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Chapter 4: Problem 56

Suppose a rocket is launched from the surface of the earth. If we disregard the effect of the sun, the work required to propel the rocket from the surface ( 3960 miles from the center of the earth) to \(x\) miles above the surface is given by the formula $$ W(x)=G M m\left(\frac{1}{3960}-\frac{1}{x}\right) $$ where \(G, M\), and \(m\) are constants. Find the work required to send the rocket from here to the end of the universe. Leave your answer in terms of \(G, M\), and \(m\).

Short Answer

Expert verified
The work required is \( \frac{G M m}{3960} \).

Step by step solution

01

Understanding the Formula

The given formula for work is \( W(x) = G M m \left( \frac{1}{3960} - \frac{1}{x} \right) \). Here, \( G, M, \) and \( m \) are constants that represent gravitational constant, mass of the Earth, and mass of the rocket, respectively. The objective is to determine the work \( W \) required to send the rocket to an infinite distance, i.e., to the end of the universe.
02

Analyzing Work to Infinity

Sending the rocket to the end of the universe implies taking the limit as \( x \) approaches infinity. Thus, we need to evaluate \( W(\infty) = G M m \left( \frac{1}{3960} - \frac{1}{\infty} \right) \). Since \( \frac{1}{\infty} = 0 \), the expression simplifies considerably.
03

Simplifying the Expression

Substituting \( \frac{1}{\infty} = 0 \) into the expression, we have:\[W(\infty) = G M m \left( \frac{1}{3960} - 0 \right) = G M m \times \frac{1}{3960}.\]Thus, the work required is simplified to \( \frac{G M m}{3960} \).

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Key Concepts

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

Gravitational Constant
The gravitational constant, denoted by \( G \), is a crucial value in the physics realm, underpinning the laws of gravitation as we know them today. This constant acts as a scaling factor in Newton's law of universal gravitation:
  • It is used to calculate the gravitational attraction between two masses.
  • The value of \( G \) is approximately \( 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \).
  • It is a measure of how strongly bodies of mass influence one another through gravitational force.
In our exercise, the presence of \( G \) in the work formula exemplifies its role in calculating the gravitational pull between Earth and a rocket. This influences the amount of energy needed to move the rocket to significant distances away from the planet. Understanding the constants involved in equations like these can provide deeper insights into the nature of forces that we contend with even beyond calculations.
Work-Energy Principle
The work-energy principle is a key concept that helps us understand the relationship between force applied, work done, and energy changes associated with motion. This principle states that the work done by all forces acting on an object equals the change in its kinetic energy.
  • Work is done when energy is transferred from one system to another, usually through force causing movement.
  • In our context, work \( W(x) \) represents the energy needed to move a rocket against Earth's gravitational pull.
  • By calculating this work, we can determine how much energy is required to reach varying heights, eventually extending to theoretically infinite distances.
This principle aids in understanding how much energy would be needed to power the rocket as it counters gravity continuously. Grasping this principle is essential to appreciate the connection between potential and kinetic energy and how they interchange.
Limits in Calculus
Limits in calculus are powerful tools allowing us to analyze behavior as variables approach certain values, including infinity. In this exercise, we used limits to explore the conditions necessary for the rocket making its journey to infinity.
  • Introducing limits allows us to deal with changes in distance represented by \( x \) as it extends towards infinity.
  • Calculating \( \lim_{x \to \infty} \left( \frac{1}{x} \right) = 0 \) showcases how the infinitesimal values diminish to zero.
  • Using limits, we simplify complex ideas, such as determining how much work \( W(\infty) \) is truly needed to overcome gravitational forces to send the rocket infinitely far.
By understanding the role of limits, students can tackle advanced concepts in calculus and physics. These evaluations are essential when working with infinite spaces and understanding the practicalities of moving bodies substantial distances across the cosmos.

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Most popular questions from this chapter

Use the formulas for the function and its first and second derivatives as an aid in sketching the graph of the given function. Note all relevant properties listed in Table 4.1. $$ \begin{aligned} &f(x)=\frac{x}{\sqrt{1-x}}, f^{\prime}(x)=\frac{2-x}{2(1-x)^{3 / 2}}, \text { and } \\ &f^{\prime \prime}(x)=\frac{4-x}{4(1-x)^{5 / 2}} \end{aligned} $$

A cable of a suspension bridge hangs in the shape of a parabola (which is the graph of a second-degree polynomial). Consider a cable that joins two points at \((0,2)\) and \((2,1)\), and passes through the point \((1,0)\), and let \(l\) be the line joining \((0,2)\) and \((2,1)\) (Figure 4.18). Find the point on the cable whose vertical distance from \(l\) is greatest.

Find the given limit. $$ \lim _{x \rightarrow \infty} e^{-x} $$

Plot the graph of \(f^{\prime}\), and then use the Newton-Raphson method to approximate all values of \(c\) for which \((c, f(c))\) is an inflection point. Continue until the output of the calculator does not change. $$ f(x)=x^{7}-x^{5}+x^{3}+3 x^{2} $$

Suppose an object is released and moves through a viscous fluid that tends to resist the motion of the object. Then the velocity \(v\) increases with time, and may approach a terminal velocity \(v_{T}\), which depends on the mass of the object and the viscosity of the fluid. If the fluid's resistance is proportional to the object's velocity, then as a function of time \(t\) the velocity is given by $$ v=v_{T}\left(1-e^{-g t / v_{T}}\right) $$ where \(g=9.8\) (meters per second per second) is the acceleration due to gravity. a. Find \(\lim _{t \rightarrow \infty} v(t)\). b. If the object is a tiny fog droplet (which is frequently on the order of \(5 \times 10^{-6}\) meters in radius), and is falling in the sky near earth, then a reasonable value for the terminal velocity is \(2.7 \times 10^{-2}\) meters per second (which is equivalent to 1 meter every 37 seconds). If the fog droplet begins falling at time \(t=0\), determine how long it takes for the velocity of the droplet to reach half of the terminal velocity.
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