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Chapter 8: Problem 48

A small rocket burns \(0.0500 \mathrm{~kg}\) of fuel per second, ejecting it as a gas with a velocity of magnitude \(1600 \mathrm{~m} / \mathrm{s}\) relative to the rocket. (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?

Short Answer

Expert verified
(a) Thrust is 80 N. (b) The rocket operates in space; steer by thrusters or gimbaling; brake by expelling gas backward.

Step by step solution

01

Understanding Thrust

Thrust is the force exerted by the rocket due to the expulsion of gas. According to the physics of rocket propulsion, the thrust \( F \) can be calculated using the formula: \( F = \dot{m} \times v \), where \( \dot{m} \) is the rate of mass expelled per second (in kg/s) and \( v \) is the exhaust velocity (in m/s).
02

Calculate Mass Flow Rate

The mass flow rate \( \dot{m} \) is given in the problem as the mass of fuel burnt per second: \( \dot{m} = 0.0500 \, \text{kg/s} \).
03

Calculate Thrust

Using the thrust formula \( F = \dot{m} \times v \), substitute the given values: \( \dot{m} = 0.0500 \, \text{kg/s} \) and \( v = 1600 \, \text{m/s} \). Therefore, the thrust \( F = 0.0500 \, \text{kg/s} \times 1600 \, \text{m/s} = 80 \, \text{N} \).
04

Rocket Operation in Outer Space

A rocket operates based on Newton's third law of motion. Since thrust is produced by expelling mass (not by air resistance), the rocket would still operate in outer space, where there is no atmosphere.
05

Steering and Braking in Space

In space, steering can be done by gimbaling the engine, using control thrusters, or by spinning the spacecraft. Braking requires expelling gas in the opposite direction of motion, using onboard fuel.

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Key Concepts

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

Thrust Calculation
To understand how rockets work, we need to start with thrust calculation. Thrust is essentially the force that propels a rocket forward. It's produced by expelling gas at high speed from the rocket's engines. The formula for calculating thrust is
  • \( F = \dot{m} \times v \),
  • where \( F \) is the thrust in Newtons, \( \dot{m} \) is the mass flow rate in kg/s, and \( v \) is the velocity of the ejected gas in m/s.
In the given exercise, the rocket expels fuel at a rate \( \dot{m} = 0.0500 \, \text{kg/s} \) with an exhaust velocity of \( 1600 \, \text{m/s} \). By substituting these values into the formula, we find the thrust to be
  • \( F = 0.0500 \, \text{kg/s} \times 1600 \, \text{m/s} = 80 \, \text{N} \).
This explains how rockets can generate force by pushing fuel out fast enough.
Newton's Third Law
Newton's third law of motion is fundamental in understanding rocket propulsion. It states: "For every action, there is an equal and opposite reaction." In the context of a rocket:
  • The action is the expulsion of gas out of the rocket engine.
  • The reaction is the movement of the rocket in the opposite direction.
This principle is what allows rockets to move forward. Regardless of whether the rocket is in the Earth's atmosphere or in outer space, the laws of physics remain consistent. Hence, the rocket can still thrust forward efficiently by expelling gas, creating necessary motion change.
Mass Flow Rate
Mass flow rate is an essential concept in calculating thrust. Represented by \( \dot{m} \), it indicates how much mass is being expelled per unit of time. In rocket science,
  • this is usually represented in kilograms per second (kg/s).
For the rocket in the exercise, the mass flow rate was given as \( 0.0500 \, \text{kg/s} \). Understanding this rate is crucial, as it directly impacts thrust. The greater the mass expelled per second, the greater the potential force exerted by the rocket. Hence, mastering control over mass flow rate is key for efficient rocket design and operation.
Space Navigation
Operating and navigating a rocket in space involves several considerations unique to the environment. In the absence of air resistance, space navigation primarily relies on controlling the direction and velocity of the spacecraft. This can be achieved through:
  • Gimbaling the rocket engine, which involves adjusting the angle of the engine to change thrust direction.
  • Using small control thrusters placed around the spacecraft to maneuver.
  • Employing reaction wheels to alter the spacecraft's orientation by conserving angular momentum.
Braking in space is typically executed by firing thrusters in the opposite direction of travel, effectively slowing down the spacecraft. This requires strategic management of onboard fuel, ensuring enough remains for both navigation and deceleration.

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Most popular questions from this chapter

On an air track, a 400.0 g glider moving to the right at \(2.00 \mathrm{~m} / \mathrm{s}\) collides elastically with a 500.0 g glider moving in the opposite direction at \(3.00 \mathrm{~m} / \mathrm{s}\). Find the velocity of each glider after the collision.

A stone with a mass of \(0.100 \mathrm{~kg}\) rests on a frictionless, horizontal surface. A bullet of mass \(2.50 \mathrm{~g}\) traveling horizontally at \(500 \mathrm{~m} / \mathrm{s}\) strikes the stone and rebounds horizontally at right angles to its original direction with a speed of \(300 \mathrm{~m} / \mathrm{s}\). (a) Compute the magnitude and direction of the velocity of the stone after it is struck. (b) Is the collision perfectly elastic?

Tennis players sometimes leap into the air to return a volley. (a) If a \(57 \mathrm{~g}\) tennis ball is traveling horizontally at \(72 \mathrm{~m} / \mathrm{s}\) (which does occur), and a \(61 \mathrm{~kg}\) tennis player leaps vertically upward and hits the ball, causing it to travel at \(45 \mathrm{~m} / \mathrm{s}\) in the reverse direction, how fast will her center of mass be moving horizontally just after hitting the ball? (b) If, as is reasonable, her racket is in contact with the ball for \(30.0 \mathrm{~ms}\), what force does her racket exert on the ball? What force does the ball exert on the racket?

Cart \(A\) has a mass of \(5 \mathrm{~kg}\) and is moving in the \(+x\) direction at \(2 \mathrm{~m} / \mathrm{s}\). Cart \(B\) has a mass of \(2 \mathrm{~kg}\) and is moving in the \(+y\) direction at \(5 \mathrm{~m} / \mathrm{s}\). (a) Do the two carts have the same momentum? Explain. (b) Is the magnitude of the momentum of each cart the same? Explain. (c) Is the kinetic energy of each cart the same? Explain.

On a frictionless, horizontal air table, puck \(A\) (with mass \(0.250 \mathrm{~kg}\) ) is moving to the right toward puck \(B\) (with mass \(0.350 \mathrm{~kg}\) ), which is initially at rest. After the collision, puck \(A\) has a velocity of \(0.120 \mathrm{~m} / \mathrm{s}\) to the left, and puck \(B\) has a velocity of \(0.650 \mathrm{~m} / \mathrm{s}\) to the right. (a) What was the speed of puck \(A\) before the collision? (b) Calculate the change in the total kinetic energy of the system that occurs during the collision.
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