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Rocket Velocity Optimization via Rocket Equation (Aerospace Engineering) Solution

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Problem Overview

We determine the instantaneous acceleration dvdt\frac{dv}{dt} of a launch vehicle governed by the Tsiolkovsky Rocket Equation with constant fuel mass burn rate:

v(m)=2500ln⁡(50,000m),m(t)=50,000−250tv(m) = 2500\ln\left(\frac{50,000}{m}\right), \qquad m(t) = 50,000 - 250t

Step-by-Step Mathematical Solution

Step 1: Pre-Process via Logarithmic Properties

Using the Quotient Property of logarithms (ln⁡(A/B)=ln⁡A−ln⁡B\ln(A/B) = \ln A - \ln B):

v(m)=2500[ln⁡(50,000)−ln⁡(m)]=2500ln⁡(50,000)−2500ln⁡(m)v(m) = 2500\left[\ln(50,000) - \ln(m)\right] = 2500\ln(50,000) - 2500\ln(m)
This converts a fractional argument into two separate terms, eliminating the need for the Quotient Rule.

Step 2: Differentiate Components

  1. Velocity with respect to mass dvdm\frac{dv}{dm}: The first term 2500ln⁡(50,000)2500\ln(50,000) is a constant scalar, so its derivative is zero:
    dvdm=0−2500(1m)=−2500m (m/s)/kg\frac{dv}{dm} = 0 - 2500\left(\frac{1}{m}\right) = -\frac{2500}{m}\text{ (m/s)/kg}
  2. Mass with respect to time dmdt\frac{dm}{dt}:
    dmdt=ddt[50,000−250t]=−250 kg/s\frac{dm}{dt} = \frac{d}{dt}[50,000 - 250t] = -250\text{ kg/s}

Step 3: Link Rates via Leibniz Notation (Instantaneous Acceleration)

dvdt=dvdm⋅dmdt=(−2500m)⋅(−250)=625,000m m/s2\frac{dv}{dt} = \frac{dv}{dm} \cdot \frac{dm}{dt} = \left(-\frac{2500}{m}\right) \cdot (-250) = \frac{625,000}{m}\text{ m/s}^2

Step 4: Evaluate Operating Conditions at t=60 secondst = 60\text{ seconds}

  1. Instantaneous rocket mass m(60)m(60):
    m(60)=50,000−250(60)=50,000−15,000=35,000 kgm(60) = 50,000 - 250(60) = 50,000 - 15,000 = 35,000\text{ kg}
  2. Instantaneous acceleration dvdt\frac{dv}{dt}:
    dvdt∣t=60=625,00035,000=62535=1257≈17.857 m/s2\frac{dv}{dt}\Big|_{t=60} = \frac{625,000}{35,000} = \frac{625}{35} = \frac{125}{7} \approx 17.857\text{ m/s}^2

Engineering Conclusion

At t=60 secondst = 60\text{ seconds}, the launch vehicle accelerates at approximately 17.86 m/s217.86\text{ m/s}^2 (roughly 1.82 g1.82\text{ g}). Even though the rocket motor burns propellant at a constant flow rate (dmdt=−250 kg/s\frac{dm}{dt} = -250\text{ kg/s}), vehicle acceleration increases over time because total mass m(t)m(t) in the denominator decreases. Aerospace propulsion engineers use this logarithmic derivative relationship to ensure structural gg-load safety limits are never breached during ascent.