Systems with Varying Mass: A Rocket — Question 6

Section 9-11 · Physics pages 1362–1363

Question 6 The ideal speed gain can ignore the throttle history Two identical rockets in force-free space burn from mi = 600 to mf = 400 kg with fixed exhaust direction and constant relative exhaust speed u = 1500 m/s. One burns steadily; the other varies its positive mass flow. +x exhaust m(t) Ideal one-dimensional rocket; exhaust is directed backward relative to the rocket. (a) Compare their velocity gains. (b) Explain which aspects of their motion need not agree, and why adding gravity changes the comparison.
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Question 6 - Solution The ideal speed gain can ignore the throttle history +x exhaust m(t) (a) Derivation and calculation For either positive throttle history, dm/dt = −q(t) and dv = uq(t) m(t) dt = −udm m . ∆v = 1500 ∫600 400 dm m = 1500 ln 1.5 = 608.20 m/s . Thus both obtain the same velocity gain in force-free space. (b) Result and interpretation Their instantaneous accelerations and displacements depend on the history: a(t) = uq(t) m(t) , ∆x = ∫tb 0 v(t) dt. In vertical uniform gravity, ∆v = u ln(mi/mf) −gtb. Different burn durations then generally produce different final velocities. Physical check Endpoint independence requires constant relative exhaust speed, fixed direction, and zero external force.

Question 6

The ideal speed gain can ignore the throttle history

Two identical rockets in force-free space burn from mi=600m_i=600 to mf=400kgm_f=400\,\mathrm{kg} with fixed exhaust direction and constant relative exhaust speed u=1500m/su=1500\,\mathrm{m/s}. One burns steadily; the other varies its positive mass flow.

Diagram shown in the original SVG worksheet.

  1. Compare their velocity gains.

  2. Explain which aspects of their motion need not agree, and why adding gravity changes the comparison.

Question 6 - Solution

The ideal speed gain can ignore the throttle history

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

For either positive throttle history, dm/dt=−q(t)dm/dt=-q(t) and dv=uq(t)m(t)dt=−udmm.dv=\frac{uq(t)}{m(t)}dt=-u\frac{dm}{m}. Δv=1500∫400600dmm=1500ln⁡1.5=608.20m/s.\boxed{\Delta v=1500\int_{400}^{600}\frac{dm}{m}=1500\ln1.5=608.20\ \mathrm{m/s}}. Thus both obtain the same velocity gain in force-free space.

(b) Result and interpretation

Their instantaneous accelerations and displacements depend on the history: a(t)=uq(t)m(t),Δx=∫0tbv(t)dt.a(t)=\frac{uq(t)}{m(t)},\qquad\Delta x=\int_0^{t_b}v(t)\,dt. In vertical uniform gravity, Δv=uln⁡(mi/mf)−gtb.\Delta v=u\ln(m_i/m_f)-gt_b. Different burn durations then generally produce different final velocities.

Physical check

Endpoint independence requires constant relative exhaust speed, fixed direction, and zero external force.