The Saturn \(V\) rocket, which was used to launch the Apollo spacecraft on their way to the Moon, has an initial mass \(M_{0}=2.80 \cdot 10^{6} \mathrm{~kg}\) and a final mass \(M_{1}=8.00 \cdot 10^{5} \mathrm{~kg}\) and burns fuel at a constant rate for \(160 .\) s. The speed of the exhaust relative to the rocket is about \(v=2700 . \mathrm{m} / \mathrm{s}\). a) Find the upward acceleration of the rocket, as it lifts off the launch pad (while its mass is the initial mass). b) Find the upward acceleration of the rocket, just as it finishes burning its fuel (when its mass is the final mass). c) If the same rocket were fired in deep space, where there is negligible gravitational force, what would be the net change in the speed of the rocket during the time it was burning fuel?

Initially, the rocket has a mass \(M_0\). While lifting off the launch pad, it will experience two forces: the weight (downward) which is \(W = M_0g\), and the thrust force (upward) which is \(T = F_{ex} = (\dot{m}v)\), where \(\dot{m}\) is the mass flow rate of the exhaust and \(v\) is the exhaust speed relative to the rocket. First, we need to find the mass flow rate, \(\dot{m}\). From the initial mass \(M_0\) and final mass \(M_1\), we know that the rocket loses \(M_0 - M_1 = 2.80 \cdot 10^6 - 8.00 \cdot 10^5 = 2.00 \cdot 10^6\,\mathrm{kg}\) in fuel. Since it takes 160 seconds to burn this fuel, we can determine the mass flow rate: \(\dot{m} = \frac{2.00\cdot 10^6 \mathrm{~kg}}{160\,\mathrm{s}} = 1.25\cdot 10^4\,\mathrm{kg/s}\) Next, we'll find the total force acting on the rocket. The thrust force \(F_{ex}\) is equal to the rate of the momentum change of the exhaust, which can be determined as: \(F_{ex} = (\dot{m}v) = (1.25\cdot 10^4\,\mathrm{kg/s})(2700\,\mathrm{m/s}) = 3.38\cdot 10^7\,\mathrm{N}\) Now, we'll calculate the gravitational force acting on the rocket at lift-off: \(W = M_0g = (2.80\cdot 10^6\,\mathrm{kg})(9.81\,\mathrm{m/s^2}) = 2.74\cdot 10^7\,\mathrm{N}\) Applying Newton's second law, the net force acting on the rocket is \(F_{net} = F_{ex} - W\). The acceleration due to this force can be determined as follows: \(a = \frac{F_{net}}{M_0} = \frac{(3.38\cdot 10^7 - 2.74\cdot 10^7)\,\mathrm{N}}{2.80\cdot 10^6\,\mathrm{kg}} \approx 2.29\,\mathrm{m/s^2}\) Thus, the upward acceleration of the rocket at lift-off is \(2.29\,\mathrm{m/s^2}\).

When the rocket finishes burning its fuel, the mass is now \(M_1\). The gravitational force acting on the rocket is: \(W = M_1g = (8.00 \cdot 10^5\,\mathrm{kg})(9.81\,\mathrm{m/s^2}) = 7.85\cdot 10^6\,\mathrm{N}\) The mass flow rate remains constant, so the thrust force is still \(F_{ex} = 3.38\cdot 10^7\,\mathrm{N}\). Applying Newton's second law, the net force acting on the rocket is again \(F_{net} = F_{ex} - W\). The acceleration due to this net force is: \(a = \frac{F_{net}}{M_1} = \frac{(3.38\cdot 10^7 - 7.85\cdot 10^6)\,\mathrm{N}}{8.00\cdot 10^5\,\mathrm{kg}} \approx 30.39\,\mathrm{m/s^2}\) Thus, the upward acceleration of the rocket just after it finishes burning its fuel is \(30.39\,\mathrm{m/s^2}\).

In deep space, there is no considerable gravitational force. Therefore, the only force acting on the rocket is the thrust force, \(F_{ex}\). We can find the net change in speed by integrating the acceleration due to the thrust force with respect to time, from \(0\) to the time when the rocket stops burning its fuel, \(t_f = 160\,\text{s}\): \(\Delta v = \int_0^{t_f} a(t) dt\) Since the mass flow rate is constant, we can assume that the acceleration is constant. Then, \(\Delta v = a \cdot t_f = a \cdot 160\,\text{s}\) Since the acceleration is the result of the thrust force acting on the varying mass, we have: \(a = \frac{F_{ex}}{M(t)}\), where \(M(t)\) is the mass of the rocket at time \(t\). As the mass is decreasing linearly, we can find the average mass during the burning: \(M_{avg} = \frac{M_0 + M_1}{2} = \frac{2.80 \cdot 10^6 + 8.00 \cdot 10^5}{2} \mathrm{kg} = 1.80\cdot 10^6\,\mathrm{kg}\) Now we can find the average acceleration: \(a_{avg} = \frac{F_{ex}}{M_{avg}} = \frac{3.38\cdot 10^7\,\mathrm{N}}{1.80\cdot 10^6\,\mathrm{kg}} \approx 18.82\,\mathrm{m/s^2}\) The net change in speed can now be calculated: \(\Delta v = a_{avg}\cdot t_f = (18.82\,\mathrm{m/s^2})(160\,\mathrm{s}) = 3009.60\,\mathrm{m/s}\) Therefore, the net change in speed of the rocket in deep space during the time it was burning fuel is \(3009.60\,\mathrm{m/s}\).