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Chapter 8: Problem 50

A rocket is fired in deep space, where gravity is negligible. In the first second, it ejects \(1 / 160\) of its mass as exhaust gas and has an acceleration of \(15.0 \mathrm{~m} / \mathrm{s}^{2}\). What is the speed of the exhaust gas relative to the rocket?

Short Answer

Expert verified
The exhaust gas speed is 2400 m/s relative to the rocket.

Step by step solution

01

Identify the Known Variables

We know the acceleration of the rocket is \(15.0 \mathrm{~m/s}^2\) and it ejects \(\frac{1}{160}\) of its mass in the first second.
02

Use the Rocket Equation

The rocket equation (Tsiolkovsky rocket equation) is given by \[ v_e = -a \times \frac{m_0}{\Delta m} \]where \(v_e\) is the exhaust velocity, \(a\) is the acceleration, \(m_0\) is the initial total mass, and \(\Delta m\) is the mass ejected. Since \(\Delta m = \frac{m_0}{160}\), substitute these values into the rocket equation.
03

Substitute Known Values and Solve

Substitute the acceleration and the mass ratio into the rocket equation: \[ v_e = -15.0 \text{ m/s}^2 \times \frac{160}{1} \]Calculate the value of \(v_e\), which gives:\[ v_e = -2400 \text{ m/s} \].
04

Interpret the Result

The result \(-2400 \text{ m/s}\) indicates that the exhaust gas is expelled at a speed of \(2400 \text{ m/s}\) relative to the rocket but in the opposite direction to the acceleration due to the negative sign, consistent with the conservation of momentum principles.

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Key Concepts

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

Exhaust Velocity
Exhaust velocity is a crucial concept when discussing rocket science and dynamics. It represents the speed at which gas is expelled from a rocket's engine. In other words, it indicates how fast the rocket is pushing against the exhaust gases to propel itself forward. This expulsion of gases causes the rocket to move in the opposite direction. The exhaust velocity directly affects the efficiency of the rocket; higher exhaust velocities generally suggest a more efficient engine.
  • This velocity is derived from the principle of conservation of momentum, where the momentum lost by the exhaust gases equals the momentum gained by the rocket.
  • In the exercise, the exhaust velocity was calculated as (-2400 m/s). The negative sign shows the direction is opposite to the rocket's acceleration.
Understanding this is essential for designing rockets that can break free from Earth's gravity and embark on interstellar missions.
Acceleration
Acceleration refers to the rate at which a rocket changes its velocity. It signifies how quickly a rocket can speed up or slow down. In the context of rockets in space, the acceleration is produced by expelling mass (fuel) as exhaust at high speeds, resulting in thrust.
  • In the given exercise, the rocket achieves an acceleration of 15.0 m/s虏, indicating it can increase its velocity by 15 meters per second every second.
  • The force behind this acceleration is due to the exhaust gases pushed out of the rocket.
Understanding the role of acceleration in rocket dynamics helps us appreciate how rockets, equipped with the right engines, can maneuver in the zero-gravity environment of space.
Mass Ejection
Mass ejection is the process by which a rocket expels part of its fuel to produce thrust. When the rocket ejects this mass, it moves in the opposite direction according to Newton's third law of motion鈥�'for every action, there is an equal and opposite reaction.'
  • In the provided scenario, the rocket ejects 1/160th of its mass in the first second.
  • This ejection leads to a forward thrust that propels the rocket.
The efficient ejection of mass is fundamental to the successful operation of a rocket, as it determines both the speed and range that the rocket can achieve.
Tsiolkovsky Rocket Equation
The Tsiolkovsky rocket equation is a fundamental formula that allows us to calculate the velocity change of a rocket, taking into account the ejected mass and exhaust velocity.
  • Also known as the "ideal rocket equation," it provides a relationship between the velocity change, the remaining mass of the rocket, the ejected mass, and the exhaust velocity.
  • In mathematical terms, the equation is expressed as: \[ \Delta v = v_e \cdot \ln \left( \frac{m_0}{m_f} \right)\]where \( \Delta v \) is the change in velocity, \( v_e \) is the exhaust velocity, \( m_0 \) is the initial mass, and \( m_f \) is the final mass after ejection.
For students studying rocket dynamics and propulsion, mastering the Tsiolkovsky rocket equation is critical, as it explains how different parameters contribute to a rocket's capability to reach its destination efficiently.

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Most popular questions from this chapter

Combining conservation laws. A \(5.00 \mathrm{~kg}\) chunk of ice is sliding at \(12.0 \mathrm{~m} / \mathrm{s}\) on the floor of an ice-covered valley when it collides with and sticks to another \(5.00 \mathrm{~kg}\) chunk of ice that is initially at rest. (See Figure \(8.39 .\) ) Since the valley is icy, there is no friction. After the collision, the blocks slide partially up a hillside and then slide back down. How fast are they moving when they reach the valley floor again? (Hint: Break this problem into two parts - the collision and the behavior after the collision - and apply the appropriate conservation law to each part.)

Archerfish are tropical fish that hunt by shooting drops of water from their mouths at insects above the water's surface to knock them into the water, where the fish can eat them. \(\mathrm{A} 65 \mathrm{~g}\) fish at rest just on the water's surface can expel a \(0.30 \mathrm{~g}\) drop of water in a short burst of \(5.0 \mathrm{~ms}\). High-speed measurements show that the water has a speed of \(2.5 \mathrm{~m} / \mathrm{s}\) just after the archerfish expels it. What is the momentum of one drop of water immediately after it leaves the fish's mouth? A. \(7.5 \times 10^{-4} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\) B. \(1.5 \times 10^{-4} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\) C. \(7.5 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\) D. \(1.5 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\)

A 4.25 g bullet traveling horizontally with a velocity of magnitude \(375 \mathrm{~m} / \mathrm{s}\) is fired into a wooden block with mass \(1.12 \mathrm{~kg},\) initially at rest on a level frictionless surface. The bullet passes through the block and emerges with its speed reduced to \(122 \mathrm{~m} / \mathrm{s} .\) How fast is the block moving just after the bullet emerges from it?

Just as it has reached an upward speed of \(5.0 \mathrm{~m} / \mathrm{s}\) during a vertical launch, a rocket explodes into two pieces. Photographs of the explosion reveal that the lower piece, with a mass one-fourth that of the upper piece, was moving downward at \(3.0 \mathrm{~m} / \mathrm{s}\) the instant after the explosion. (a) Find the speed of the upper piece just after the explosion. (b) How high does the upper piece go above the point where the explosion occurred?

The speed of the fastest-pitched baseball was \(45 \mathrm{~m} / \mathrm{s},\) and the ball's mass was \(145 \mathrm{~g}\). (a) What was the magnitude of the momentum of this ball, and how many joules of kinetic energy did it have? (b) How fast would a 57 gram ball have to travel to have the same amount of (i) kinetic energy, and (ii) momentum?
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