Question

A spacecraft engine creates \(53.2 \mathrm{MN}\) of thrust with a propellant velocity of \(4.78 \mathrm{~km} / \mathrm{s}\). a) Find the rate \((d m / d t)\) at which the propellant is expelled. b) If the initial mass is \(2.12 \cdot 10^{6} \mathrm{~kg}\) and the final mass is \(7.04 \cdot 10^{4} \mathrm{~kg},\) find the final speed of the spacecraft (assume the initial speed is zero and any gravitational fields are small enough to be ignored). c) Find the average acceleration till burnout (the time at which the propellant is used up; assume the mass flow rate is constant until that time).

Short answer

Question: Calculate the rate at which the propellant is expelled, the final speed of the spacecraft, and the average acceleration until burnout for a given thrust force of \(53.2 \times 10^{6}\,\text{N}\), propellant velocity of \(4.78 \times 10^{3}\,\text{m/s}\), initial mass of \(2.12 \times 10^6\, \text{kg}\), and final mass of \(7.04 \times 10^4\, \text{kg}\). Answer: The rate at which the propellant is expelled is approximately \(11.13\times 10^{3} \,\text{kg/s}\), the final speed of the spacecraft is about \(17,306\, \text{m/s}\), and the average acceleration until burnout is roughly \(97.03\, \text{m/s}^{2}\).

Step-by-step solution

Calculating the rate at which the propellant is expelled

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: $$\frac{dm}{dt} = \frac{F}{v_{propellant}} = \frac{53.2 \times 10^{6}\,\text{N}}{4.78 \times 10^{3}\,\text{m/s}} \approx 11.13\times 10^{3} \,\text{kg/s}$$

Calculate the final speed of the spacecraft

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 \(17,306\, \text{m/s}\).

Calculate the average acceleration till burnout

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 \(97.03\, \text{m/s}^{2}\).

Key concepts

Thrust Force

In simple terms, thrust force is what propels the rocket forward. It is like the push that makes the rocket move. This force is a result of the high-speed expulsion of gases from the rocket engine. The greater the thrust, the faster the rocket can go. In our exercise, the rocket's engine provides a thrust force of 53.2 million Newtons (MN).
Thrust force is calculated using the relation:

• \( F = \frac{d(mv)}{dt} \), where \( mv \) is the momentum of the rocket.

Since the velocity at which the gases (or propellant) are expelled is constant, we can simplify the equation to \( F = v_{propellant} \times \frac{dm}{dt} \). This means thrust force is a product of the exhaust velocity and the rate at which the rocket burns fuel. Understanding this concept is crucial in determining how efficiently a rocket can travel in space.

Propellant Velocity

Propellant velocity, also known as exhaust velocity, is the speed at which gases are expelled from the rocket engine. It's a key factor in determining the rocket's efficiency. The higher the velocity, the more efficient the rocket is at converting fuel into forward motion.
In our scenario, the propellant is expelled at a velocity of 4.78 km/s (or 4780 m/s).
To understand why this matters, think of it this way:

• High propellant velocity means a larger change in momentum per unit of fuel.
• This results in more thrust for less fuel, making the rocket more efficient.
  

In a practical sense, a high propellant velocity helps spacecraft achieve greater speeds while conserving fuel, which is essential for long missions where carrying extra fuel is not feasible.

Conservation of Momentum

The principle of conservation of momentum is fundamental in rocket propulsion. It states that the total momentum of a system remains constant if no external forces are acting on it. For rockets, this principle is applied during fuel expulsion.
As the rocket expels gas out of its engine at high speed, it gains forward momentum in the opposite direction. This is due to the symmetrical nature of momentum exchange:

• When the mass of expelled gases increases their velocity, the rocket itself experiences an equal and opposite change in momentum.
• This behavior is explained by Newton’s third law: every action has an equal and opposite reaction.
  

Understanding this principle helps us predict the movement of rockets based on the speed and mass of the expelled propellant.

Rocket Equation

The rocket equation, also known as Tsiolkovsky's rocket equation, is a vital formula used to calculate a rocket's final velocity. It relates the velocity change to the exhaust velocity and the mass ratio of the rocket.
The equation is:

• \( v_{f} = v_{propellant} \ln \left(\frac{m_{initial}}{m_{final}}\right) \)

This tells us that the final speed of a rocket depends on:

• How fast the propellant is expelled: higher \( v_{propellant} \) leads to higher final speed.
• The initial and final mass of the rocket: a larger mass ratio results in greater speed.
  

In the given exercise, the initial mass is 2.12 million kg, and the final mass is 70,400 kg. By plugging these values into the rocket equation, the calculated final speed was approximately 17,306 meters per second. This equation is significant for planning space missions, determining how much fuel a spacecraft needs to achieve desired speeds, and defining payload capacities.