Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

An ion thruster mounted in a satellite uses electric forces to eject xenon ions with a speed of \(21.45 \mathrm{~km} / \mathrm{s}\). The ion thruster produces a thrust of \(1.187 \cdot 10^{-2} \mathrm{~N}\). What is the rate of fuel consumption of the thruster?

Short Answer

Expert verified
Answer: The rate of fuel consumption is \(5.533 \times 10^{-7} \ \text{kg/s}\).

Step by step solution

01

List the given information

We are given the following information: - Speed of xenon ions \(v = 21.45 \ \text{km/s} = 21,\!450 \ \text{m/s}\) - Thrust of ion thruster \(T = 0.01187 \ \text{N}\)
02

Use Newton's second law to find the mass flow rate

According to Newton's second law, force equals mass times acceleration (\(F = ma\)). However, in our case, force is the thrust produced by the ion thruster and "acceleration" refers to the speed of the xenon ions multiplied by the mass flow rate (rate of fuel consumption). Therefore, we have: \(T = \dot{m}v\) Where \(\dot{m}\) is the mass flow rate (rate of fuel consumption).
03

Solve for the mass flow rate

Now we need to solve for the mass flow rate \(\dot{m}\) by using the given values for thrust and speed of xenon ions: \(\dot{m} = \frac{T}{v} = \frac{0.01187 \ \text{N}}{21,\!450 \ \text{m/s}}\)
04

Calculate the mass flow rate

Plugging in the given values, we get the mass flow rate: \(\dot{m} = \frac{0.01187 \ \text{N}}{21,\!450 \ \text{m/s}} = 5.533 \times 10^{-7} \ \text{kg/s}\)
05

Write the final answer

The rate of fuel consumption of the ion thruster is \(5.533 \times 10^{-7} \ \text{kg/s}\).

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 Vaia!

Key Concepts

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

Newton's Second Law
Newton's second law is a fundamental principle in physics that describes how the velocity of an object changes when it is subjected to an external force. The law states that the force applied on an object is equal to the mass of the object multiplied by the acceleration it experiences, often written as the equation:
\[\begin{equation} F = ma \[10pt] \[10pt]\end{equation}\]
In the context of ion thruster fuel consumption, Newton's second law helps us understand the relationship between the thrust produced by the ion thruster (the force) and how the mass of the fuel, being expelled at a certain speed (the acceleration), relates to this force. Essentially, the ion thruster's efficiency and its capability to propel a spacecraft are derived from this basic principle.
Mass Flow Rate
Mass flow rate, often denoted by \(\dot{m}\), is a measure of the mass of substance passing through a given surface per unit time. It is a critical component in the equation for thrust in propulsion systems like ion thrusters, defined as:
\[\begin{equation} \( \dot{m} = \frac{m}{\Delta t} \) \[10pt] \[10pt]\end{equation}\]
where \(m\) is the mass and \(\Delta t\) is the time interval. In regards to ion thrusters, the mass flow rate describes the rate of fuel consumption, which reflects how much mass is lost as the xenon ions are expelled from the thruster over time. This rate is crucial for calculating not only the thrust but also for determining how long the spacecraft can operate before it runs out of fuel.
Thrust Calculation

Understanding Thrust

Thrust is the force that moves a spacecraft through space and is generated by the propulsion system, such as an ion thruster. We calculate thrust using the relationship:
\[\begin{equation} T = \dot{m} v \[10pt] \[10pt]\end{equation}\]
where \(T\) is the thrust produced, \(\dot{m}\) is the mass flow rate, and \(v\) is the exhaust velocity of the ions. This fundamental equation allows us to determine how efficiently a spacecraft can be propelled given a specific fuel consumption rate and ion ejection speed. By rearranging the equation to solve for \(\dot{m}\), as shown in the provided solution, engineers can determine the amount of fuel needed for a mission or the duration the spacecraft can operate with its fuel reserves.
Xenon Ion Propulsion

What is Xenon Ion Propulsion?

Xenon ion propulsion is a form of electric spacecraft propulsion system where xenon gas is ionized, then accelerated to high speeds using an electric field, effectively pushing the spacecraft in the opposite direction according to Newton's third law of motion. This advanced technology offers high efficiency and a better thrust-to-weight ratio compared to conventional chemical rockets.

Xenon as a Propellant

Xenon is often chosen as the propellant for ion thrusters due to its inert nature, high atomic mass, and the ease with which it can be ionized and accelerated. These factors make xenon an ideal candidate to produce a sufficient thrust while maintaining low fuel consumption, extending the life of space missions without the need for excessive amounts of propellant.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider a rocket firing in the vacuum of outer space. Which of the following statements is (are) true? a) The rocket will not produce any thrust because there is no air to push against. b) The rocket will produce the same thrust in vacuum that it can produce in air. c) The rocket will produce half the thrust in vacuum that it can produce in air. d) The rocket will produce twice the thrust in vacuum that it can produce in air.

A toy car of mass \(2.00 \mathrm{~kg}\) is stationary, and a child rolls a toy truck of mass \(3.50 \mathrm{~kg}\) straight toward it with a speed of \(4.00 \mathrm{~m} / \mathrm{s}\). The toy car and truck then undergo an elastic collision. a) What is the velocity of the center of mass of the system consisting of the two toys? b) What are the velocities of the truck and the car with respect to the center of mass of the system consisting of the two toys before and after the collision?

A man with a mass of \(55.0 \mathrm{~kg}\) stands up in a 65.0 - \(\mathrm{kg}\) canoe of length \(4.00 \mathrm{~m}\) floating on water. He walks from a point \(0.750 \mathrm{~m}\) from the back of the canoe to a point \(0.750 \mathrm{~m}\) from the front of the canoe. Assume negligible friction between the canoe and the water. How far does the canoe move?

The density of a \(1.00-\mathrm{m}\) long rod can be described by the linear density function \(\lambda(x)=100 . \mathrm{g} / \mathrm{m}+10.0 x \mathrm{~g} / \mathrm{m}^{2}\) One end of the rod is positioned at \(x=0\) and the other at \(x=1.00 \mathrm{~m} .\) Determine (a) the total mass of the rod, and (b) the center-of-mass coordinate.

An artillery shell is moving on a parabolic trajectory when it explodes in midair. The shell shatters into a very large number of fragments. Which of the following statements is (are) true (select all that apply)? a) The force of the explosion will increase the momentum of the system of fragments, and so the momentum of the shell is not conserved during the explosion. b) The force of the explosion is an internal force and thus cannot alter the total momentum of the system. c) The center of mass of the system of fragments will continue to move on the initial parabolic trajectory until the last fragment touches the ground. d) The center of mass of the system of fragments will continue to move on the initial parabolic trajectory until the first fragment touches the ground. e) The center of mass of the system of fragments will have a trajectory that depends on the number of fragments and their velocities right after the explosion.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free