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Chapter 8: Problem 63

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 m/s and you want the rocket's speed eventually to be 1.00\(\times\) 10\(^{-3}c\), where \(c\) is the speed of light in vacuum, 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 m/s?

Short Answer

Expert verified
(a) The fraction is nearly 0, indicating impracticality. (b) The fraction is 22.3%.

Step by step solution

01

Understand the Rocket Equation

The rocket equation, also known as the Tsiolkovsky rocket equation, is fundamental in understanding how a rocket propels itself. It relates the velocity of the rocket to the velocity of the ejected fuel and the initial and final masses of the rocket. The equation is given by:\[ v_f = v_e \ln\left(\frac{m_0}{m_f}\right) \]where \( v_f \) is the final velocity of the rocket, \( v_e \) is the exhaust velocity, \( m_0 \) is the initial total mass (rocket plus fuel), and \( m_f \) is the final total mass.
02

Calculate for Final Speed of 1.00脳10^{-3}c

Given that \( v_f = 1.00 \times 10^{-3}c \) and \( v_e = 2000 \, \text{m/s} \), we convert \( v_f \) to meters per second using \( c = 3.00 \times 10^8 \, \text{m/s} \). Therefore, \( v_f = 3.00 \times 10^5 \, \text{m/s} \).Substitute into the rocket equation:\[ 3.00 \times 10^5 = 2000 \ln\left(\frac{m_0}{m_f}\right) \]This rearranges to:\[ \ln\left(\frac{m_0}{m_f}\right) = \frac{3.00 \times 10^5}{2000} \]Solve for \( \frac{m_0}{m_f} \):\[ \frac{m_0}{m_f} = e^{150} \]Calculate the value for \( m_f / m_0 \):This step would need actual computation, which shows that this fraction is extremely small (virtually negligible compared to computational norms), indicating a virtually impossible scenario without advanced rocket technology.
03

Calculate for Final Speed of 3000 m/s

For \( v_f = 3000 \, \text{m/s} \) and \( v_e = 2000 \, \text{m/s} \), substitute into the rocket equation:\[ 3000 = 2000 \ln\left(\frac{m_0}{m_f}\right) \]This rearranges to:\[ \ln\left(\frac{m_0}{m_f}\right) = \frac{3000}{2000} = 1.5 \]Solve for \( \frac{m_0}{m_f} \):\[ \frac{m_0}{m_f} = e^{1.5} \approx 4.48 \]Thus, the fraction \( \frac{m_f}{m_0} \) is:\[ \frac{m_f}{m_0} = \frac{1}{4.48} \approx 0.223 \]This indicates that about 22.3% of the initial mass is not fuel.

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Key Concepts

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

Rocket Propulsion
Rocket propulsion is the force that moves a rocket through space. This force is generated by expelling gas out of the rocket's engine at high speeds. According to Newton's Third Law of Motion, for every action, there is an equal and opposite reaction. When a rocket expels gas downward, an upward thrust is generated. This thrust propels the rocket forward.

The propulsion method is essential for overcoming Earth's gravity or moving through space where gravity is negligible. It determines how fast and how far a rocket can travel. The Tsiolkovsky rocket equation is crucial for predicting the movement of rockets by relating the velocity change needed with the characteristics of the rocket and the speed of the expelled gas.

- Rocket propulsion relies on ejecting mass backward to move forward.
- Thrust is directly influenced by the rate of mass expulsion and the velocity of the expelled gas.
Exhaust Velocity
Exhaust velocity is the speed at which exhaust leaves the rocket engine. It is a vital factor in the efficiency and effectiveness of rocket propulsion. A higher exhaust velocity means that the propellant is expelled at greater speed, providing more thrust.

For example, if a rocket's exhaust velocity is 2000 m/s, it means that the gases are being expelled at a rate of 2000 meters per second. This speed is crucial when calculating how fast you can make a rocket travel. It directly affects the rocket's final velocity according to the rocket equation.

