Problem 64 - University Physics with Modern Physics

Question

Obviously, we can make rockets to go very fast, but what is a reasonable top speed? Assume that a rocket is fired from rest at a space station in deep space, where gravity is negligible. (a) If the rocket ejects gas at a relative speed of 2000 \(\mathrm{m} / \mathrm{s}\) and you want the rocket's speed eventually to be \(1.00 \times 10^{-3} \mathrm{c}\) , where \(c\) is the speed of light, what fraction of the initial mass of the rocket and fuel is not fuel? (b) What is this fraction if the final speed is to be 3000 \(\mathrm{m} / \mathrm{s} ?\)

Step-by-Step Solution

Answer

Part (a): Unfeasible with chemical rockets. Part (b): 77.7% of the mass is not fuel.

Step 1: Understand the Problem

We are dealing with a rocket problem involving its speed and mass. Part (a) asks us to calculate the mass fraction for a speed of \(1.00 \times 10^{-3} c \), while part (b) requires the mass fraction for a speed of 3000 \( \mathrm{m/s} \).

Step 2: Apply the Rocket Equation

The equation we use is \( v_f = v_e \ln\left( \frac{m_i}{m_f} \right) \), where \( v_f \) is the final velocity, \( v_e \) is the effective exhaust velocity, \( m_i \) is the initial total mass, and \( m_f \) is the final mass. We'll solve for \( \frac{m_f}{m_i} \).

Step 3: Calculate Fraction for Part (a)

Given \( v_f = 1.00 \times 10^{-3} c = 1.00 \times 10^{-3} \times 3.00 \times 10^8 \text{ m/s} = 3.00 \times 10^5 \text{ m/s} \), and \( v_e = 2000 \text{ m/s} \):\[ 3.00 \times 10^5 = 2000 \ln\left( \frac{m_i}{m_f} \right) \]Solve for \( \ln\left( \frac{m_i}{m_f} \right) = \frac{3.00 \times 10^5}{2000} = 150 \), thus \( \frac{m_i}{m_f} = e^{150} \). Calculate \( \frac{m_f}{m_i} \) and subtract from 1 for the fuel fraction.

Step 4: Estimate for Part (a)

Because \( e^{150} \) is impractically large, this implies it is not feasible for current technology to reach such speeds with chemical rockets using this effective exhaust velocity.

Step 5: Calculate Fraction for Part (b)

For part (b), \( v_f = 3000 \text{ m/s} \), and continue using \( v_e = 2000 \text{ m/s} \):\[ 3000 = 2000 \ln\left( \frac{m_i}{m_f} \right) \]Solve for \( \ln\left( \frac{m_i}{m_f} \right) = 1.5 \), thus \( \frac{m_i}{m_f} = e^{1.5} \approx 4.48 \).

Step 6: Calculate Mass Fraction for Part (b)

So, \( \frac{m_f}{m_i} = \frac{1}{4.48} \approx 0.223 \) and the non-fuel mass fraction is \( 1 - 0.223 \approx 0.777 \).

Step 7: Conclude Solution

For part (a), the required energy for achieving the high speed is unfeasible with chemical propulsion. For part (b), about 77.7% of the initial mass is not fuel.

Key Concepts

Effective Exhaust Velocity

The effective exhaust velocity (\(v_e\)) is a crucial factor in determining how fast a rocket can travel. It represents the speed at which exhaust gases are expelled from the rocket engine relative to the rocket itself. This parameter directly influences the rocket's ability to change momentum, which in turn affects its final velocity.

- **Definition**: It is the speed of gas exit relative to the rocket.- **Importance**: Higher effective exhaust velocity means greater efficiency as more thrust is produced per unit of fuel.

In the exercise, we are given a constant effective exhaust velocity of 2000 \(\text{m/s}\). This informs us about the propulsion efficiency of the rocket design being considered. Having a high effective exhaust velocity allows the rocket to achieve greater final velocities given the same amount of fuel.

Mass Fraction

The mass fraction in rocketry describes the proportion of the initial total mass that is made up of fuel. This is an essential concept when calculating how much of the rocket's mass is consumed to achieve a certain velocity.

The **Rocket Equation**:
\[v_f = v_e \ln\left( \frac{m_i}{m_f} \right)\]directly relates the mass fraction to the final velocity (\(v_f\)) and the effective exhaust velocity (\(v_e\)). Here:

• \(m_i\) is the initial mass (rocket + fuel).
• \(m_f\) is the final mass (rocket without burned fuel).

In the exercise for part (b), we have computed that about 77.7% of the mass is non-fuel. Understanding mass fraction helps in designing rockets that maximize the payload while providing sufficient fuel for reaching desired speeds.

Chemical Propulsion

Chemical propulsion is the most common method to propel rockets, utilizing chemical reactions to produce thrust. It generally relies on the combustion of propellant (fuel and oxidizer), and the resulting gas expansion pushes the rocket forward.

- **Mechanism**: Combustion inside the rocket engine produces high-pressure, high-temperature gas ejected through the rocket's nozzle.- **Advantages**: Well-understood technology with established engineering practices.- **Limitations**: Limited effective exhaust velocity and the need for carrying oxidizer restrict the maximum achievable speeds.

In the context of the exercise, attempting to reach extremely high velocities, like a fraction of the speed of light (\(c\)), highlights the constraints of chemical propulsion systems. They simply cannot provide the necessary energy-to-thrust ratio due to physical and material limitations.

Final Velocity

Final velocity (\(v_f\)) is the ultimate speed reached by the rocket after all fuel has been expended. Achieving the desired final velocity depends on several factors including the effective exhaust velocity and the mass fraction.

- **Determining Factors**: - Effective exhaust velocity (\(v_e\)) - Mass fraction (\(m_f/m_i\))
- **Rocket Equation Application**: Used to calculate the final velocity achievable by comparing the initial and final masses of the rocket.

In the exercise, two scenarios are explored:

• Scenario (a) targets a final velocity of 0.1% of the speed of light (\(1.00 \times 10^{-3} c\)), highlighting the infeasibility with chemical propulsion.
• Scenario (b) considers a final velocity of 3000 \(\text{m/s}\), showing a feasible non-fuel mass fraction.

Understanding final velocity is pivotal for mission planning, ensuring that the rocket can reach the required speed for its intended purpose.