/*! 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 16 By 1619 Johannes Kepler had comp... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 16

By 1619 Johannes Kepler had completed the first accurate mathematical model describing the motion of planets around the sun. His model consisted of three laws that, for the first time in history, made possible the accurate prediction of future locations of planets. Kepler's third law related the period (the length of time required for a planet to complete a single trip around the sun) to the mean distance \(D\) from the planet to the sun. In particular, he stated that the period \(P\) is proportional to \(D^{1.5}\). a. Neptune is about 30 times as far from the sun as is the Earth. How long does it take Neptune to complete an orbit around the sun? (Hint: The period for the Earth is 1 year. If the distance is increased by a factor of 30 , by what factor will the period be increased?) b. The period of Mercury is about 88 days. The Earth is about 93 million miles from the sun. How far is Mercury from the sun? (Hint: The period of Mercury is different from that of the Earth by a factor of \(\frac{88}{365}\).)

Short Answer

Expert verified
Neptune's orbit takes about 164.3 years. Mercury is approximately 36.27 million miles from the sun.

Step by step solution

01

Understanding Kepler's Third Law

Kepler's Third Law states that the square of the period \( P \) of orbit is proportional to the cube of the semi-major axis of its orbit, i.e., the average distance \( D \) from the planet to the sun. The relation can be expressed as \( P^2 \propto D^3 \) or \( P \propto D^{1.5} \). This means if the distance is multiplied by a factor, the period is multiplied by the square root of the same factor cubed.
02

Determine the Period of Neptune

Earth takes 1 year to orbit the sun, with \( D = 1 \). For Neptune, \( D = 30 \). Using the relationship \( P \propto D^{1.5} \), Neptune's period will be \( P_{\text{Neptune}} = P_{\text{Earth}} \times 30^{1.5} = 1 \times 30^{1.5} \). Therefore, \( P_{\text{Neptune}} = 30^{1.5} = \sqrt{30^3} \). Calculate the value of \( \sqrt{30^3} \) to find Neptune's orbital period in years.
03

Calculate \( 30^{1.5} \)

Calculate 30 raised to the power of 1.5 by first finding \( 30^3 \). Use the relation \( 30^3 = 30 \times 30 \times 30 = 27000 \). Then find the square root: \( \sqrt{27000} \approx 164.3 \). Thus, the orbital period for Neptune is approximately 164.3 years.
04

Understanding Mercury's Situation

For Mercury, we need to find \( D \) knowing \( P = 88 \) days. First, express \( P \) in years: \( \frac{88}{365} \approx 0.24 \) years. Now use \( P^2 \propto D^3 \) to express the distance relative to Earth. Thus, \( (0.24)^2 \propto D^3 \).
05

Calculate Mercury's Average Distance

Using the proportion, \( (0.24)^2 = D^3 \times (1)^3 \), solve for \( D \): \( D^3 = (0.24)^2 = 0.0576 \). Therefore, \( D = \sqrt[3]{0.0576} \approx 0.39 \) times the Earth's distance to the sun. Thus, Mercury is about \( 93 \times 0.39 \approx 36.27 \) million miles from the sun.

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 91影视!

Key Concepts

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

Planetary Motion
Planetary motion refers to the movement of planets as they revolve around a star, such as our Sun, following specific paths known as orbits. Johannes Kepler was a pioneer in providing a mathematical foundation to this concept in the early 17th century. He proposed three laws that described how planets orbit the Sun. This was groundbreaking because it allowed for accurate predictions of planetary positions, when previously, such movements were clouded in mystery and speculation.
  • First Law (Law of Ellipses): Planets move in elliptical orbits with the Sun at one focus of the ellipse.
  • Second Law (Law of Equal Areas): A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time, which implies that a planet travels faster when closer to the Sun.
  • Third Law (Law of Harmonies): The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.
These laws fundamentally changed our understanding of planetary motion, moving away from a geocentric view of the cosmos to a more accurate heliocentric model.
Orbital Period
The term "orbital period" refers to the amount of time a planet takes to complete a full orbit around the Sun. It gives us a key insight into a planet's distance from the Sun and invariably, its speed in space. Kepler's Third Law, which focuses on the relationship between an orbit's size and its period, is crucial here. The understanding of this law is relevant for calculating how long planets take to orbit based on their distances from the Sun.
  • For instance, Earth's orbital period is one year, and it serves as a reference point for calculating the periods of other planets.
  • In our example, Neptune's period is influenced strongly by its distance; being 30 times further from the Sun than Earth, we use the equation derived from Kepler's Third Law: \( P \propto D^{1.5} \) to find it takes about 164.3 Earth years to complete one orbit.
Understanding orbital periods are integral for space exploration and predicting celestial events.
Distance from the Sun
The distance from the Sun greatly impacts a planet's orbital characteristics, such as its period and speed. In the realm of planetary motion, the mean distance from a planet to the Sun is often expressed in terms of the astronomical unit (AU), which is approximately the distance from the Earth to the Sun. Kepler's Third Law highlights the importance of this distance in determining the time a planet takes to orbit the Sun.
  • Distance is not just a measure of space but also of time, as it directly affects the orbital period through the relationship \( P^2 \propto D^3 \).
  • This relationship helps us uncover details about planets like Mercury, which orbits physically closer to the Sun than Earth, taking only about 88 days or 0.24 Earth years to complete one orbit.
  • By understanding \( D \) and using mathematical calculations, it's possible to ascertain that Mercury is approximately 36.27 million miles from the Sun.
Grasping these concepts assists in the broader understanding of the dynamics and timings involved in our solar system.

