/*! 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 9 Suppose that a satellite is in f... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 9

Suppose that a satellite is in flight on the line between a planet of mass \(M_{1}\) and its moon of mass \(M_{2}\), which are a constant distance \(R\) apart. The distance \(x\) between the satellite and the planet satisfies the nonlinear second order equation \(x^{\prime \prime}=-g M_{1} x^{-2}+\) \(g M_{2}(R-x)^{-2}\) where \(g\) is the gravitational constant. Transform this equation into a system of first order equations. Find and classify the equilibrium point of the linearized system.

Short Answer

Expert verified
The linearized system's equilibrium point is a center, with the position given by \( x = \frac{\sqrt{M_{1}} R}{\sqrt{M_{1}} + \sqrt{M_{2}}} \).

Step by step solution

01

- Express the second-order equation as a system of first order equations

Introduce a new variable to help convert the second-order differential equation into a system of first-order equations. Let \( x_1 = x \) and \( x_2 = x' \). This gives us the system: \[ \begin{cases} x_1' = x_2 ewline x_2' = -g M_{1} x_1^{-2} + g M_{2} (R - x_1)^{-2} \end{cases} \]
02

- Identify the equilibrium points

To find the equilibrium points, set \( x_1' = 0 \) and \( x_2' = 0 \). This leads to: \[ \begin{cases} x_2 = 0 ewline -g M_{1} x_1^{-2} + g M_{2} (R - x_1)^{-2} = 0 \end{cases} \] Solving the second equation for \( x_1 \), we obtain \[ M_{1} x_1^{-2} = M_{2} (R - x_1)^{-2} \]
03

- Solve for the equilibrium distance

Solve the equation \( M_{1} x_1^{-2} = M_{2} (R - x_1)^{-2} \) for \( x_1 \):\[ \sqrt{M_{1}} (R - x_1) = \sqrt{M_{2}} x_1 \Rightarrow \sqrt{M_{1}} R = x_1 (\sqrt{M_{1}} + \sqrt{M_{2}}) \Rightarrow x_1 = \frac{\sqrt{M_{1}} R}{\sqrt{M_{1}} + \sqrt{M_{2}}} \]
04

- Linearize the system around the equilibrium point

Linearize the system around the equilibrium point \( x_1 = \frac{\sqrt{M_{1}} R}{\sqrt{M_{1}} + \sqrt{M_{2}}} \). Let the perturbations be \( \tilde{x_1} = x_1 - \frac{\sqrt{M_{1}} R}{\sqrt{M_{1}} + \sqrt{M_{2}}} \) and \( \tilde{x_2} = x_2 \). The linearized system becomes:\[ \frac{d}{dt} \begin{pmatrix} \tilde{x_1} ewline \tilde{x_2} \end{pmatrix} = \begin{pmatrix} 0 & 1 ewline \frac{d}{dx_1}(-g M_{1} x_1^{-2} + g M_{2} (R - x_1)^{-2})|_{x_1 = \frac{\sqrt{M_{1}} R}{\sqrt{M_{1}} + \sqrt{M_{2}}}} & 0 \end{pmatrix} \begin{pmatrix} \tilde{x_1} ewline \tilde{x_2} \end{pmatrix} \]
05

- Calculate the Jacobian matrix

Calculate the partial derivative \( \frac{d}{dx_1}(-g M_{1} x_1^{-2} + g M_{2} (R - x_1)^{-2}) \) evaluated at the equilibrium point. After computing the derivatives, the Jacobian matrix \( J \) is:\[ J = \begin{pmatrix} 0 & 1 ewline \frac{2 g M_{1}}{x_1^3} + \frac{2 g M_{2}}{(R - x_1)^3} & 0 \end{pmatrix} \]
06

