Systems with Varying Mass: A Rocket — Question 2

Section 9-11 · Physics pages 1354–1355

Question 2 Mass ratio grows exponentially with required velocity change An ideal rocket in force-free space has constant relative exhaust speed u = 2000 m/s. Its final mass after a burn is 500 kg. It must gain 1000 m/s along a fixed direction. mi/mf ∆v/u 1 e 1 Logarithmic ideal velocity gain; the physical domain starts at mass ratio 1. (a) Find its initial mass and expelled propellant mass. (b) If the required gain doubles, does the propellant mass simply double?
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Question 2 - Solution Mass ratio grows exponentially with required velocity change mi/mf ∆v/u 1 e 1 (a) Derivation and calculation Integrate the differential momentum equation: dv = −udm m ⇒∆v = u ln mi mf . mi = mfe∆v/u = 500e0.5 = 824.36 kg , mp = mi −mf = 324.36 kg . (b) Result and interpretation Doubling the required velocity gain gives m′ p = 500(e2000/2000 −1) = 859.14 kg . m′ p > 2mp = 648.72 kg. The required mass ratio grows exponentially with velocity gain, so propellant does not scale linearly. Physical check The final 500 kg must be accelerated throughout the burn.

Question 2

Mass ratio grows exponentially with required velocity change

An ideal rocket in force-free space has constant relative exhaust speed u=2000m/su=2000\,\mathrm{m/s}. Its final mass after a burn is 500kg500\,\mathrm{kg}. It must gain 1000m/s1000\,\mathrm{m/s} along a fixed direction.

Diagram shown in the original SVG worksheet.

  1. Find its initial mass and expelled propellant mass.

  2. If the required gain doubles, does the propellant mass simply double?

Question 2 - Solution

Mass ratio grows exponentially with required velocity change

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Integrate the differential momentum equation: dv=−udmm⇒Δv=uln⁡mimf.dv=-u\frac{dm}{m}\Rightarrow\Delta v=u\ln\frac{m_i}{m_f}. mi=mfeΔv/u=500e0.5=824.36kg,\boxed{m_i=m_fe^{\Delta v/u}=500e^{0.5}=824.36\ \mathrm{kg}}, mp=mi−mf=324.36kg.\boxed{m_p=m_i-m_f=324.36\ \mathrm{kg}}.

(b) Result and interpretation

Doubling the required velocity gain gives mp′=500(e2000/2000−1)=859.14kg.m_p'=500(e^{2000/2000}-1)=\boxed{859.14\ \mathrm{kg}}. mp′>2mp=648.72kg.m_p'>2m_p=648.72\ \mathrm{kg}. The required mass ratio grows exponentially with velocity gain, so propellant does not scale linearly.

Physical check

The final 500 kg must be accelerated throughout the burn.