91Ó°ÊÓ

We are given the thrust force \(F = 53.2 \times 10^{6}\,\text{N}\) and the propellant velocity \(v_{propellant} = 4.78 \times 10^{3}\,\text{m/s}\). To calculate the mass flow rate \(\frac{dm}{dt}\), we'll make use of the second law relationship: $$F = \frac{d(mv)}{dt}$$ Here, \(mv\) represents the change in momentum of the rocket. Since the propellant velocity is constant, we can write this equation as: $$F = v_{propellant} \times \frac{dm}{dt}$$ Solving for the mass flow rate \(\frac{dm}{dt}\), we have: $$\frac{dm}{dt} = \frac{F}{v_{propellant}} = \frac{53.2 \times 10^{6}\,\text{N}}{4.78 \times 10^{3}\,\text{m/s}} \approx 11.13\times 10^{3} \,\text{kg/s}$$

Now we'll make use of the conservation of momentum and the Rocket Propulsion Equation to calculate the final speed of the spacecraft (\(v_f\)). The equation is given by: $$v_{f} = v_{propellant} \ln \left(\frac{m_{initial}}{m_{final}}\right)$$ Plugging in the values given, we obtain: $$v_{f} = 4780\, \text{m/s} \times \ln \left(\frac{2.12 \times 10^6\, \text{kg}}{7.04 \times 10^4\, \text{kg}}\right) \approx 17,306\, \text{m/s}$$ So the final speed of the spacecraft is \(17,306\, \text{m/s}\).

Now we'll calculate the average acceleration of the spacecraft until burnout, assuming a constant mass flow rate. Since the mass flow rate is constant, we can use the following equation to calculate the burnout time \(t_{burnout}\): $$t_{burnout} = \frac{\Delta m}{\frac{dm}{dt}} = \frac{2.12 \times 10^6\,\text{kg} - 7.04 \times 10^4\,\text{kg}}{11.13\times 10^{3} \,\text{kg/s}} \approx 178.35\, \text{s}$$ To calculate the average acceleration (\(a_{avg}\)), we'll consider the initial and final velocities (\(v_{i} = 0\,\text{m/s}\) and \(v_{f} = 17,306\, \text{m/s}\)) and the burnout time, and then use the equation: $$a_{avg} = \frac{v_{f} - v_{i}}{t_{burnout}} = \frac{17,306\,\text{m/s}}{178.35\,\text{s}} \approx 97.03\, \text{m/s}^{2}$$ So the average acceleration till burnout is about \(97.03\, \text{m/s}^{2}\).