- Find and classify the equilibrium point

The eigenvalues of the Jacobian matrix determine the stability of the equilibrium point. Solve the characteristic equation of \( J \):\[ \det(J - \lambda I) = \begin{vmatrix} -\lambda & 1 ewline \frac{2 g (M_{1} + M_{2})}{\left( \frac{\sqrt{M_{1}} R}{\sqrt{M_{1}} + \sqrt{M_{2}}} \right)^3} & -\lambda \end{vmatrix} = \lambda^2 - \frac{2 g (M_{1} + M_{2})}{\left( \frac{\sqrt{M_{1}} R}{\sqrt{M_{1}} + \sqrt{M_{2}}} \right)^3} = 0 \]The roots of the characteristic equation are \( \lambda = \pm i \sqrt{\frac{2 g (M_{1} + M_{2})}{\left( \frac{\sqrt{M_{1}} R}{\sqrt{M_{1}} + \sqrt{M_{2}}} \right)^3}} \). Since the eigenvalues are purely imaginary, the equilibrium point is a center.

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Key Concepts

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

Gravitational System
A gravitational system consists of objects interacting with each other through the force of gravity. In our problem, we consider a planet of mass \(M_1\) and its moon of mass \(M_2\). The satellite is influenced by both the planet's and the moon's gravitational forces. We use Newton's law of universal gravitation, which states that every point mass attracts every other point mass by a force pointing along the line intersecting both points and proportional to the product of the two masses and inversely proportional to the square of the distance between them. The equation of motion for the satellite is given by: \[ x^{\text{''}} = -g M_1 x^{-2} + g M_2 (R - x)^{-2} \], where \(g\) is the gravitational constant.
Equilibrium Point
An equilibrium point occurs when all forces and changes acting on the system balance out, causing the system to remain in a steady state. For the satellite, this means finding positions \(x\) where it experiences no net acceleration. Mathematically, we set the first-order derivatives to zero: \[ x_1' = 0, \ x_2' = 0 \]. This transforms into: \[ -g M_1 x_1^{-2} + g M_2 (R - x_1)^{-2} = 0 \]. Solving for \(x_1\), we find the equilibrium distance between the planet and the moon where the gravitational forces are balanced.
Linearization
Linearization is a method used to approximate a nonlinear system near an equilibrium point with a linear one. This simplification helps in analyzing the system's behavior around that point. After finding the equilibrium distance, \( x_1^* = \frac{\sqrt{M_1} R}{\sqrt{M_1} + \sqrt{M_2}} \), we express small deviations from this equilibrium as \( \tilde{x_1} \) and \( \tilde{x_2} \). These perturbations lead us to the linearized system defined by: \[ \frac{d}{dt} \begin{pmatrix} \tilde{x_1} \ \tilde{x_2} \end{pmatrix} = \begin{pmatrix} 0 & 1 \ \frac{d}{dx_1}(-g M_1 x_1^{-2} + g M_2 (R - x_1)^{-2}) |_{x_1^*} & 0 \end{pmatrix} \begin{pmatrix} \tilde{x_1} \ \tilde{x_2} \end{pmatrix}. \]
Jacobian Matrix
The Jacobian matrix captures the first-order partial derivatives of a vector system of first-order equations. For our linearized system, the Jacobian matrix \(J\) provides the local dynamics around the equilibrium point. Calculating it involves finding the partial derivatives of our vector functions. After some computation, the Jacobian matrix for our problem is given by: \[ J = \begin{pmatrix} 0 & 1 \ \frac{2 g M_1}{x_1^3} + \frac{2 g M_2}{(R - x_1)^3} & 0 \end{pmatrix}. \] This matrix helps us understand the system's stability and behavior near the equilibrium.
Stability Analysis
Stability analysis examines whether an equilibrium point remains stable (without changing its state) or unstable (drifting away) when a small disturbance is introduced. We use the eigenvalues of the Jacobian matrix to determine this. Solving the characteristic equation: \( \det(J - \lambda I) = 0\), we find the eigenvalues as: \( \lambda = \pm i \sqrt{\frac{2 g (M_1 + M_2)}{\left( \frac{\sqrt{M_1} R}{\sqrt{M_1} + \sqrt{M_2}} \right)^3}} \). Since the eigenvalues are purely imaginary, the equilibrium point is classified as a center. This means the satellite will exhibit periodic motion around the equilibrium point without drifting away.

