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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 \(1.00 \times 10^{-3} c\) , 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 \(\mathrm{m} / \mathrm{s} ?\)

Short Answer

Expert verified
(a) The fraction is nearly 0. (b) The fraction is about 0.2231.

Step by step solution

01

Define Variables and Constants

Let the speed of light be \( c = 3.00 \times 10^8 \ \mathrm{m/s} \). Define the initial mass of the rocket with fuel as \( m_0 \) and the final mass of the rocket without fuel as \( m_f \). Let the effective exhaust velocity be \( v_e = 2000 \ \mathrm{m/s} \).
02

Calculate Rocket's Speed as a Fraction of Light Speed

For part (a), the desired final speed of the rocket \( v \) is given as \( 1.00 \times 10^{-3} c \), which means \( v = 1.00 \times 10^{-3} \times 3.00 \times 10^8 = 3.00 \times 10^5 \ \mathrm{m/s} \).
03

Use the Rocket Equation

The rocket equation, also known as the Tsiolkovsky rocket equation, is given by: \[ v = v_e \ln \left( \frac{m_0}{m_f} \right) \], where \( \ln \) denotes the natural logarithm.
04

Solve for Mass Fraction for Top Speed

For part (a), substitute \( v = 3.00 \times 10^5 \ \mathrm{m/s} \) and \( v_e = 2000 \ \mathrm{m/s} \) into the rocket equation: \[ 3.00 \times 10^5 = 2000 \ln \left( \frac{m_0}{m_f} \right) \] Solving, \[ \ln \left( \frac{m_0}{m_f} \right) = \frac{3.00 \times 10^5}{2000} = 150 \] Taking the exponential of both sides: \[ \frac{m_0}{m_f} = e^{150} \] The fraction of the initial mass which is not fuel is \( \frac{m_f}{m_0} = \frac{1}{e^{150}} \), which is nearly 0, practically meaning the entire mass is fuel.
05

Calculate for Final Speed of 3000 m/s

For part (b), substitute \( v = 3000 \ \mathrm{m/s} \) and \( v_e = 2000 \ \mathrm{m/s} \) into the rocket equation: \[ 3000 = 2000 \ln \left( \frac{m_0}{m_f} \right) \] Solving, \[ \ln \left( \frac{m_0}{m_f} \right) = \frac{3000}{2000} = 1.5 \] Taking the exponential of both sides gives: \[ \frac{m_0}{m_f} = e^{1.5} \] Thus, \( \frac{m_f}{m_0} = \frac{1}{e^{1.5}} \approx 0.2231 \).

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Key Concepts

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

Tsiolkovsky Rocket Equation
The Tsiolkovsky Rocket Equation is a fundamental principle in rocket physics, named after the Russian scientist Konstantin Tsiolkovsky. It describes the motion of rockets and allows us to calculate the change in velocity that a rocket can achieve given its propellant mass. The equation is given by the formula \( v = v_e \ln \left( \frac{m_0}{m_f} \right) \), where \( v \) is the change in velocity, \( v_e \) is the effective exhaust velocity, \( m_0 \) is the initial total mass of the rocket, and \( m_f \) is the final mass after the fuel has been burned.
The beauty of this equation lies in its simplicity; it highlights the exponential relationship between the mass of the rocket and its speed. Simply put, the more fuel you have relative to the rocket's structural mass, the more velocity you can achieve.
  • This equation is crucial for designing efficient rockets because it helps engineers determine the amount of fuel needed for a desired speed.
  • It also reveals one of the challenges in rocketry: the need to carry large amounts of propellant to achieve high speeds.
Exhaust Velocity
Exhaust velocity, denoted as \( v_e \), refers to the speed at which exhaust gases leave the rocket engine. It's an essential parameter in determining a rocket's performance and efficiency. The higher the exhaust velocity, the more efficient the rocket's engine, as it signifies greater thrust production.
There are some key points about exhaust velocity:
  • Exhaust velocity is measured in meters per second (m/s), and a typical value for chemical rockets is between 2000 and 4500 m/s.
  • It is directly proportional to the force exerted by the rocket (thrust) due to momentum exchange between the expelled gas and the rocket.
  • Higher exhaust velocities mean the rocket can accelerate more effectively, decreasing the amount of fuel required to reach a certain speed.
In the provided exercise, the rocket has an exhaust velocity of 2000 m/s, which directly affects the calculations for the final speed and mass fraction.
Space Propulsion
Space propulsion is the science of moving a spacecraft in outer space, where traditional propulsion systems used on Earth, such as wheels and propellers, are ineffective. In the vacuum of space, propulsion relies on Newton’s third law of motion: for every action, there is an equal and opposite reaction.
Various methods exist for achieving propulsion in space, including:
  • Chemical Rockets: Utilize the combustion of propellant to produce hot gases expelled at high speed, providing thrust.
  • Ion Thrusters: Use electrically charged particles (ions) to create thrust with high exhaust velocities but low thrust output.
  • Nuclear Propulsion: Employs nuclear reactions to heat the propellant, creating powerful thrust with potentially high efficiency.
Understanding propulsion is essential for mission planning, as it determines how a spacecraft can maneuver and reach distant destinations. The type of propulsion used affects the mission's speed, range, and payload capacity.
Mass Fraction Calculation
Mass fraction calculation is a vital part of rocket design, helping engineers understand how much of the rocket's total mass is available as payload versus fuel. The mass fraction refers to the proportion of the initial rocket mass that is propellant. In the context of the Tsiolkovsky Rocket Equation, it allows for calculating the required fuel to reach a desired speed.
To find the mass fraction, consider:
  • Initial mass of the rocket (\( m_0 \)), which includes both the structure and the propellant.
  • Final mass of the rocket (\( m_f \)), which is its mass without fuel.
The mass fraction \( \frac{m_f}{m_0} \) is crucial for determining how efficient a rocket is and can be calculated using the equation \( \frac{m_f}{m_0} = \frac{1}{e^{\ln(\frac{m_0}{m_f})}} \). In the exercise, this calculation illustrates how much of a rocket's mass must be fuel to achieve certain speeds, demonstrating the practical limitations of rocket travel and the importance of efficient design.

