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Chapter 9: Problem 107

A \(6100 \mathrm{~kg}\) rocket is set for vertical firing from the ground. If the exhaust speed is \(1200 \mathrm{~m} / \mathrm{s}\), how much gas must be cjected cach second if the thrust (a) is to equal the magnitude of the gravitational force on the rocket and (b) is to give the rocket an initial upward acceleration of \(21 \mathrm{~m} / \mathrm{s}^{2} ?\)

Short Answer

Expert verified
(a) 49.82 kg/s, (b) 156.57 kg/s

Step by step solution

01

Understanding the problem

We are given a rocket with a mass of \( 6100 \, \text{kg} \) and an exhaust speed of \( 1200 \, \text{m/s} \). We want to find the amount of gas ejected each second for two situations: (a) when the thrust equals the gravitational force on the rocket, and (b) when the thrust provides an initial upward acceleration of \( 21 \, \text{m/s}^2 \).
02

Calculate gravitational force

To find the thrust needed to counteract gravity, calculate the gravitational force on the rocket using: \[ f_g = m \cdot g \] where \( m = 6100 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \). Substitute: \[ f_g = 6100 \times 9.8 = 59780 \, \text{N} \].
03

Use exhaust speed for thrust equation

For part (a), where thrust equals gravitational force, the thrust \( T \) is equal to the product of the mass flow rate \( \dot{m} \) and exhaust speed \( v_e \): \[ f_g = \dot{m} \cdot v_e \] Rearrange to find \( \dot{m} \): \[ \dot{m} = \frac{f_g}{v_e} = \frac{59780}{1200} \approx 49.82 \, \text{kg/s} \].
04

Calculate total thrust for desired acceleration

For part (b), find the total thrust using the formula: \[ T = m \cdot (a + g) \] where \( a = 21 \, \text{m/s}^2 \). Substitute: \[ T = 6100 \times (21 + 9.8) = 6100 \times 30.8 = 187880 \, \text{N} \].
05

Calculate mass flow rate for desired acceleration

Using the thrust equation \( T = \dot{m} \times v_e \) for part (b), rearrange to solve for \( \dot{m} \): \[ \dot{m} = \frac{T}{v_e} = \frac{187880}{1200} \approx 156.57 \, \text{kg/s} \].

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Key Concepts

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

Gravitational Force
Gravitational force is the pull that the Earth exerts on objects, keeping them grounded. It's what gives objects weight. For a rocket, which is essentially an object on Earth, this force acts downwards, opposing any attempt to lift off. The formula used to calculate this force is:
  • \[ f_g = m \cdot g \] where
    • \(f_g\) is the gravitational force,
    • \(m\) is the mass of the object, and
    • \(g\) is the acceleration due to gravity, typically \(9.8 \, \text{m/s}^2\).
For instance, with a rocket mass of \(6100 \, \text{kg}\), we get a gravitational force of \(59780 \, \text{N}\). This means that in order to just hover, the rocket needs a force equal to the gravitational force acting on it, otherwise it won't lift off the ground.
Thrust
Thrust is the force which moves the rocket through the air, pushing it upwards against gravity. It must at least equal the gravitational force for the rocket to take off. In our exercise, this condition relates to the thrust needed for the rocket to just balance the gravitational pull without additional upward acceleration.
  • Thrust is achieved by expelling gas at high speed from the rocket nozzles.
  • This expulsion action produces a reaction force, known as thrust, according to Newton's third law of motion.
To maintain that thrust, our rocket has to balance the gravitational force, calculated as \(59780 \text{ N}\). This indicates the rocket's engines must generate this same amount of force continuously, putting into action both the expulsion of gases and the rocket's design to reach and maintain lift-off.
Mass Flow Rate
Mass flow rate is a measure of how much mass is expelled from the rocket every second. It's a vital part of achieving and maintaining the calculated thrust.
  • Expressed as \(\dot{m}\), you can find it through the formula:
  • \[\dot{m} = \frac{T}{v_e}\]
For our first scenario, where thrust equals the gravitational force, the mass flow rate needed is approximately \(49.82 \text{ kg/s}\). This means the engines must pump out nearly 50 kilograms of gas every second just to keep the rocket hovering.
  • In a different case, where the rocket also needs to accelerate upwards at \(21 \, \text{m/s}^2\), the mass flow increases to \(156.57 \, \text{kg/s}\). This adjustment ensures the additional force needed for upward acceleration.
Exhaust Speed
Exhaust speed is an essential element in rocket science. It refers to the speed at which gases are expelled from the rocket. The faster these gases exit, the more thrust is generated, due to increased momentum transfer.
  • In our exercise, the exhaust speed was given as \(1200 \, \text{m/s}\).
  • This implies that every kilogram of gas expelled contributes significantly to the total thrust generated.
A higher exhaust speed enables a rocket to generate more thrust with a given mass flow rate, improving efficiency. Thus, the speed at which the rocket expels gas is pivotal to achieving the necessary thrust for lifting off against Earth's gravitational pull. The chosen exhaust speed is a balance between fuel efficiency and thrust capability, optimizing the rocket's performance during launch.

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Most popular questions from this chapter

A steel ball of mass \(0.500 \mathrm{~kg}\) is fastened to a cord that is \(70.0 \mathrm{~cm}\) long and fixed at the far end. The ball is then released when the cord is horizontal (Fig. \(9-65\) ). At the bottom of its path, the ball strikes a \(2.50 \mathrm{~kg}\) steel block initially at rest on a frictionless surface. The collision is elastic. Find (a) the speed of the ball and (b) the speed of the block, both just after the collision.

block 1 of mass \(m_{1}\) slides from rest along a frictionless ramp from height \(h=2.50 \mathrm{~m}\) and then collides with stationary block \(2,\) which has mass \(m_{2}=2.00 m_{1}\). After the collision, block 2 slides into a region where the coefficient of kinetic friction \(\mu_{k}\) is 0.500 and comes to a stop in distance \(d\) within that region. What is the value of distance \(d\) if the collision is (a) elastic and (b) completely inelastic?

A cue stick strikes a stationary pool ball, with an average force of 32 N over a time of \(14 \mathrm{~ms}\). If the ball has mass \(0.20 \mathrm{~kg}\), what speed does it have just after impact?

A \(6090 \mathrm{~kg}\) space probe moving nose-first toward Jupiter at \(105 \mathrm{~m} / \mathrm{s}\) relative to the Sun fires its rocket engine, ejecting \(80.0 \mathrm{~kg}\) of exhaust at a speed of \(253 \mathrm{~m} / \mathrm{s}\) relative to the space probe. What is the final velocity of the probe?

particle 1 of mass \(m_{1}=0.30 \mathrm{~kg}\) slides rightward along an \(x\) axis on a frictionless floor with a speed of \(2.0 \mathrm{~m} / \mathrm{s}\). When it reaches \(x=0,\) it undergocs a one-dimensional elastic collision with stationary particle 2 of mass \(m_{2}=0.40 \mathrm{~kg}\). When particle 2 then reaches a wall at \(x_{n}=70 \mathrm{~cm},\) it bounces from the wall with no loss of speed. At what position on the \(x\) axis does particle 2 then collide with particle \(1 ?\)
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