. Check whether a 200kg200\,\mathrm{kg} payload meets the requirement, and explain why the ideal logarithm cannot diverge for this rocket.

Question 10 - Solution

A dry-mass constraint limits the deliverable velocity change

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

The full-burn capability must satisfy 2000ln⁡400+L100+L≥1500.2000\ln\frac{400+L}{100+L}\geq1500. With R=e0.75>1R=e^{0.75}>1, 400+L≥R(100+L)⇒0≤L≤400−100RR−1=168.58kg.400+L\geq R(100+L)\Rightarrow\boxed{0\leq L\leq\frac{400-100R}{R-1}=168.58\ \mathrm{kg}}.

(b) Result and interpretation

For L=200L=200 kg, Δv=2000ln⁡(600/300)=1386.29m/s<1500m/s.\boxed{\Delta v=2000\ln(600/300)=1386.29\ \mathrm{m/s}<1500\ \mathrm{m/s}}. The final mass obeys mf=100+L≥100m_f=100+L\geq100 kg. In fact Δv≤2000ln⁡4=2772.59m/s,\Delta v\leq2000\ln4=2772.59\ \mathrm{m/s}, so the logarithm cannot diverge within the stated physical model.

Physical check

Adding payload decreases the mass ratio and therefore decreases the achievable velocity gain.