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Most popular questions from this chapter

A small rocket burns 0.0500 \(\mathrm{kg}\) of fuel per second, ejecting it as a gas with a velocity relative to the rocket of magnitude 1600 \(\mathrm{m} / \mathrm{s}\) . (a) What is the thrust of the rocket? (b) Would the rocket operate in outer space where there is no atmosphere? If so, how would you steer it? Could you brake it?

When two hydrogen atoms of mass \(m\) combine to form a diatomic hydrogen molecule \(\left(H_{2}\right),\) the potential energy of the system after they combine is \(-\Delta,\) where \(\Delta\) is a positive quantity called the binding energy of the molecule. (a) Show that in a collision that involves only two hydrogen atoms, it is impossible to form an \(\mathrm{H}_{2}\) molecule because momentum and energy cannot simultaneously be conserved. (Hint: If you can show this to be true in one frame of reference, then it is true in all frames of reference. Can you see why?) (b) An \(\mathrm{H}_{2}\) molecule can be formed in a collision that involves three hydrogen atoms. Suppose that before such a collision, each of the three atoms has speed \(1.00 \times 10^{3} \mathrm{m} / \mathrm{s}\) , and they are approaching at \(120^{\circ}\) angles so that at any instant, the atoms lie at the comers of an equilateral triangle. Find the speeds of the \(\mathrm{H}_{2}\) molecule and of the single hydrogen atom that remains after the collision. The binding energy of \(\mathrm{H}_{2}\) is \(\Delta=7.23 \times 10^{-19} \mathrm{J},\) and the mass of the bindrogen atom is \(1.67 \times 10^{-27} \mathrm{kg}\) .

At one instant, the center of mass of a system of two particles is located on the \(x\) -axis at \(x=2.0 \mathrm{m}\) and has a velocity of \((5.0 \mathrm{m} / \mathrm{s})\) ) One of the particles is at the origin. The other particle has a mass of 0.10 \(\mathrm{kg}\) and is at rest on the \(x\) -axis at \(x=8.0 \mathrm{m}\) . (a) What is the mass of the particle at the origin? (b) Calculate the total momentum of this system. (c) What is the velocity of the particle at the origin?

A \(0.150-\mathrm{kg}\) frame, when suspended from a coil spring, stretches the spring 0.050 \(\mathrm{m}\) . A \(0.200-\mathrm{kg}\) lump of putty is dropped from rest onto the frame from a height of 30.0 \(\mathrm{cm}\) (Fig. 8.42\()\) . Find the maximum distance the frame moves downward from its initial position.

A \(5.00-\mathrm{g}\) bullet is shot through a \(1.00-\mathrm{kg}\) wood block suspended on a string 2.00 \(\mathrm{m}\) long. The center of mass of the block rises a distance of 0.45 \(\mathrm{cm} .\) Find the speed of the bullet as it emerges from the block if its initial speed is 450 \(\mathrm{m} / \mathrm{s}\) .

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