Rocket Propellant Mass Flow Rate (Aerospace Engineering) Solution

Problem Overview

We are given a rocket motor mass profile $M(t) = 800 - 12t^2\text{ kg}$ . We calculate $\frac{dM}{dt}$ using the limit definition and determine the instantaneous mass flow rate and thrust at $t = 5\text{ s}$ .

Step-by-Step Mathematical Solution

Step 1: Set up the Difference Quotient Limit

\frac{dM}{dt} = \lim_{h \to 0} \frac{M(t+h) - M(t)}{h} = \lim_{h \to 0} \frac{[800 - 12(t+h)^2] - (800 - 12t^2)}{h}

Step 2: Expand the Binomial $(t+h)^2 = t^2 + 2th + h^2$

\frac{dM}{dt} = \lim_{h \to 0} \frac{800 - 12(t^2 + 2th + h^2) - 800 + 12t^2}{h}

\frac{dM}{dt} = \lim_{h \to 0} \frac{800 - 12t^2 - 24th - 12h^2 - 800 + 12t^2}{h}

Step 3: Cancel Non- $h$ Terms and Factor out $h$

\frac{dM}{dt} = \lim_{h \to 0} \frac{-24th - 12h^2}{h} = \lim_{h \to 0} \frac{h(-24t - 12h)}{h}

\frac{dM}{dt} = \lim_{h \to 0} (-24t - 12h)

Step 4: Direct Substitution ( $h = 0$ )

\frac{dM}{dt} = -24t - 12(0) = -24t\text{ kg/s}

Step 5: Evaluate at $t = 5\text{ seconds}$

M'(5) = -24(5) = -120\text{ kg/s}

The mass flow rate is

\dot{m} = \left|\frac{dM}{dt}\right| = 120\text{ kg/s}

(the negative sign indicates system mass loss).

Step 6: Compute Rocket Thrust

F_{\text{thrust}} = c \cdot \dot{m} = 2500\text{ m/s} \times 120\text{ kg/s} = 300{,}000\text{ N} = 300\text{ kN}

Engineering Conclusion

At $t = 5\text{ seconds}$ , propellant mass is exiting the nozzle throat at $120\text{ kg/s}$ , generating a reaction thrust of $300\text{ kN}$ ( $67{,}440\text{ lbf}$ ). Tracking instantaneous $\frac{dM}{dt}$ through calculus allows aerospace propulsion engineers to predict vehicle acceleration curves and verify structural stage separation timing.