Question

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?

Short answer

The upward acceleration of the rocket at lift-off is 2.29 m/s², and the upward acceleration after finishing burning its fuel is 30.39 m/s². The net change in speed of the rocket in deep space during the time it was burning fuel is 3009.60 m/s.

Step-by-step solution

a) Finding the upward acceleration when the rocket lifts off

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 \(2.29\,\mathrm{m/s^2}\).

b) Finding the upward acceleration when the rocket finishes burning its fuel

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 \(30.39\,\mathrm{m/s^2}\).

c) Net change in speed of the rocket in deep space

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 \(3009.60\,\mathrm{m/s}\).

Key concepts

Newton's Second Law

One of the fundamental principles of rocket physics is Newton's second law, which states that the force acting on an object is equal to the rate of change of its momentum. Mathematically, this is expressed as \( F = ma \), where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration. In the context of rockets, this law explains how the change in momentum due to the expulsion of exhaust gases results in the rocket's movement. The key insight is that as the rocket ejects mass in one direction (via exhaust), it propels itself in the opposite direction. This reciprocal action is a direct illustration of Newton's second law in action. For rockets, understanding this law is crucial to determining their acceleration and velocity changes during flight.

Thrust Force

Thrust force is the "push" experienced by a rocket due to the rapid ejection of exhaust gases. It is the primary force that propels the rocket upward and is generated by the engine burning fuel. Thrust can be quantified using the formula \( F_{thrust} = \dot{m}v \), where \( \dot{m} \) is the mass flow rate of fuel, and \( v \) is the exhaust velocity.

• This formula highlights two essential facets controlling thrust: the rate of fuel burned and the speed at which gases leave the engine.
• Thus, the greater the mass flow rate or the exhaust speed, the higher the thrust.

Rockets are designed to optimize both of these parameters, ensuring efficient use of fuel and achieving the desired lift-off thrust necessary to overcome gravitational forces.

Mass Flow Rate

Mass flow rate refers to the rate at which fuel is consumed and expelled by the rocket engine. It is a critical variable because it directly affects the thrust and, consequently, the rocket's acceleration. The mass flow rate, denoted as \( \dot{m} \), can be calculated as \( \dot{m} = \frac{M_0 - M_1}{t} \), where \( M_0 \) is the initial mass, \( M_1 \) is the final mass, and \( t \) is the time duration of fuel burning.

• In the Saturn V scenario, the mass flow rate determines how quickly the rocket can generate the necessary thrust to overcome gravitational and other resisting forces.
• A high mass flow rate would result in a rapid decrease in mass, increasing acceleration as the reduction in mass means less force is needed to accelerate the rocket.

Gravitational Force

Gravitational force, often simply referred to as "gravity," is the force exerted by the Earth on the rocket, pulling it downward. This is calculated as \( F_{gravity} = mg \), where \( m \) is the mass of the rocket and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \mathrm{m/s^2} \) on Earth's surface). During launch, the rocket must generate sufficient thrust to not only lift itself but also to overcome this gravitational pull.

• A larger gravitational force implies a greater thrust requirement for achieving lift-off.
• As the rocket consumes fuel, its mass decreases, which in turn reduces the gravitational force acting on it, allowing for higher acceleration even when thrust remains constant.

Understanding gravitational force is vital in computing the net force acting on the rocket and its subsequent acceleration.

Exhaust Velocity

Exhaust velocity is the speed at which exhaust gases are expelled from the rocket engine. It is a crucial factor in determining the rocket's efficiency and performance. A higher exhaust velocity means the rocket will have greater thrust for a given mass flow rate, as calculated in the thrust formula \( F_{thrust} = \dot{m}v \).

• The concept of exhaust velocity is tied to the efficiency of the engine itself; engines designed to expel gases at higher speeds can achieve greater propulsion.
• In practical terms, maximizing the exhaust velocity allows rockets to gain higher speeds and improve overall efficiency during space travel by more effectively converting the chemical energy of the fuel into kinetic energy.

Therefore, achieving a high exhaust velocity is one of the key goals in rocket engine design, as it directly impacts the rocket's ability to reach and escape orbit.