Question

To illustrate the use of a multistage rocket consider the following: (a) A certain rocket carries \(\left.60 \% \text { of its initial mass as fuel. (That is, the mass of fuel is } 0.6 m_{\mathrm{o}} .\right)\) What is the rocket's final speed, accelerating from rest in free space, if it burns all its fuel in a single stage? Express your answer as a multiple of \(v_{\mathrm{ex}} .\) (b) Suppose instead it burns the fuel in two stages as follows: In the first stage it burns a mass \(0.3 m_{\mathrm{o}}\) of fuel. It then jettisons the first-stage fuel tank, which has a mass of \(0.1 m_{\mathrm{o}}\), and then burns the remaining \(0.3 m_{\mathrm{o}}\) of fuel. Find the final speed in this case, assuming the same value of \(v_{\mathrm{ex}}\) throughout, and compare.

Short answer

In one stage: \( v_f = v_{ex} \ln(2.5) \). In two stages: not calculated fully yet.

Step-by-step solution

Calculate final speed in one stage

For a single-stage rocket that starts from rest and uses all its fuel, the rocket equation is given by: \[ v_f = v_{ex} \ln\left(\frac{m_0}{m_0 - m_f}\right) \] where \( m_f \) is the mass of the fuel. In this case, \( m_f = 0.6m_0 \), thus the final speed is: \[ v_{f1} = v_{ex} \ln\left(\frac{m_0}{0.4m_0}\right) = v_{ex} \ln\left(2.5\right) \] This is the final speed when the fuel is burned in a single stage.

Calculate speed after first stage

In the first stage of two-stage operation, the rocket burns \( 0.3 m_{0} \) of fuel. The relevant equation for this stage is: \[ v_{1} = v_{ex} \ln\left(\frac{m_0}{m_0 - 0.3m_0}\right) \] which simplifies to: \[ v_{1} = v_{ex} \ln\left(\frac{m_0}{0.7m_0}\right) = v_{ex} \ln\left(1.4286\right) \]

Key concepts

Rocket Equation

The rocket equation, also known as the Tsiolkovsky rocket equation, is an essential formula for calculating a rocket's velocity after burning a certain amount of fuel. It encapsulates the relationship between the velocity, the mass of the rocket, and the velocity of the expelled gas. The equation is given by: \ \( v_f = v_{ex} \ln\left(\frac{m_0}{m_f}\right) \) \ where

• \( v_f \) is the final velocity of the rocket,
• \( v_{ex} \) is the exhaust velocity or the speed at which gas leaves the rocket,
• \( m_0 \) is the initial total mass of the rocket (including fuel), and
• \( m_f \) is the final mass after fuel is burned.

Understanding the rocket equation is crucial for designing efficient propulsion systems. It shows how the expulsion of mass (fuel) at high speeds propels the rocket forward. By maximizing the exhaust velocity and optimizing fuel mass, a rocket can achieve greater final speeds.

Final Speed

Final speed in rocketry refers to the velocity a rocket achieves once it has expended its designated fuel. Using the rocket equation, we determine this speed from the mass of the fuel burned and the initial mass of the rocket.For a rocket starting from rest and burning all its fuel in a single stage, the final speed is given by \ \( v_{f1} = v_{ex} \ln\left(2.5\right) \), where 60% of the initial mass is fuel. This reflects how much velocity can be gained by burning all the fuel continuously.In contrast, by staging the burn, as in a two-stage operation, you can achieve a higher final speed since each stage optimizes the remaining mass after spending fuel, reducing the weight while maintaining a high energy output. This approach effectively maximizes the velocity gained from the remaining fuel.

Fuel Mass

Fuel mass significantly influences the effectiveness of a rocket's thrust. In the context of the rocket equation, it's represented as a fraction of the initial mass. For instance, if \( 60\% \) of the initial mass \( m_0 \) is fuel, then \( m_f = 0.6 m_0 \).Fuel mass is crucial because:

• More fuel allows for a longer burn time, providing more thrust and, hence, a higher velocity.
• A larger fuel mass means increased total weight, making the initial stages of flight slower until most of the fuel is consumed.
• Optimizing fuel mass is vital in multi-stage operations where fuel efficiency is enhanced by jettisoning empty tanks, reducing mass radically and enhancing speed.

Harmonizing the amount of fuel with the structure and mission goals of the rocket ensures optimal performance.

Two-Stage Operation

Two-stage operation harnesses the benefit of staging, a design strategy where the rocket sheds parts of its mass at different stages during its ascent. This approach significantly enhances the rocket's efficiency and final velocity.For a rocket using a two-stage operation:

• In the first stage, some fuel is burned \( 0.3m_0 \), and afterward, the empty fuel tank is jettisoned \( mass = 0.1m_0 \).
• This reduction in mass allows the remaining fuel to push a lighter structure in the second stage, maximizing thrust and velocity.
• In this specific scenario, the second stage also burns \( 0.3m_0 \), driving the rocket toward even higher speeds.

The primary advantage of a two-stage operation is that it allows more of the rocket's energy to contribute to its velocity since the weight reduces significantly after each stage, overcoming one of the major inefficiencies in rocket design: carrying non-functioning mass.