- Higher exhaust velocities lead to greater propulsion efficiency.
- It is a key variable in determining a rocket's top speed.
- It affects fuel consumption; higher velocities might require more efficient fuel management.
Mass Ratio
The mass ratio of a rocket is defined as the ratio between its initial total mass, including fuel, and its final mass, after the fuel has been used. It is often represented as \( \frac{m_0}{m_f} \). A larger mass ratio means more of the rocket's mass is fuel, which can lead to a higher change in velocity.

In the rocket equation, the mass ratio determines how much the rocket's speed can increase. With a higher ratio, the propulsion system can provide greater acceleration, as more fuel is available to be burnt and expelled. However, there are limits as increasing fuel also increases the mass that needs to be lifted initially.

- A high mass ratio can facilitate high-speed travel.
- It is key in the calculation of a rocket鈥檚 potential speed increase.
- Too high a mass ratio may not be feasible with current technology.
Speed of Light
The speed of light, denoted as \( c \), is a universal physical constant approximately equal to \( 3.00 \times 10^8 \, \text{m/s} \). It is commonly used as a benchmark in physics, especially in relativity where it represents the maximum speed at which information can travel.

In the context of the Tsiolkovsky rocket equation, achieving speeds close to a small fraction of the speed of light (\( 1.00 \times 10^{-3}c \)) can illustrate the challenges of space travel. Rockets attempting to reach these speeds would require incredible technological advancements due to the enormous demands on fuel and structural integrity.

- The speed of light presents physical limits for travel speeds in a vacuum.
- Rocket speeds, in comparison to the speed of light, underscore the need for extensive technological evolution.
- Achieving speeds approaching the speed of light introduces relativistic effects that are significant in spacecraft engineering.

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Most popular questions from this chapter

Experimental tests have shown that bone will rupture if it is subjected to a force density of 1.03 \(\times\) 10\(^8\) N/m\(^2\). Suppose a 70.0-kg person carelessly rollerskates into an overhead metal beam that hits his forehead and completely stops his forward motion. If the area of contact with the person's forehead is 1.5 cm\(^2\), what is the greatest speed with which he can hit the wall without breaking any bone if his head is in contact with the beam for 10.0 ms?

Two skaters collide and grab on to each other on frictionless ice. One of them, of mass 70.0 kg, is moving to the right at 4.00 m/s, while the other, of mass 65.0 kg, is moving to the left at 2.50 m/s. What are the magnitude and direction of the velocity of these skaters just after they collide?

Two ice skaters, Daniel (mass 65.0 kg) and Rebecca (mass 45.0 kg), are practicing. Daniel stops to tie his shoelace and, while at rest, is struck by Rebecca, who is moving at 13.0 m/s before she collides with him. After the collision, Rebecca has a velocity of magnitude 8.00 m/s at an angle of 53.1\(^\circ\) from her initial direction. Both skaters move on the frictionless, horizontal surface of the rink. (a) What are the magnitude and direction of Daniel's velocity after the collision? (b) What is the change in total kinetic energy of the two skaters as a result of the collision?

Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracting the cavity to force out the water through an opening. A 6.50-kg squid (including the water in the cavity) at rest suddenly sees a dangerous predator. (a) If the squid has 1.75 kg of water in its cavity, at what speed must it expel this water suddenly to achieve a speed of 2.50 m/s to escape the predator? Ignore any drag effects of the surrounding water. (b) How much kinetic energy does the squid create by this maneuver?

To keep the calculations fairly simple but still reasonable, we model a human leg that is 92.0 cm long (measured from the hip joint) by assuming that the upper leg and the lower leg (which includes the foot) have equal lengths and are uniform. For a 70.0-kg person, the mass of the upper leg is 8.60 kg, while that of the lower leg (including the foot) is 5.25 kg. Find the location of the center of mass of this leg, relative to the hip joint, if it is (a) stretched out horizontally and (b) bent at the knee to form a right angle with the upper leg remaining horizontal.
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