91Ó°ÊÓ

A rocket ejects exhaust with an exhaust velocity \(u\). The rate at which the exhaust mass is used (mass per unit time) is \(b\). We assume that the rocket accelerates in a straight line starting from rest, and that no external forces act on it. Let the rocket's initial mass (fuel plus the body and payload) be \(m_{i}\), and \(m_{f}\) be its final mass, after all the fuel is used up. (a) Find the rocket's final velocity, \(v\), in terms of \(u, m_{i}\), and \(m_{f} .\) Neglect the effects of special relativity. (b) A typical exhaust velocity for chemical rocket engines is \(4000 \mathrm{~m} / \mathrm{s}\). Estimate the initial mass of a rocket that could accelerate a one-ton payload to \(10 \%\) of the speed of light, and show that this design won't work. (For the sake of the estimate, ignore the mass of the fuel tanks. The speed is fairly small compared to \(c\), so it's not an unreasonable approximation to ignore relativity.) (answer check available at lightandmatter.com)

Step by step solution

01

Understand the Rocket Equation

The rocket's motion can be described by the Tsiolkovsky rocket equation:\[ v = u \ln \left( \frac{m_i}{m_f} \right) \]where \( v \) is the final velocity, \( u \) is the exhaust velocity, \( m_i \) is the initial mass, and \( m_f \) is the final mass of the rocket.

02

Solve for Final Velocity (Part a)

We are asked to find the final velocity \( v \) in terms of \( u, m_i, \) and \( m_f \). Using the rocket equation derived in Step 1, we have:\[ v = u \ln \left( \frac{m_i}{m_f} \right) \]This formula provides the final velocity of the rocket based solely on the given parameters.

03

Identify Constraints (Part b)

In part b, we need to estimate the initial mass \( m_i \) for a rocket to accelerate a one-ton payload (i.e., the final mass \( m_f = 1000 \, \text{kg} \)) to 10% of the speed of light (\( v = 0.1c \)), with \( u = 4000 \, \text{m/s} \). The speed of light \( c \) is approximately \( 3 \times 10^8 \, \text{m/s} \). Thus, \( v = 3 \times 10^7 \, \text{m/s} \).

04

Calculate Initial Mass (Part b)

Rearrange the rocket equation to solve for \( m_i \):\[ 3 \times 10^7 = 4000 \ln \left( \frac{m_i}{1000} \right) \]Dividing both sides by 4000 gives:\[ \ln \left( \frac{m_i}{1000} \right) = \frac{3 \times 10^7}{4000} \approx 7500 \]Thus, exponentiating both sides gives:\[ \frac{m_i}{1000} = e^{7500} \]\[ m_i = 1000 \times e^{7500} \]

05

Assess the Plausibility

The calculated initial mass involves \( e^{7500} \), a number that is astronomically large, greater than the number of particles in the observable universe, which is physically unrealizable. Therefore, this design will not work with conventional chemical rockets.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rocket Propulsion

Rocket propulsion is a fascinating and complex process. It's all about Newton's Third Law: for every action, there is an equal and opposite reaction. When a rocket engine expels mass in one direction, it generates thrust in the opposite direction. This is what propels the rocket forward. No external forces mean the rocket's motion is primarily dependent on its own mechanics and the movement of its exhaust gases.
The key components of rocket propulsion include:

• Thrust: The force that moves the rocket.
• Exhaust velocity: The speed at which exhaust leaves the rocket engine.
• Fuel: The source of mass that is expelled.

These elements work together to push the rocket. It's a self-contained system that doesn't rely on air or external substances. As long as the rocket has fuel to expel, it can move forward, which is especially important in the vacuum of space where external forces like air resistance are absent.

Exhaust Velocity

Exhaust velocity is one of the most critical factors in rocket propulsion. It determines how efficiently the thrust is generated. High exhaust velocity means that the rocket can gain more speed for the amount of fuel burned.
The exhaust velocity, symbolized by \( u \), is essentially how fast the combustion gases leave the rear of the rocket. The higher this speed, the more "kick" each particle of fuel imparts to the rocket.

• Measured typically in meters per second (m/s).
• Dependent on the type of fuel and engine design.
• An average exhaust velocity for chemical rockets is around 4000 m/s.

Understanding and optimizing exhaust velocity is crucial for designing efficient rockets. Whether launching a payload into orbit or sending a spacecraft on a deep space journey, maximizing exhaust velocity can significantly reduce the amount of fuel needed.

Chemical Rocket

Chemical rockets are the most common type of rockets in use today. They operate based on the combustion of propellants that generate large volumes of hot gas, which are expelled through a nozzle to produce thrust.
Here's how a chemical rocket works:

• Fuel and oxidizer are combined and burned inside the combustion chamber.
• This chemical reaction produces hot gases.
• The gases expand rapidly and are expelled at high speed.

Because they use highly energetic chemical reactions to produce thrust, chemical rockets provide huge amounts of power. However, they require a lot of fuel to achieve high speeds, which makes them impractical for long-distance space travel without advanced efficiency improvements.

Speed of Light

The speed of light, denoted as \( c \), is an essential concept in understanding the upper limits of speed in physics. It's approximately \( 3 \times 10^8 \, ext{m/s} \) and represents the fasted speed at which information or matter can travel through space.
In the context of rockets:

• A speed of 0.1c would be 10% of the speed of light, a velocity that is enormous compared to current technological capabilities.
• Achieving a fraction of the speed of light is a theoretical challenge due to the massive amounts of energy required.
• In our exercise, trying to achieve such speed with conventional chemical rockets demonstrates the limitations of current propulsion technology.

The speed of light imposes a natural limit on how fast we can travel using any propulsion method that relies on known physics, including chemical rockets.