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Chapter 3: Problem 6

In the early stages of the Saturn V rocket's launch, mass was ejected at about \(15,000 \mathrm{kg} / \mathrm{s}\), with a speed \(v_{\mathrm{ex}} \approx 2500 \mathrm{m} / \mathrm{s}\) relative to the rocket. What was the thrust on the rocket? Convert this to tons (1 ton \(\approx 9000\) newtons) and compare with the rocket's initial weight (about 3000 tons).

Short Answer

Expert verified
The thrust is 37,500,000 N, equivalent to about 4167 tons, greater than the rocket's initial weight of 3000 tons.

Step by step solution

01

Understanding Thrust Equation

Thrust produced by a rocket is calculated using the formula \( F = \dot{m} \times v_{\text{ex}} \), where \( \dot{m} \) is the mass ejection rate and \( v_{\text{ex}} \) is the speed of the ejected mass. Here, \( \dot{m} = 15,000 \, \text{kg/s} \) and \( v_{\text{ex}} = 2500 \, \text{m/s} \).
02

Calculating the Thrust Force

Substitute the given values into the thrust formula: \( F = 15,000 \times 2500 = 37,500,000 \, \text{N} \). This is the thrust force in newtons.
03

Converting Thrust to Tons

Given that 1 ton is approximately equal to 9000 newtons, we convert the thrust to tons by dividing by 9000: \( \frac{37,500,000}{9000} \approx 4166.67 \) tons.
04

Comparing with Rocket's Initial Weight

The initial weight of the rocket is about 3000 tons. Compare this with the calculated thrust in tons, which is approximately 4166.67 tons.

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Key Concepts

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

Saturn V rocket
The Saturn V rocket was a marvel of engineering, designed to carry astronauts to the moon during the Apollo missions. At its heart, the rocket was built to overcome Earth's gravitational pull and reach space. This colossal vehicle stood about 110 meters tall and had multiple stages to ensure it could reach the high speeds required for leaving Earth's orbit. Each stage would burn its fuel, and then separate, shedding weight and allowing the next stage to take over. The most impressive aspect of the Saturn V was its massive thrust force, which helped carry astronauts safely to the moon and back. By understanding the Saturn V's basic parameters, we can better comprehend the calculations related to its thrust during launch.
mass ejection rate
In rocketry, the mass ejection rate is a critical factor. It refers to how quickly the fuel is expelled from the rocket's engines. The Saturn V rocket, like many others, relied on rapid mass ejection to generate the necessary thrust for takeoff.
  • For the Saturn V, the mass ejection rate during its initial launch phase was about 15,000 kilograms per second.
  • This high rate is indicative of the powerful engines working hard to produce the necessary force to overcome gravity and atmospheric drag.
The speed at which this mass is ejected, combined with the rate itself, directly influences how much thrust is generated. Rocket engineers meticulously calculate and optimize this rate to ensure that the rocket maintains an efficient and safe launch trajectory.
newton to ton conversion
The conversion between newtons and tons is often used when discussing large forces like rocket thrust.
  • In physics, a newton is a unit of force that results from accelerating a one-kilogram mass by one meter per second squared.
  • When dealing with massive forces like those in rocketry, tons offer a simpler way to understand the scale.
For the Saturn V, the thrust calculated in newtons was around 37,500,000 N. By dividing this number by 9,000, since one ton is approximately equivalent to 9,000 newtons, we convert the force into tons: \[ rac{37,500,000 ext{ N}}{9,000 ext{ N/ton}} \ \approx 4166.67 ext{ tons} \] This conversion provides a more tangible figure, allowing for easier comparison with other massive objects, such as the Saturn V's total weight.
rocket thrust formula
The rocket thrust formula is central to understanding how rockets move. It relates the force generated by the rocket engine to the mass ejection rate and the velocity at which the mass is expelled. The formula is: \[ F = \ ext{mass ejection rate} imes ext{velocity of ejection}, \ \text{or} \ F = \ ext{ḿ} \times ext{v}_{ ext{ex}} \]
  • The mass ejection rate (\( ext{ḿ} \)) is a measure of how quickly mass exits the rocket.
  • The ejection velocity (\( v_{ ext{ex}} \)) is the speed of this expelled mass relative to the rocket.
Let's illustrate with the Saturn V: - Using a mass ejection rate of 15,000 kg/s and an ejection speed of 2,500 m/s, the calculated thrust force becomes: \[ F = 15,000 \ imes 2,500 = 37,500,000 ext{ N} \] This formula highlights the importance of both components in producing sufficient force to propel a rocket skyward.

