The differential equation for the path of a planet around the sun (or any object in an inverse square force field) is, in polar coordinates, $$ \frac{1}{r^{2}} \frac{d}{d \theta}\left(\frac{1}{r^{2}} \frac{d r}{d \theta}\right)-\frac{1}{r^{3}}=-\frac{k}{r^{2}} $$ Make the substitution \(u=1 / r\) and solve the equation to show that the path is a conic section.

Short Answer

Expert verified
The path is a conic section described by \( u = A\cos(\theta - \theta_0) + k \) where \(u = \frac{1}{r}\).

Step by step solution

01

Substitute and Simplify

Let's make the given substitution first: let \(u = \frac{1}{r}\), then \(r = \frac{1}{u}\). We need to express \(\frac{d r}{d \theta}\) and higher derivatives in terms of \(u\). Use the chain rule, recognizing \( \frac{d r}{d \theta} = \frac{d}{d \theta}\left(\frac{1}{u}\right) = -\frac{1}{u^2} \frac{d u}{d \theta} \).
02

Find Second Derivative

Differentiate \( \frac{d r}{d \theta} \) again to find \(\frac{d^2 r}{d \theta^2} \). We have: \( \frac{d^2 r}{d \theta^2} = \frac{d}{d \theta} \left( -\frac{1}{u^2} \frac{d u}{d \theta} \right) \). Using the product rule: \( \frac{d^2 r}{d \theta^2} = \frac{d}{d \theta} \left( -\frac{1}{u^2} \right) \cdot \frac{d u}{d \theta} + \left( -\frac{1}{u^2} \right) \cdot \frac{d^2 u}{d \theta^2} \).
03

Apply the Derivatives

Simplify \(\frac{d^2 r}{d \theta^2} \). The first term becomes \(-\left( -\frac{2}{u^3} \cdot \frac{d u}{d \theta} \right) \cdot \frac{d u}{d \theta} = \frac{2}{u^3} \left( \frac{d u}{d \theta} \right)^2 \). The second term remains \( -\frac{1}{u^2} \cdot \frac{d^2 u}{d \theta^2} \), thus: \[ \frac{d^2 r}{d \theta^2} = \frac{2}{u^3} \left( \frac{d u}{d \theta} \right)^2 - \frac{1}{u^2} \frac{d^2 u}{d \theta^2} \]
04

Substitute Derivatives into Original Equation

Substitute the expressions for \( \frac{1}{r} \) and \( \frac{d^2 r}{d \theta^2} \) into the original differential equation. The original equation simplifies to: \[ \frac{1}{\left(\frac{1}{u}\right)^2} \frac{d}{d \theta} \left( \frac{1}{\left(\frac{1}{u}\right)^2} \frac{d \left(\frac{1}{u}\right)}{d \theta} \right) - \frac{1}{\left(\frac{1}{u}\right)^3} = -\frac{k}{\left(\frac{1}{u}\right)^2} \].
05

Simplify Equation

Simplify the equation further: \[ u^2 \left[ \frac{d}{d \theta} \left( u^2 \left( -\frac{1}{u^2} \frac{d u}{d \theta} \right) \right) \right] - u^3 = -\frac{k}{u^2} \].
06

Solve Differential Equation

Recognize that \( \frac{d}{d\theta} \left( -\frac{du}{d\theta} \right) = -\frac{d^2u}{d\theta^2} \). Thus, the equation simplifies to: \[ -u^2 \frac{d^2u}{d\theta^2} + 2\frac{du}{d\theta}\frac{du}{d\theta} - u^3 = -\frac{k}{u^2} \].
07

Rearrange Equation

Group like terms: \[\frac{d^2 u}{d \theta^2} + u = k \]. This is a simple harmonic oscillator equation.
08

General Solution

The general solution to this differential equation is given by: \[ u = A\cos(\theta - \theta_0) + k \] where \(A\) and \(\theta_0\) are constants determined by initial conditions.
09

Conclusion

Since \(u = \frac{1}{r}\), substituting back, \[ \frac{1}{r} = A\cos(\theta - \theta_0) + k \]. Multiplying everything by \(r\), one gets: \[ 1 = Ar\cos(\theta - \theta_0) + kr \], which describes a conic section.

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 Vaia!

