Problem 61 - Calculus Early Transcendentals

Question

A rocket, fired upward from rest at time \(t=0\), has an initial mass of \(m_{0}\) (including its fuel). Assuming that the fuel is consumed at a constant rate \(k,\) the mass \(m\) of the rocket, while fuel is being burned, will be given by \(m=m_{0}-k t\). It can be shown that if air resistance is neglected and the fuel gases are expelled at a constant speed \(c\) relative to the rocket, then the velocity \(v\) of the rocket will satisfy the equation $$ m \frac{d v}{d t}=c k-m g $$ where \(g\) is the acceleration due to gravity. (a) Find \(v(t)\) keeping in mind that the mass \(m\) is a function of \(t .\) (b) Suppose that the fuel accounts for \(80 \%\) of the initial mass of the rocket and that all of the fuel is consumed in 100 s. Find the velocity of the rocket in meters per second at the instant the fuel is exhausted. [Note: Take \(\left.g=9.8 \mathrm{m} / \mathrm{s}^{2} \text { and } c=2500 \mathrm{m} / \mathrm{s} .\right]\)

Step-by-Step Solution

Answer

The velocity of the rocket when the fuel is exhausted is approximately 5934.5 m/s.

Step 1: Identify the equation to solve

We need to find the velocity \(v(t)\) of the rocket. The given differential equation is \( m \frac{d v}{d t} = c k - m g \). Since the mass \(m\) is given by \(m(t) = m_0 - kt\), we substitute this into the differential equation.

Step 2: Substitute mass expression into the equation

Replace \(m\) with \(m(t) = m_0 - kt\) in the equation \(m \frac{dv}{dt} = ck - mg\) to get:\[(m_0 - kt) \frac{dv}{dt} = ck - (m_0 - kt)g.\] Simplify this to: \[\frac{dv}{dt} = \frac{ck - m_0 g + kgt}{m_0 - kt}.\]

Step 3: Solve the differential equation

This is a separable differential equation. Rewriting gives:\[\frac{dv}{dt} = \frac{ck - m_0 g}{m_0 - kt} + \frac{kg t}{m_0 - kt}.\] Integrating both sides with respect to \(t\) results in:\[v(t) = \int \left(\frac{ck - m_0 g}{m_0 - kt} + g\right) dt.\] Resulting in:\[v(t) = -c \ln(m_0 - kt) - g t + C,\] where \(C\) is the constant of integration.

Step 4: Determine the constant of integration

To find \(C\), use the initial condition \(v(0) = 0\) because the rocket starts from rest. This gives:\[0 = -c \ln(m_0) + C.\] Therefore, \(C = c \ln(m_0).\)

Step 5: Find the velocity equation

Substitute \(C = c \ln(m_0)\) back into the velocity equation:\[v(t) = -c \ln(m_0 - kt) + c \ln(m_0) - g t.\] Simplify using logarithmic properties:\[v(t) = c \ln\left(\frac{m_0}{m_0 - kt}\right) - g t.\]

Step 6: Calculate velocity at fuel exhaustion

First, calculate time to fuel exhaustion: \(t_f = \frac{80}{100} m_0/k = 0.8m_0/k\). Substitute \(t = 100\) seconds when all fuel is exhausted:\[v(100) = c \ln\left(\frac{m_0}{m_0 - 100k}\right) - g \times 100.\] Use the given values: \(c = 2500 \text{ m/s}\), \(g = 9.8 \text{ m/s}^2\). Since \(m_0 = 5m_f\) and \(m_f = 0.2m_0\), \(k = 0.8m_0/100\), resulting in:\[v(100) = 2500 \ln(5) - 980.\] Calculate numerically for final result.

Key Concepts

Differential Equations

Differential equations are mathematical tools we use to describe the rate at which something changes. In the case of rocket motion, we're particularly interested in how the velocity of the rocket changes over time. A differential equation involves derivatives, which are expressions for these rates of change, like how fast position changes with time (velocity), or how fast the velocity itself changes (acceleration).
In our problem, the differential equation reads \(m \frac{dv}{dt} = ck - mg\), where \(dv/dt\) represents the change in velocity over time, \(c\) is the speed at which fuel gases are expelled, \(k\) is the rate at which fuel mass decreases, and \(g\) is the acceleration due to gravity. Understanding how this equation encapsulates the forces at play is key to solving for the rocket's velocity.

Separable Equations

When a differential equation allows us to manipulate it so that each variable can be handled separately, it is identified as separable. This characteristic simplifies solving them. We can rearrange the terms to have all expressions involving one variable (say \(v\)) on one side of the equation, and those involving the other variable (usually \(t\)) on the opposite side.
In rocket motion, after expressing the mass \(m(t) = m_0 - kt\), we rewrote our equation to isolate \(dv/dt\) on one side and expressions involving time \(t\) on the other. This separability allows us to integrate both sides individually, thereby finding a function \(v(t)\) for the rocket's velocity. By simplifying intricate interactions into more manageable terms, separable equations make finding solutions more approachable.

Integration Techniques

Integration is the mathematical technique that allows us to determine a whole from its parts, much like assembling a puzzle. In the rocket problem, integration is employed to solve the simplified differential equation in terms of \(v(t)\), which represents the rocket's velocity over time. Our main task is to integrate functions of the form \(\frac{dv}{dt}\) to derive a velocity expression.
During integration, one common technique is partial fraction decomposition, which simplifies complex fractions, making their integration feasible. Another important technique, used in our solution, involves integrating logarithms which arise when we separate variables, provided 〈u〉 using \(\ln\) functions. Calculating definite integrals with these techniques is essential for understanding motion physics thoroughly.

Physics Applications in Calculus

In physics, calculus is the backbone of solving real-world problems involving rates of change in various physical quantities. For the rocket, understanding how forces affect velocity is vital. We utilized the rocket equation to relate forces to motion through gravitational force (gravity \(g\), acting down) and propulsion (fuel gases expelled at speed \(c\)).
Excluding air resistance and external forces emphasizes the primary principles of motion. Using calculus, specifically differential equations and integration, we model complex systems simply and clearly. These mathematical tools are invaluable when predicting physical phenomena, allowing us to calculate the rocket’s speed when the fuel runs out, considering initial and changing conditions like fuel consumption and gravity.