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Most popular questions from this chapter

Many applications of conservation of momentum involve conservation of energy as well, and we haven't yet begun our discussion of energy. Nevertheless, you know enough about energy from your introductory physics course to handle some problems of this type. Here is one elegant example: An elastic collision between two bodies is defined as a collision in which the total kinetic energy of the two bodies after the collision is the same as that before. (A familiar example is the collision between two billiard balls, which generally lose extremely little of their total kinetic energy.) Consider an elastic collision between two equal mass bodies, one of which is initially at rest. Let their velocities be \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}=0\) before the collision, and \(\mathbf{v}_{1}^{\prime}\) and \(\mathbf{v}_{2}^{\prime}\) after. Write down the vector equation representing conservation of momentum and the scalar equation which expresses that the collision is elastic. Use these to prove that the angle between \(\mathbf{v}_{1}^{\prime}\) and \(\mathbf{v}_{2}^{\prime}\) is \(90^{\circ} .\) This result was important in the history of atomic and nuclear physics: That two bodies emerged from a collision traveling on perpendicular paths was strongly suggestive that they had equal mass and had undergone an elastic collision.

A particle of mass \(m\) is moving on a frictionless horizontal table and is attached to a massless string, whose other end passes through a hole in the table, where I am holding it. Initially the particle is moving in a circle of radius \(r_{\mathrm{o}}\) with angular velocity \(\omega_{\mathrm{o}},\) but I now pull the string down through the hole until a length \(r\) remains between the hole and the particle. What is the particle's angular velocity now?

Find the position of the center of mass of three particles lying in the \(x y\) plane at \(\mathbf{r}_{1}=(1,1,0)\) \(\mathbf{r}_{2}=(1,-1,0),\) and \(\mathbf{r}_{3}=(0,0,0),\) if \(m_{1}=m_{2}\) and \(m_{3}=10 m_{1} .\) Illustrate your answer with a sketch and comment.

Consider a uniform solid disk of mass \(M\) and radius \(R\), rolling without slipping down an incline which is at angle \(\gamma\) to the horizontal. The instantaneous point of contact between the disk and the incline is called \(P\). (a) Draw a free-body diagram, showing all forces on the disk. (b) Find the linear acceleration \(\dot{v}\) of the disk by applying the result \(\dot{\mathbf{L}}=\Gamma^{\text {ext }}\) for rotation about \(P .\) (Remember that \(L=I \omega\) and the moment of inertia for rotation about a point on the circumference is \(\frac{3}{2} M R^{2} .\) The condition that the disk not slip is that \(v=R \omega \text { and hence } \dot{v}=R \dot{\omega} .)\) (c) Derive the same result by applying \(\dot{\mathbf{L}}=\mathbf{\Gamma}^{\mathrm{ext}}\) to the rotation about the CM. (In this case you will find there is an extra unknown, the force of friction. You can eliminate this by applying Newton's second law to the motion of the CM. The moment of inertia for rotation about the \(\mathrm{CM}\) is \(\frac{1}{2} M R^{2}\).)

A shell traveling with speed \(v_{\mathrm{o}}\) exactly horizontally and due north explodes into two equal-mass fragments. It is observed that just after the explosion one fragment is traveling vertically up with speed \(v_{\mathrm{o}} .\) What is the velocity of the other fragment?
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