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Most popular questions from this chapter

the differential equation \(x^{\prime \prime}+\sin x=0\) can be used to describe the motion of a pendulum. In this case, \(x(t)\) represents the displacement from the position \(x=0\). Represent this second order equation as a system of first order equations. Find and classify the equilibrium points of this system. Describe the physical significance of these points. Graph several paths in the phase plane of the system.

Consider the nonlinear autonomous system $$ \begin{aligned} &d x / d t=-y+x\left(1-x^{2}-y^{2}\right) \\ &d y / d t=x+y\left(1-x^{2}-y^{2}\right) \end{aligned} $$ (a) Show that \((0,0)\) is an equilibrium point of this system. Classify \((0,0)\). (b) Show that \(x=r \cos \theta, y=r \sin \theta\) transforms the system to the (uncoupled) system $$ \begin{aligned} &d r / d t=r\left(1-r^{2}\right) \\ &d \theta / d t=1 \end{aligned} $$ (c) Show that this system has solution \(r(t)=\left(1+a e^{-2 t}\right)^{-1 / 2}, \theta(t)=t+b\) so that \(x(t)=\left(1+a e^{-2 t}\right)^{-1 / 2} \cos (t+b)\), \(y(t)=\left(1+a e^{-2 t}\right)^{-1 / 2} \sin (t+b) .\) (d) Calculate \(\lim _{t \rightarrow \infty} x(t)\) and \(\lim _{t \rightarrow \infty} y(t)\) to show that all solutions approach the circle \(x^{2}+y^{2}=1\) as \(t \rightarrow \infty\). Thus \(x^{2}+\) \(y^{2}=1\) is a limit cycle.

(See Exercise 11.) Li茅nard's theorem states that if (i) \(F(x)\) is an odd function, (ii) \(F(x)\) is zero only at \(x=0, x=a, x=-a\) (for some \(a>0\) ), (iii) \(F(x) \rightarrow \infty\) monotonically for \(x>a\), and (iv) \(g(x)\) is an odd function where \(g(x)>0\) for all \(x>0\); then Li茅nard's equation has a unique limit cycle. Use Li茅nard's theorem to determine which of the following equations has a unique limit cycle: $$ \begin{aligned} &x^{\prime \prime}+\epsilon\left(1-x^{2}\right) x^{\prime}+x=0, \quad \epsilon>0 \\ &x^{\prime \prime}+3 x^{2} x^{\prime}+x^{3}=0 . \end{aligned} $$

Solve the initial-value problem to find the concentration of a substance on each side of a permeable membrane modeled by the system $$ \begin{aligned} &\frac{d x}{d t}=P\left(\frac{1}{V_{2}} y-\frac{1}{V_{1}} x\right) \\ &\frac{d y}{d t}=P\left(\frac{1}{V_{1}} x-\frac{1}{V_{2}} y\right) \end{aligned} \quad, x(0)=a, y(0)=b . $$ (a) Find \(\lim _{t \rightarrow \infty} x\) ( \(t\) ) and \(\lim _{t \rightarrow \infty} y(t)\). (b) Determine a condition so that \(x(t)>\) \(y(t)\) as \(t \rightarrow \infty\). When does \(\lim _{t \rightarrow \infty} x(t)\) \(=\lim _{t \rightarrow \infty} y(t)\) ? (c) Find a condition so that \(x(t)\) is an increasing function. Find a condition so that \(y(t)\) is an increasing function. Can these functions increase simultaneously? (d) If \(V_{1}=V_{2}\) and \(a=b\), describe what eventually happens to \(x(t)\) and \(y(t)\).

Suppose that a radioactive substance \(X\) decays into another unstable substance \(Y\), which in turn decays into a stable substance Z. Show that we can model this situation through the system of differential equations \(\left\\{\begin{array}{l}d x / d t=-a x \\ d y / d t=a x-b y \quad, \text { where } a \text { and } b \text { are pos- } \\ d z / d t=b y\end{array}\right.\) itive constants. (Assume that one unit of \(X\) decomposes into one unit of \(Y\), and one unit of \(Y\) decomposes into one unit of \(Z\).)
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