91Ó°ÊÓ

Rocket Science. A rocket has total mass \(M_{i}=360 \mathrm{kg}\) , including 330 \(\mathrm{kg}\) of fuel and oxidizer. In interstellar space it starts from rest, turns on its engine at time \(t=0\) , and puts out exhaust with relative speed \(v_{e}=1500 \mathrm{m} / \mathrm{s}\) at the constant rate \(k=2.50 \mathrm{kg} / \mathrm{s}\) . The fuel will last for an actual burn time of \(330 \mathrm{kg} /(2.5 \mathrm{kg} / \mathrm{s})=132 \mathrm{s},\) but define a "projected depletion time" as \(T_{p}=M_{i} / k=\) 144 s. (This would be the burn time if the rocket could use its payload and fuel tanks as fuel, and even the walls of the combustion chamber.) (a) Show that during the burn the velocity of the rocket is given as a function of time by $$ v(t)=-v_{e} \ln \left[1-\left(t / T_{p}\right)\right] $$ (b) Make a graph of the velocity of the rocket as a function of time for times running from 0 to 132 s. (c) Show that the acceleration of the rocket is $$ a(t)=v_{e} /\left(T_{p}-t\right) $$ (d) Graph the acceleration as a function of time. (e) Show that the position of the rocket is $$ x(t)=v_{e}\left(T_{p}-t\right) \ln \left[1-\left(t / T_{p}\right)\right]+v_{e} t $$ (f) Graph the position during the burn.

Step by step solution

01

Derive rocket velocity as a function of time

Use the rocket equation, which states that the change in rocket velocity is equal to the exhaust velocity times the natural logarithm of the initial mass over the remaining mass. Since we have the mass loss rate (\(k\to 2.5 \text{kg/s}\) and the projected depletion time \(T_p = M_i/k = 144\text{s}\), we can express the remaining mass as a function of time \(M(t) = M_i - kt\). The total change in rocket velocity over the burn time is then expressed as \(v(t) = v_e \ln \left(\frac{M_i}{M_i-kt}\right) = v_e \ln \left[1 - \left(\frac{t}{T_p}\right)\right]\).

02

Graph rocket velocity as a function of time

Create a graph where time (in seconds, from 0 to 132 s) is on the x-axis, and rocket velocity (in m/s) is on the y-axis. Use the velocity function derived in Step 1 to plot the graph.

03

Derive rocket acceleration as a function of time

We get the acceleration by differentiating the velocity function with respect to time. Differentiating the velocity equation \(v(t) = -v_e \ln \left[1 - \left(\frac{t}{T_p}\right)\right]\) with respect to time \(t\), we get the acceleration \(a(t) = \frac{dv}{dt} = \frac{v_e}{T_p - t}\).

04

Graph rocket acceleration as a function of time

Create a graph where time (in seconds, from 0 to 132 s) is on the x-axis, and rocket acceleration (in m/s^2) is on the y-axis. Use the acceleration function obtained in Step 3 to plot the graph.

05

Derive rocket position as a function of time

The position is found by integrating the velocity function. Integrate the velocity equation to obtain the position function \[ x(t) = \int v(t) dt = v_e \int \left[- \ln \left[1 - \left(\frac{t}{T_p}\right)\right]\right] dt + \int dt\]. The integration yields the position function \(x(t) = v_e\left(T_p - t\right) \ln \left[1 - \left(\frac{t}{T_p}\right)\right] + v_et\).

06

Graph rocket position as a function of time

Create a graph where time (in seconds, from 0 to 132 s) is on the x-axis, and rocket position (in meters) is on the y-axis. Use the position function obtained in Step 5 to plot the graph.

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Key Concepts

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

Rocket Equation

The rocket equation is integral in understanding how rockets accelerate in space. This principle, also known as the Tsiolkovsky rocket equation, characterizes how the speed of a rocket changes as it burns fuel. The fundamental idea is that as a rocket burns its fuel, it loses mass, but it also generates thrust by expelling exhaust gases in the opposite direction.

In our exercise, the rocket equation is used to derive the velocity of the rocket as a function of time. Specifically, as the rocket consumes fuel at a constant rate, its mass decreases, which can be expressed as the initial mass (\( M_i \) ) minus the product of mass loss rate (\( k \) ) and time (\( t \) ), or \( M(t) = M_i - kt \) . The change in velocity (\( v(t) \) ) is then directly related to the exhaust velocity (\( v_e \) ), and the proportion of the remaining mass relative to the initial mass, given by the natural logarithm function.

Exhaust Velocity

The exhaust velocity (\( v_e \) ) is an essential parameter in rocketry—it defines the speed at which the propellant is ejected from the rocket's engine. Higher exhaust velocities mean that the rocket’s engine is more efficient at converting fuel into kinetic energy, leading to a more effective propulsion system.

In the context of our problem, \( v_e \) is the fixed speed of 1500 meters per second, which is a significant factor in calculating the rocket's velocity and position over time. The exhaust velocity not only impacts the rocket's acceleration directly but also indirectly affects the rocket's changing mass throughout the burn, which is why it appears in all three of the equations for velocity, acceleration, and position we are working with.

Mass Loss Rate

The mass loss rate (\( k \) ) in rocketry details the rate at which a rocket loses mass due to fuel consumption. It quantifies the amount of fuel burned by the rocket per unit time and is a crucial metric for determining how the mass of the rocket changes throughout its burn. A high mass loss rate signifies that the rocket expends its fuel quickly, while a lower rate indicates a longer burn time.

The importance of the mass loss rate is highlighted in our exercise by demonstrating that it is part of the equation that describes the rocket's velocity as a function of time. It is used to determine the 'projected depletion time' (\( T_p \)), which represents the time it would take for the rocket to consume all its mass if it could use all parts of itself as fuel. This concept helps explain the dynamics of rocket motion and is pivotal for calculating the rocket’s velocity, acceleration, and displacement during the fuel burn phase.