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Chapter 10: Problem 9

Rocket in resisting medium A rocket of initial mass } M , \text { of which } M - m \text { is fuel, burns } its fuel at a constant rate \(k\) and ejects the exhaust gases with constant speed \(u\). The rocket starts from rest and moves through a medium that exerts the resistance force \(-\epsilon k v\), where \(v\) is the forward velocity of the rocket, and \(\epsilon\) is a small positive constant. Gravity is absent. Find the maximum speed \(V\) achieved by the rocket. Deduce a two term approximation for \(V\), valid when \(\epsilon\) is small.

Short Answer

Expert verified
The maximum speed \(V\) achieved by the rocket, given a small positive constant \(\epsilon\), is approximately the exhaust gas speed \(u\). So, \(V \approx u\) when \(\epsilon\) is small. This short answer is derived from applying the law of motion and resistance formula in the absence of gravity, solving the presenting differential equation, and refining the solution with a two term Taylor series approximation to the exponential function for small \(\epsilon\).

Step by step solution

01

Set Up the Equation of Motion

The total external force acting on the rocket is the thrust from the burning fuel (rate \(k\) with speed \(u\)) and the resistance force. Applying Newton's second law, produce the equation of motion for the rocket: \(M \frac{dv}{dt} = ku - \epsilon k v\) where \(M\) is the rocket mass, \(v\) is the rocket's velocity, \(k\) is the fuel burn rate, \(u\) is the exhaust gas speed, and \(\epsilon\) is the constant of resistance.
02

Rewrite the Equation

Rewrite the earlier equation into a form suitable for separable variable integration. The equation becomes \(\frac{dV}{ku-\epsilon kv} = \frac{dt}{M}\). This equation implies that M is reducing at the rate \(k\), and can be replaced by \(M-m+kt\). The equation now becomes \(\frac{dV}{k(u-\epsilon v)} = \frac{dt}{M-m+kt}\)
03

Solve the Separated Differential Equation

Solving the integral, then substituting \(M-m\) by \(M'\) for convenience and \(k\) time as \(t\), the solution yields \(ln|u-\epsilon v| = -\epsilon \frac{t}{M'} + C\), which can be simplified to \(u-\epsilon v = ke^{-\epsilon \frac{t}{M'}}\). At \(t = 0\), \(v = 0\). So substituting these values gives the constant \(k = u\), thus the solution of the differential equation becomes \(u-\epsilon v = ue^{-\epsilon \frac{t}{M'}}\). From this equation, find the maximum speed \(V\) by setting the derivative of \(v\) with respect to \(t\) equal to zero, which gives the maximum velocity \(V = \frac{u}{e}\). However, this solution is valid when \(\epsilon\) is small, as specified in the exercise. Gravity is not considered here as it is mentioned in the question that the effect of gravity is absent.
04

Two Term Approximation for V

When \(\epsilon\) is very small which is the case, a two term approximation is needed. The first two terms of the Taylor series of \(e^x\) are \(1 + x\). So, apply this near \(x = 0\) for \(e^{-\epsilon t/M'}\). Hence, \(e^{-\epsilon t/M'} = 1 - \epsilon t/M'\). Substituting back to \(V = \frac{u}{e}\) and applying the approximation gives \(V = \frac{u}{1 - \epsilon t/M'}\). As \(\epsilon\) is very small, \(V\) approximately equals \(u\), which is the maximum velocity.

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Key Concepts

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

Equation of Motion
Understanding the equation of motion for a rocket traveling in a resisting medium is critical for evaluating its performance and capabilities. The equation of motion describes the relationship between the forces acting upon an object and its velocity change over time. In the context of a rocket, these forces include the thrust from the expelled fuel and the resistance force due to the medium it's moving through.

In the absence of gravity, the equation of motion for the rocket is set by balancing the internal force (rocket's thrust) and the external forces (resistance). A key aspect to remember is that this equation is based on Newton's second law, which articulates that the force acting on an object is equal to the change in momentum with respect to time. The equation is represented as:
\[M \frac{dv}{dt} = ku - \epsilon k v\]
where each symbol has a specific physical meaning:
  • \(M\) represents the rocket's mass,
  • \(v\) is the velocity,
  • \(k\) signifies the fuel burn rate,
  • \(u\) denotes the exhaust gas speed,
  • and finally, \(\epsilon\) is a constant representing the resistance of the medium.
This equation eloquently captures the dynamics of a rocket in a resistive environment and forms the foundation for analyzing rocket propulsion.
Differential Equations in Physics
Differential equations play a pivotal role in physics, allowing us to describe the variation of physical quantities over time and space. They provide a mathematical framework to model the behavior of systems under the influence of various interacting factors.

In our rocket example, the motion of the rocket can be accurately described with a differential equation, encapsulating the principles of thrust and the resistance faced. By setting up a differential equation based on the forces involved, we can solve for the rocket's velocity as a function of time. The equation of motion initially given is essentially a differential equation that has been separated for ease of integration:
\[\frac{dV}{k(u-\epsilon v)} = \frac{dt}{M-m+kt}\]
Solving this differential equation allows us to predict the rocket's behavior under the specified conditions, which is an invaluable tool for engineers and physicists working on aerospace projects. It's crucial to be adept at handling such equations to grasp the complexities of motion in resisting mediums.
Resistance Force
In rocketry, the resistance force is the opposing force that acts against the rocket's motion as it travels through a medium such as air or fluid. This force is particularly important in the study of motion because it directly affects the acceleration and final velocity that the rocket can achieve. In the given problem, the resistance force is denoted as \( -\epsilon k v \), which implies that it is proportional to the velocity \(v\) of the rocket and the rate \(k\) at which fuel is burned.

The coefficient \(\epsilon\) represents the medium's resistive properties and is a measure of how much the medium opposes the rocket's motion. It's essential to include this resistance force in the equation of motion to accurately simulate real-world scenarios. In practice, overcoming resistance is a significant engineering challenge, making it a critical factor in the design and operation of rockets.
Thrust and Rocket Propulsion
Thrust is the force that propels a rocket forward and is generated by the reaction of ejecting high-speed exhaust gases from the rocket's engine. This is described by Newton鈥檚 third law of motion: for every action, there is an equal and opposite reaction. Rocket propulsion relies on this principle, where the action is the expulsion of gas molecules at high speed, and the reaction is the thrust force that propels the rocket forward.

In mathematical terms, the thrust \(T\) produced by a rocket engine is given by the product of the mass flow rate of the exhaust \(k\) and the velocity \(u\) at which it is expelled, represented as \(T = ku\). It's this thrust that is responsible for overcoming the resistance force and increasing the velocity of the rocket.

Understanding thrust is essential for designing rockets that can achieve the desired acceleration and reach their target speeds, making it one of the most crucial aspects in the study of rocket physics. By carefully managing the fuel burn rate and exhaust velocity, engineers can optimize the thrust to ensure efficient and effective rocket propulsion.

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Most popular questions from this chapter

Show that, if a system moves periodically, then the average of the total external force over a period of the motion must be zero. A juggler juggles four balls of masses \(M, 2 M, 3 M\) and \(4 M\) in a periodic manner. Find the time average (over a period) of the total force he applies to the balls. The juggler wishes to cross a shaky bridge that cannot support the combined weight of the juggler and his balls. Would it help if he juggles his balls while he crosses?

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