Key Concepts

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

inverse square law
The inverse square law is fundamental in understanding the motion of planets and other celestial bodies. It states that a force (such as gravity) between two masses is inversely proportional to the square of the distance between them. In formula terms, if the force is denoted as F and the distance as r, we can write: \( F \propto\frac{1} {r^2} \). This means that if the distance between two objects doubles, the gravitational force becomes four times weaker.

The inverse square law is pivotal in deriving the differential equations describing planetary motion. For example, Newton used this concept to explain why planets orbit the sun. By substituting the distances in Newton's Law of Universal Gravitation into our planetary motion equation, we align our understanding of forces with observed celestial paths.
conic sections
Conic sections are the curves obtained by intersecting a plane with a cone. These include circles, ellipses, parabolas, and hyperbolas. When dealing with planetary motion, especially in polar coordinates, we often find that the paths taken by celestial bodies are conic sections.

The equation derived from our differential equation simplifies into the form of conic sections. When we say that the path of a planet is a conic section, it means it can be a circle, an ellipse, a parabola, or a hyperbola, depending on the energy and conditions of the orbit. For example, if the energy of a planetary body relative to the central mass (like the sun) is negative, the path is an ellipse.

Understanding conic sections helps us predict the behavior of orbits and plan space missions as they reflect natural laws governing gravitational forces.
simple harmonic oscillator
A simple harmonic oscillator is a system where the restoring force is directly proportional to the displacement from equilibrium, leading to oscillatory motion. In our differential equation context, it appears when we rearrange terms to get the form \(\frac{d^2u}{d\theta^2}\ + \ u = k\)\.

This form closely resembles the simple harmonic oscillator equation: \( \frac{d^2x}{dt^2} + \omega^2 x = 0 \), where \[ \omega \] is the angular frequency. Solving it gives us sinusoidal solutions like \( x(t) = A\cos(\theta - \theta_0) + k \), indicating repetitive orbital paths.

Recognizing our planetary motion equation as a simple harmonic oscillator helps identify the sinusoidal nature of the orbit, harmonizing the planets' positions with predictable, periodic motion. The constants in the solution's general form are determined by initial conditions, aligning the theoretical paths with real-world observations.

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

If \(10 \mathrm{~kg}\) of rock salt is placed in water, it dissolves at a rate proportional to the amount of salt still undissolved. If \(2 \mathrm{~kg}\) dissolve during the first 10 minutes, how long will it be until only 2 kg remain undissolved?

Identify each of the differential equations as to type (for example, separable, linear first order, linear second order, etc.), and then solve it. \(\frac{d^{2} r}{d t^{2}}-6 \frac{d r}{d t}+9 r=0\)

Find the "general solution" of the equation \(y^{\prime}=\sqrt{y}\) by separation of variables. Find a particular solution satisfying \(y=0\) when \(x=0\). Show that the singular solution \(y=0\) cannot be obtained from the general solution. Sketch graphs of the "general solution" for several values of the arbitrary constant, and observe that each of them is tangent to the singular solution. Thus there are two solutions passing through any point on the \(x\) axis; in particular, there are two solutions satisfying \(x=y=0 .\) Problems 17 and 18 are physical problems leading to this differential equation.

Find the "general solution" (that is, a solution containing an arbitrary constant) of each of the following differential equations, by separation of variables. Then find a particular solution of each equation satisfying the given boundary conditions. \(x \sqrt{1-y^{2}} d x+y \sqrt{1-x^{2}} d y=0, \quad y=\frac{1}{2}\) when \(x=\frac{1}{2}\).

Consider an equation for damped forced vibrations (mechanical or electrical) in which the right-hand side is a sum of several forces or emfs of different frequencies. For example, in (6.32) let the right-hand side be $$ F_{1} e^{i \omega_{1}^{\prime} t}+F_{2} e^{\operatorname{ior}_{2} l}+F_{3} e^{i \operatorname{cog} T} $$ Write the solution by the principle of superposition. Suppose, for given \(\omega_{1}^{\prime}, \omega_{2}^{\prime}, \omega_{3}^{\prime}\), that we adjust the system so that \(\omega=\omega_{1}^{\prime}\); show that the principal term in the solution is then the first one. Thus the system acts as a "filter" to select vibrations of one frequency from a given set (for example, a radio tuned to one station selects principally the vibrations of the frequency of that station).

See all solutions

Recommended explanations on Combined Science 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.

Sign-up for free