Rocket Velocity Optimization via Rocket Equation (Aerospace Engineering) Solution

Problem Overview

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

Step-by-Step Mathematical Solution

Step 1: Pre-Process via Logarithmic Properties

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

Step 2: Differentiate Components

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

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

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

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

Engineering Conclusion

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