/*! 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 82 Use the fact that the orbit of a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 82

Use the fact that the orbit of a planet about the Sun is an ellipse, with the Sun at one focus. The aphelion of a planet is its greatest distance from the Sun, and the perihelion is its shortest distance. The mean distance of a planet from the Sun is the length of the semimajor axis of the elliptical orbit. The perihelion of Pluto is 4551 million miles, and the distance from the center of its elliptical orbit to the Sun is 897.5 million miles. Find the aphelion of Pluto. What is the mean distance of Pluto from the Sun? Find an equation for the orbit of Pluto about the Sun.

Short Answer

Expert verified
Aphelion: 6346 million miles, Mean distance: 5448.5 million miles. Orbit equation: \[ \frac{x^2}{5448.5^2} + \frac{y^2}{5376.4^2} = 1 \].

Step by step solution

01

Understand the Problem

You need to find the aphelion (greatest distance from the Sun), the mean distance (semimajor axis), and the equation of the orbit of Pluto. Use the given perihelion, the closest distance to the Sun, and the distance from the center to the Sun.
02

Identify Given Values

Perihelion (closest distance) is given as 4551 million miles and the distance from the center to the Sun is 897.5 million miles.
03

Find the Semimajor Axis

Use the fact that the distance from the center to the focus (Sun) is the same as the difference between the semimajor axis (a) and the distance to the perihelion (r_p). Therefore, we have:\[ a - r_p = 897.5 \text{ million miles} \]Since the perihelion (r_p) is 4551 million miles, solve for the semimajor axis (a):\[ a = 4551 + 897.5 = 5448.5 \text{ million miles} \]
04

Calculate the Aphelion Distance

Aphelion distance (r_a) can be found using the relation:\[ r_a = 2a - r_p \]Substitute the known values:\[ r_a = 2 \times 5448.5 - 4551 = 6346 \text{ million miles} \]
05

Write the Equation of the Orbit

For an ellipse centered at the origin with the semimajor axis a and the semiminor axis b, the foci are located at \( \pm c \). Here, c is given as 897.5 million miles. The relationship \( c^2 = a^2 - b^2 \) must be used to find b.\[ c = 897.5 \rightarrow c^2 = 805052.25 \]\[ a = 5448.5 \rightarrow a^2 = 29685482.25 \]\[ b^2 = a^2 - c^2 = 28880430 \]So, the standard form of the elliptical equation is:\[ \frac{x^2}{5448.5^2} + \frac{y^2}{5376.4^2} = 1 \]

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.

Aphelion
The aphelion is the point in the orbit of a planet where it is furthest from the Sun.
For Pluto, the aphelion can be computed by first finding the mean distance (semimajor axis) and using the perihelion information.
The formula for the aphelion distance \(r_a\) is: \[ r_a = 2a - r_p \]
Here, \(a\) represents the semimajor axis, and \(r_p\) is the perihelion distance.
After substituting in the given values and calculations, we find Pluto's aphelion to be 6346 million miles.
This long distance from the Sun signifies that at aphelion, Pluto receives less solar radiation and experiences much colder temperatures compared to when it is at perihelion.
Perihelion
The perihelion is the point where a planet is closest to the Sun during its orbit.
For Pluto, this closest distance is given as 4551 million miles.
At perihelion, Pluto is much closer to the Sun compared to at aphelion, receiving more solar radiation and potentially experiencing slightly warmer temperatures.
This proximity to the Sun is an important factor in understanding how different parts of Pluto鈥檚 orbit can affect its climate and solar energy intake.
Semimajor Axis
The semimajor axis is half of the longest diameter of an ellipse.
It's essentially the average distance from a planet to the Sun.
For Pluto, we calculated this distance by adding the given perihelion distance (4551 million miles) and the distance from the center of the orbit to the Sun (897.5 million miles):\[ a = 4551 + 897.5 = 5448.5 \text{ million miles} \]
Hence, 5448.5 million miles is the semimajor axis of Pluto鈥檚 orbit.
This semimajor axis helps in understanding the scale and size of Pluto's elliptical orbit around the Sun.
Elliptical Equation
The elliptical equation represents the path of a planet's orbit.
For an ellipse with the Sun at one focus, we determined key elements such as the semimajor axis (a) and the distance from the center to the focus (c).
Using the relationship \[ c^2 = a^2 - b^2 \], we computed the semiminor axis \(b\).
Here, \[ c = 897.5, \ a = 5448.5 \rightarrow a^2 = 29685482.25 \]
Substituting these values: \[ b^2 = a^2 - c^2 = 28880430 \]
Therefore, the standard form of Pluto's elliptical equation is: \[ \frac{x^2}{5448.5^2} + \frac{y^2}{5376.4^2} = 1 \]
This equation characterizes the two-dimensional shape of Pluto's orbit around the Sun.

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 bridge is built in the shape of a semielliptical arch. The arch has a span of 120 feet and a maximum height of 25 feet. Choose a suitable rectangular coordinate system and find the height of the arch at distances of \(10,30,\) and 50 feet from the center.

Are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Solve the equation \(\log _{5} x+\log _{5}(x-4)=1\)

An Explosion Two recording devices are set 2400 feet apart, with the device at point \(A\) to the west of the device at point \(B\). At a point between the devices 300 feet from point \(B,\) a small amount of explosive is detonated. The recording devices record the time until the sound reaches each. How far directly north of point \(B\) should a second explosion be done so that the measured time difference recorded by the devices is the same as that for the first detonation?

Are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Find the horizontal asymptote for the graph of $$ f(x)=4 e^{x+1}-5 $$

Show that an equation of the form $$A x^{2}+C y^{2}+F=0 \quad A \neq 0, C \neq 0, F \neq 0$$ where \(A\) and \(C\) are of the same sign and \(F\) is of opposite sign, (a) is the equation of an ellipse with center at (0,0) if \(A \neq C\) (b) is the equation of a circle with center (0,0) if \(A=C\).
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Statistics

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.