/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 Rocket Science. A rocket has tot... [FREE SOLUTION] | 91Ó°ÊÓ

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.

Short Answer

Expert verified
From the given conditions, the velocity, acceleration and position of the rocket are derived as functions of time. The equations obtained are \(v(t) = - v_e * ln[1 - (t/T_p)]\), \(a(t) = v_e/(T_p - t)\), and \(x(t) = v_e*(T_p - t) * ln[1 - (t/T_p)] + v_e*t\) respectively. These equations can then be graphed against time.

Step by step solution

01

Understand the Given Variables

Ensure to understand each variable provided. \(M_i\) represents the rocket's initial total mass, including fuel, which is 360 kg. \(v_e\) represents the exhaust velocity, which is 1500 m/s. \(k\) is the rate at which the rocket expels exhaust, which is 2.50 kg/s. \(T_p\) represents the projected depletion time, calculated as \(M_i / k\), which comes to 144 s.
02

Derivation of Rocket's Velocity Over Time

To find the rocket's velocity over time, we need to integrate the Tsiolkovsky rocket equation, which says that \(v(t) = v_e * ln(M_i/(M_i - k*t))\). This simplifies to \(v(t) = - v_e * ln[1 - (t/T_p)]\) as required. Notice that the equation changes format after taking into account that the change in mass equals the initial mass minus the burnt fuel.
03

Graph of Rocket's Velocity Over Time

Plot the equation \(v(t) = - v_e * ln[1 - (t/T_p)]\) from t=0 to t=132 s to get a graph representing the rocket's velocity against time.
04

Derivation of Rocket's Acceleration Over Time

Acceleration is the derivative of velocity with respect to time. Therefore, differentiate \(v(t) = - v_e*ln[1 - (t/T_p)]\) with respect to time to get \(a(t) = v_e/(T_p - t)\).
05

Graph of Rocket's Acceleration Over Time

Plot the equation \(a(t) = v_e/(T_p - t)\) from t=0 to t=132 s to graphically represent the rocket's acceleration against time.
06

Derivation of Rocket's Position Over Time

As position is the integral of velocity with respect to time, integrate the velocity equation \(v(t) = - v_e * ln[1 - (t/T_p)]\) with respect to time to get \(x(t) = v_e*(T_p - t) * ln[1 - (t/T_p)] + v_e*t\).
07

Graph of Rocket's Position During Burn

Plot the equation \(x(t) = v_e*(T_p - t) * ln[1 - (t/T_p)] + v_e*t\) from t=0 to t=132 s to graphically represent the rocket's position during the burn.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Tsiolkovsky Rocket Equation
The Tsiolkovsky rocket equation is a fundamental principle in rocket science that describes how a rocket's velocity changes as it expels fuel. It allows us to calculate the velocity of a rocket as a function of the mass of fuel burnt. This equation is essential because it lays the theoretical foundation for understanding rocket motion and efficiency. In our exercise, we started with a total mass of the rocket, including the fuel. By integrating the Tsiolkovsky equation, we derived the formula:
  • \(v(t) = -v_e \ln \left[1-\left(t / T_p\right)\right]\)
This particular form tells us how the velocity of the rocket increases as the rocket consumes more fuel, understanding that the velocity is dependent on the natural logarithm of the mass ratio. The negative sign indicates the initial rest state.
Velocity-Time Graph
A velocity-time graph illustrates how a rocket's speed changes over time during the burn. Using the formula derived from the Tsiolkovsky rocket equation,
  • \(v(t) = - v_e \ln[1 - (t/T_p)]\)
we can plot the rocket's velocity from the start (t=0) until the fuel is depleted (t=132s). The curve shows an increasing velocity as the mass of the rocket decreases due to burning of fuel. The rate of increase decreases over time because as more fuel is burnt, the incremental gain in speed per unit time reduces. Observing this graph, one can see the characteristic hyperbolic nature, tracing velocity rise through continuous fuel consumption.
Acceleration-Time Graph
Acceleration-time graphs provide a visual understanding of how a rocket's acceleration changes during the burn. Acceleration is the derivative of velocity with respect to time. From our earlier velocity equation, differentiating gives us the acceleration equation:
  • \(a(t) = v_e/(T_p - t)\)
By plotting this from t=0 to t=132 seconds, we see that the acceleration increases over time. This may seem counterintuitive; however, as the rocket becomes lighter with less fuel, it responds more to the constant thrust provided by the engines. Hence, the acceleration goes to infinity as the fuel approaches depletion, underlining the inverse relationship between time and rocket acceleration on this graph.
Position-Time Graph
The position-time graph helps us visualize the path taken by the rocket as a function of time during the burn. To find the rocket's position, we integrate the velocity equation over time. This gives the formula:
  • \(x(t) = v_e(T_p - t) \ln[1 - (t/T_p)] + v_e t\)
When plotted, this graph shows the rocket’s total displacement over time, starting at zero and increasing continuously. The curvature reflects how the rocket gains distance progressively, at a pace influenced by decreasing mass and increasing acceleration. The position increases steadily, suggesting a nonlinear path influenced by the changing velocity function throughout the fuel consumption process.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A ball of mass \(0.200 \mathrm{kg}\) has a velocity of \(150 \hat{\mathbf{i}} \mathrm{m} / \mathrm{s} ;\) a ball of mass \(0.300 \mathrm{kg}\) has a velocity of \(-0.400 \hat{\mathbf{i}} \mathrm{m} / \mathrm{s} .\) They meet in a head-on elastic collision. (a) Find their velocities after the collision. (b) Find the velocity of their center of mass before and after the collision.

A railroad car of mass \(2.50 \times 10^{4} \mathrm{kg}\) is moving with a speed of \(4.00 \mathrm{m} / \mathrm{s} .\) It collides and couples with three other coupled railroad cars, each of the same mass as the single car and moving in the same direction with an initial speed of \(2.00 \mathrm{m} / \mathrm{s} .\) (a) What is the speed of the four cars after the collision? (b) How much mechanical energy is lost in the collision?

A billiard ball moving at \(5.00 \mathrm{m} / \mathrm{s}\) strikes a stationary ball of the same mass. After the collision, the first ball moves, at \(4.33 \mathrm{m} / \mathrm{s},\) at an angle of \(30.0^{\circ}\) with respect to the original line of motion. Assuming an elastic collision (and ignoring friction and rotational motion), find the struck ball's velocity after the collision.

In a slow-pitch softball game, a \(0.200-\mathrm{kg}\) softball crosses the plate at \(15.0 \mathrm{m} / \mathrm{s}\) at an angle of \(45.0^{\circ}\) below the horizontal. The batter hits the ball toward center field, giving it a velocity of \(40.0 \mathrm{m} / \mathrm{s}\) at \(30.0^{\circ}\) above the horizontal. (a) Determine the impulse delivered to the ball. (b) If the force on the ball increases linearly for \(4.00 \mathrm{ms}\), holds constant for \(20.0 \mathrm{ms},\) and then decreases to zero linearly in another \(4.00 \mathrm{ms},\) what is the maximum force on the ball?

A rod of length \(30.0 \mathrm{cm}\) has linear density (mass-perlength) given by $$ \lambda=50.0 \mathrm{g} / \mathrm{m}+20.0 x \mathrm{g} / \mathrm{m}^{2} $$ where \(x\) is the distance from one end, measured in meters. (a) What is the mass of the rod? (b) How far from the \(x=0\) end is its center of mass?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.