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

Show that the following data can be modeled by a quadratic function. $$ \begin{array}{|l|c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 & 4 \\ \hline P(x) & 6 & 5 & 8 & 15 & 26 \\ \hline \end{array} $$

If we view a star now, and then view it again 6 months later, our position will have changed by the diameter of the Earth's orbit around the sun. (See Figure 5.68.) For stars within about 100 light-years of Earth, the change in viewing location is sufficient to make the star appear to be in a different location in the sky. Half of the angle from one location to the next is known as the parallax angle. Even for nearby stars, the parallax angle is very \(\operatorname{small}^{36}\) and is normally measured in seconds of arc. The distance to a star can be determined from the parallax angle. The table below gives parallax angle \(p\) measured in seconds of arc and the distance \(d\) from the sun measured in light-years. $$ \begin{array}{|l|c|c|} \hline \text { Star } & \text { Parallax angle } & \text { Distance } \\ \hline \text { Markab } & 0.030 & 109 \\ \hline \text { Al Na'ir } & 0.051 & 64 \\ \hline \text { Alderamin } & 0.063 & 52 \\ \hline \text { Altair } & 0.198 & 16.5 \\ \hline \text { Vega } & 0.123 & 26.5 \\ \hline \text { Rasalhague } & 0.056 & 58 \\ \hline \end{array} $$ a. Make a plot of \(\ln d\) against \(\ln p\), and determine whether it is reasonable to model the data with a power function. b. Make a power function model of the data for \(d\) in terms of \(p\). c. If one star has a parallax angle twice that of a second, how do their distances compare? d. The star Mergez has a parallax angle of \(0.052\) second of arc. Use functional notation to express how far away Mergez is, and then calculate that value. e. The star Sabik is 69 light-years from the sun. What is its parallax angle?

Assume that a long horizontal pipe connects the bottom of a reservoir with a drainage area. Cox's formula provides a way of determining the velocity \(v\) of the water flowing through the pipe: $$ \frac{H d}{L}=\frac{4 v^{2}+5 v-2}{1200} . $$ Here \(H\) is the depth of the reservoir in feet, \(d\) is the pipe diameter in inches, \(L\) is the length of the pipe in feet, and the velocity \(v\) of the water is in feet per second. (See Figure \(5.103\) on the following page.) a. Graph the quadratic function \(4 v^{2}+5 v-2\) using a horizontal span from 0 to 10 . b. Judging on the basis of Cox's formula, is it possible to have a velocity of \(0.25\) foot per second? c. Find the velocity of the water in the pipe if its diameter is 4 inches, its length is 1000 feet, and the reservoir is 50 feet deep. d. If the water velocity is too high, there will be erosion problems. Assuming that the pipe length is 1000 feet and the reservoir is 50 feet deep, determine the largest pipe diameter that will ensure that the water velocity does not exceed 10 feet per second.

When seeds of a plant are sown at high density in a plot, the seedlings must compete with each other. As time passes, individual plants grow in size, but the density of the plants that survive decreases. \({ }^{33}\) This is the process of selfthinning. In one experiment, horseweed seeds were sown on October 21 , and the plot was sampled on successive dates. The results are summarized in Table \(5.8\), which gives for each date the density \(p\), in number per square meter, of surviving plants and the average dry weight \(w\), in grams, per plant. a. Explain how the table illustrates the phenomenon of self-thinning. b. Find a formula that models \(w\) as a power function of \(p\). c. If the density decreases by a factor of \(\frac{1}{2}\), what happens to the weight? d. The total plant yield y per unit area is defined to be the product of the average weight per plant and the density of the plants: \(y=w \times p\). As time goes on, the average weight per plant increases while the density decreases, so it's unclear whether the total yield will increase or decrease. Use the power function you found in part b to determine whether the total yield increases or decreases with time. Check your answer using the table. $$ \begin{array}{|l|r|c|} \hline \text { Date } & \text { Density } p & \text { Weight } w \\ \hline \text { November 7 } & 140,400 & 1.6 \times 10^{-4} \\ \hline \text { December } 16 & 36,250 & 7.7 \times 10^{-4} \\ \hline \text { January 30 } & 22,500 & 0.0012 \\ \hline \text { April 2 } & 9100 & 0.0049 \\ \hline \text { May 13 } & 4510 & 0.018 \\ \hline \text { June } 25 & 2060 & 0.085 \\ \hline \end{array} $$

By comparing the surface area of a sphere with its volume and assuming that air resistance is proportional to the square of velocity, it is possible to make a heuristic argument to support the following premise: For similarly shaped objects, terminal velocity varies in proportion to the square root of length. Expressed in a formula, this is $$ T=k L^{0.5}, $$ where \(L\) is length, \(T\) is terminal velocity, and \(k\) is a constant that depends on shape, among other things. This relation can be used to help explain why small mammals easily survive falls that would seriously injure or kill a human. a. A 6-foot man is 36 times as long as a 2-inch mouse (neglecting the tail). How does the terminal velocity of a man compare with that of a mouse? b. If the 6-foot man has a terminal velocity of 120 miles per hour, what is the terminal velocity of the 2 -inch mouse? c. Neglecting the tail, a squirrel is about 7 inches long. Again assuming that a 6-foot man has a terminal velocity of 120 miles per hour, what is the terminal velocity of a squirrel?
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation
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.