Systems with Varying Mass: A Rocket — Question 3

Section 9-11 · Physics pages 1356–1357

Question 3 Constant thrust gives increasing acceleration A rocket in force-free space starts with mass 1000 kg and expels propellant at constant q = 5 kg/s with relative exhaust speed u = 1800 m/s. The burn lasts 40 s. t a (m/s2) 9 11.25 40 s Constant thrust produces increasing acceleration as the rocket mass decreases. (a) Find thrust, initial acceleration, final acceleration and velocity gain. (b) Compare with using the initial acceleration for the whole burn.
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Question 3 - Solution Constant thrust gives increasing acceleration t a (m/s2) 9 11.25 40 s (a) Derivation and calculation Mass and thrust during the burn are m(t) = 1000 −5t, T = uq = 1800(5) = 9000 N . ai = 9000/1000 = 9.0 , af = 9000/800 = 11.25 m/s2 . ∆v = 1800 ln(1000/800) = 401.66 m/s . (b) Result and interpretation Holding acceleration at its initial value would give ∆vest = 9(40) = 360 m/s, ∆v −∆vest = 41.66 m/s . Because a(t) = 9000/(1000 −5t) rises monotonically, the initial-acceleration estimate is a lower bound. Physical check The exact velocity gain lies between 9(40) = 360 and 11.25(40) = 450 m/s.

Question 3

Constant thrust gives increasing acceleration

A rocket in force-free space starts with mass 1000kg1000\,\mathrm{kg} and expels propellant at constant q=5kg/sq=5\,\mathrm{kg/s} with relative exhaust speed u=1800m/su=1800\,\mathrm{m/s}. The burn lasts 40s40\,\mathrm s.

Diagram shown in the original SVG worksheet.

  1. Find thrust, initial acceleration, final acceleration and velocity gain.

  2. Compare with using the initial acceleration for the whole burn.

Question 3 - Solution

Constant thrust gives increasing acceleration

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Mass and thrust during the burn are m(t)=1000−5t,T=uq=1800(5)=9000N.m(t)=1000-5t,\qquad\boxed{T=uq=1800(5)=9000\ \mathrm N}. ai=9000/1000=9.0,af=9000/800=11.25m/s2.\boxed{a_i=9000/1000=9.0},\qquad\boxed{a_f=9000/800=11.25\ \mathrm{m/s^2}}. Δv=1800ln⁡(1000/800)=401.66m/s.\boxed{\Delta v=1800\ln(1000/800)=401.66\ \mathrm{m/s}}.

(b) Result and interpretation

Holding acceleration at its initial value would give Δvest=9(40)=360m/s,\Delta v_{\mathrm{est}}=9(40)=360\ \mathrm{m/s}, Δv−Δvest=41.66m/s.\boxed{\Delta v-\Delta v_{\mathrm{est}}=41.66\ \mathrm{m/s}}. Because a(t)=9000/(1000−5t)a(t)=9000/(1000-5t) rises monotonically, the initial-acceleration estimate is a lower bound.

Physical check

The exact velocity gain lies between 9(40)=3609(40)=360 and 11.25(40)=45011.25(40)=450 m/s.